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CVR_Pre U_STPM_2022 Mathematics T (Semester 3).indd 1-3
CVR_Pre U_STPM_2022 Mathematics T (Semester 3).indd 1-3 28/10/2021 3:33 PM
CONTENTS
Chapter
1 DATA DESCRIPTION 1
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
1.1 Data Representation 2
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1.2 Measures of Central Tendency 15
1.3 Measures of Dispersion 28
1.4 The Shape of a Distribution 49
Chapter
2 PROBABILITY 71
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
2.1 Counting Techniques 72
2.2 Probability 81
Chapter
3 PROBABILITY DISTRIBUTIONS 114
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
3.1 Discrete Random Variables 115
3.2 Continuous Random Variables 128
3.3 Binomial Distribution 148
3.4 Poisson Distribution 156
3.5 Normal Distribution 164
Chapter
4 SAMPLING AND ESTIMATION 194
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
4.1 Sampling 195
4.2 Estimation 213
Chapter
5 HYPOTHESIS TESTING 237
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
5.1 Hypothesis Tests 238
5.2 Testing Population Mean 244
5.3 Testing Population Proportion 250
Chapter
6 CHI-SQUARED TESTS 260
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
6.1 The Chi-Squared Distribution 261
6.2 Goodness-Of-Fit Tests 265
6.3 Tests of Independence 272
STPM Model Paper (954/3) 283
Answers 285
iii iii
Mathematics Semester 3 STPM Chapter 2 Probability
CHAPTER
2 PROBABILITY
2
Learning Outcome
(a) Apply the addition principle and the multiplication principle.
(b) Use the formulae for combinations and permutations in simple cases.
(c) Identify a sample space, and calculate the probability of an event.
(d) Identify complementary, exhaustive and mutually exclusive events.
(e) Use the formula P(A B) = P(A) + P(B) − P(A B).
(f) Calculate conditional probabilities, and identify independent events.
(g) Use the formulae P(A B) = P(A) × P(B | A)
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= P(B) × P(A | B).
(h) Use the rule of total probability.
Bilingual Keywords
addition principle – prinsip penambahan independent – tak bersandar
combination – gabungan multiplication principle – prinsip pendaraban
complementary – pelengkap mutually exclusive – saling eksklusif
conditional probability – kebarangkalian bersyarat outcome – hasil
dependent – bersandar permutation – pilih atur
disjoint – tak bercantum relative frequency – kekerapan relatif
equally likely – sama kemungkinannya sample space – ruang sampel
event – peristiwa total probability – jumlah kebarangkalian
exhaustive event – peristiwa habisan
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Mathematics Semester 3 STPM Chapter 2 Probability
2.1 Counting Techniques
Addition principle of counting
Let A , A , ..., A be disjoint events with n , n , …, n possible outcomes, respectively. Then the total number
k
k
2
1
2
1
of outcomes for the event “A or A or ... or A ” is n + n + … + n .
1 2 k 1 2 k
Suppose that we want to buy a fruit from one of two stalls A and A . Suppose also that those stalls have 10
1
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2
and 15 different types of fruits, respectively. How many types of fruits are there altogether to choose from?
2 Choosing one from given types of fruits from either stalls is called an event and the choices for either
event are called the outcomes of the event. Thus the event “selecting one from stall A ”, for example, has 10
1
outcomes. Essentially, the addition principle says that if we want to count the number of ways either one
case could happen or another case could happen, then we should add the number of ways each individual
case could happen. Thus, we can choose one of 10 types of fruits from stall A or one of 15 types of fruits
1
from stall A , there are altogether 10 + 15 = 25 types of fruits to choose from.
2
Note that the events must be disjoint, that is they must not have common outcomes for this principle to
be applicable.
Example 1
Suppose there are 4 different flavours of noodle dishes and 7 different ingredients of fry rice dishes. How
many selections does a customer have?
Solution: An event is “selecting a dish of either kind”.
There are 4 outcomes for the noodle event and 7 outcomes for the rice event.
According to the addition principle, there are 4 + 7 = 11 possible selections.
Multiplication principle of counting
Let A , A , ..., A be events with n , n , ..., n possible outcomes, respectively. Then the total number of
1
k
k
2
2
1
outcomes for the sequence of these k events is n × n × … × n .
k
1
2
If we are buying a cup of ice cream that comes in a choice of three flavours from vanilla, chocolate or
mango, and two sizes either small cup or large cup, how many different types of ice creams can be ordered?
We have three choices for the flavours, for each choice of flavour; there are two choices of sizes. Selecting one
of three choices is called an event, and a specific size is called the outcome of the event. The multiplication
principle tells us that if we want to count the number of ways that one case could happen and another case
could happen, then we should multiply the number of ways that each individual case could happen. Thus,
we could order 3 × 2 = 6 different types of ice cream.
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02 STPM Math(T) T3.indd 72 28/10/2021 10:21 AM
Mathematics Semester 3 STPM Chapter 2 Probability
Example 2
A password consists of three symbols. If the first symbol is a character, the second and the third symbols
are digits, how many three-symbol passwords could be formed?
Solution: Let A be the set of characters, B and C be the sets of digits. Then A × B × C is
the set of all passwords fulfilled the requirement. There are 26 elements in the
set of characters, 10 elements in the set of digits. We thus have
n(A) × n(B) × n(C)
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= 26 × 10 × 10
= 2 600
Thus, a total of 2 600 three-symbol passwords could be formed. 2
Example 3
Find the number of different 5 digit numbers. How many of these numbers are even?
Solution: The first digit could be any numbers from 1 to 9. Each of the next four digits
could be any digits. By the multiplication principle, there are
4
9 × 10 = 90 000 such numbers.
Now the final digit must be one of the numbers 0, 2, 4, 6 or 8, i.e. 5 ways. Again
by the multiplication principle, there are
3
9 × 10 × 5 = 45 000 such numbers.
Note: If no letters can be duplicated in a label, then the first letter of a label can be selected from all 26
characters. The second letter, however, must be selected from 25 characters because one letter has
been selected for the first position and that letter cannot be used for the second position. Similarly
the third letter is now selected from the remaining 24 characters and the fourth from 23 characters.
Permutations
Assume that you try to arrange three persons A, B and C sitting in a row for dinner, how many ways can
you line up the three persons? As the number is small, it is not difficult to make such arrangement. We
could write down all the possibilities: ABC, ACB, BCA, BAC, CAB and CBA. But what if there were eight
persons? Now, the arrangement is getting more complicated. We must come up a new counting method
to handle this problem. The way to do this is to work in steps and then use the multiplication principle.
Back to the case of three persons, at the initial stage we have three ways to fill the first seat, then two persons
remain to fill the second seat and finally just one person left for the last seat. So, the number of ways to
arrange the persons in order is 3 × 2 × 1 = 6.
This multiplication principle works with any finite number of persons. For eight persons, the number of
distinct arrangements is 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 40 320.
The above discussion introduces another basic counting principle.
A permutation of a set of elements is a way of arranging all or part of the elements in a definite order.
For n elements of a set to be orderly arranged, we can do this in n steps by line up one element at a time.
There are n choices for the first place. Once the first place is filled, any one of the n – 1 elements can be
filled in the second place, and so on. For each step, there is one less than the total elements in the previous
step. By the multiplication principle, the number of permutations is
.
.
n(n – 1)(n – 2) … 3 2 1
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02 STPM Math(T) T3.indd 73 28/10/2021 10:21 AM
Mathematics Semester 3 STPM Chapter 2 Probability
Whenever we calculate permutation, product like this comes up very frequently. We represent this product
by the notation n!, which is read “n factorial”. Thus,
.
.
n! = n(n – 1)(n – 2) … 3 2 1
Note: By definition 1! = 1 and 0! = 1.
First permutation rule
The number of permutations of n distinct elements is n!.
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Example 4
2
There is a photo taking session in a birthday party. If 6 people line up for taking a photo, how many
different ways can they be arranged from the left to the right?
Solution: Any of the 6 people can be placed in the first position from the left. Once the first
position is taken, there are 5 people left for the second position. After the first two
positions are taken, there are only 4 people to choose for the third position, and so on.
It is observed that there is one less person to choose from each time a position
is taken. Thus, the number of ways the 6 people could be arranged is
= 6 × 5 × 4 × 3 × 2 × 1
= 6!
= 720
Thus, they can line up in any of the 720 possible ways.
Sometimes, we may consider only a certain number of elements in a set to be arranged in order instead
of all of them. For example, if you were to arrange ten students to sit in the first row but you only have
three chairs, how many ways the chairs could be occupied? Although this problem is slightly different from
previous examples, the approach in getting the solution is similar. Imagine there are three slots and the slots
are to be filled one at a time. Any of 10 students may fill the first slot. After the first student is selected, any
of 9 students may fill the second slot, and any of 8 students in the last slot.
Possible ways: 10 9 8
The number of possible ways of placing 3 of the 10 students to sit in the first row is = 10 × 9 × 8
= 720
10
10
10
This product is commonly denoted by the symbol P . So, we have P = 10 × 9 × 8 = 720. P is read as
3
3
3
“the number of permutations of 10 objects taken 3 at a time”.
n
In general, P represents the number of ways r elements being selected from a set of n elements and placing
r
them in order. By following the similar procedure as the above example, we have
n P = n(n – 1)(n – 2) … [n – (r – 1)] = n(n – 1)(n – 2) … (n – r + 1)
r 1444442444443
r factors
.
This expression can be simplified by multiplying (n – r)(n – r – 1) … 2 1 , which is just equal to 1.
… .
(n – r)(n – r – 1) 2 1
74
02 STPM Math(T) T3.indd 74 28/10/2021 10:21 AM
Mathematics Semester 3 STPM Chapter 2 Probability
.
n P = n(n – 1)(n – 2) … (n – r + 1) (n – r)(n – r – 1) … 2 1
r (n – r)(n – r – 1) … .
2 1
.
n(n – 1)(n – 2) … (n – r + 1)(n – r)(n – r – 1) … 2 1
=
2 1
(n – r)(n – r – 1) … .
= n!
(n – r)!
Second permutation rule
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The number of permutations of n distinct elements taken r at a time is
n!
n P = (n – r)! . 2
r
Example 5
Evaluate each of the following permutations.
6
(a) 8 P 2 (b) 12 P 9 (c) 5 P 5 (d) P 1
Solution: (a) 8 P = 8!
2 (8 – 2)!
= 8!
6!
= 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
6 × 5 × 4 × 3 × 2 × 1
= 8 × 7
= 56
or
8 P = 8 × 7 = 56 by using multiplication principle.
2
(b) P = 12!
12
9 (12 – 9)!
= 12!
3!
= 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
= 79 833 600 3 × 2 × 1
5
(c) P = 5!
5 (5 – 5)!
= 5!
0! 0! = 1
= 5!
= 120
(d) P = 6!
6
1 5!
= 6
Note: 1. Example 5(c) demonstrates the first permutation rule, P = n!.
n
n
2. Example 5(d) illustrates the case for selecting one element from a set of n elements, i.e. P = n.
n
1
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02 STPM Math(T) T3.indd 75 28/10/2021 10:21 AM
Mathematics Semester 3 STPM Chapter 2 Probability
Example 6
A badminton team has 5 players. How many ways can a coach select the first and second singles?
Solution: Here we are choosing 2 from 5 and arranging them in order. So, the number of
ways are
5 P = 5 × 4 = 20
2
Thus, there are 20 ways for the coach to select the first and second singles.
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So far we have considered permutations of distinct elements. If the letters B and C are both equal to X, then
2 the 6 permutations of the letters A, B and C becomes AXX, AXX, XXA, XAX, XAX, XXA, of which only 3
are different. Thus, with 3 letters, 2 being the same, there are 3! = 3 distinct permutations.
2!
In general, the number of distinct permutations of n elements of which n are of one kind, n of a second
1
2
kind, …, n of a kth kind is given by the formula
k
n!
n ! n ! … n !
1 2 k
Example 7
Find the number of different ways to arrange 2 yellow, 3 red and 4 green bulbs in a string of Christmas
tree lights with 9 sockets.
Solution: The total number of different arrangements is
9!
= 1260
2! 3! 4!
Example 8
Find the number of arrangements that can be formed from all the letters of the word MATHEMATICS, if
(a) there is no restrictions,
(b) the two M’s are separated.
Solution: For the word MATHEMATICS, there are 11 letters with 2 A’s, 2 M’s, 2 T’s, 1 H,
1 E, 1 I, 1 C and 1 S.
(a) If there is no restrictions, the total number of different arrangements
= 11! = 4 989 600
2! 2! 2! 1! 1! 1! 1! 1!
(b) If the two M’s are together, they can be considered as a single letter and
the total number of different arrangements
= 10! = 907 200
2! 2! 1! 1! 1! 1! 1! 1!
Thus, if the two M’s are separated, the total number of different arrangements
= 4 989 600 – 907 200
= 4 082 400
Note: This is an example of permutation with restriction.
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02 STPM Math(T) T3.indd 76 28/10/2021 10:21 AM
Mathematics Semester 3 STPM Chapter 2 Probability
Combinations
In dealing with permutations the order of the elements does matter. For instance, in Example 6 the coach’s
selection of the first and second singles from the 5 players, the arrangement would be different if we exchanged
the positions of the 2 selected players.
Now, let us consider different type of selections. If the coach wished to pick a doubles from the 5 players,
the order of choosing the first and the second player does not really matter. We begin with how many ways
we can choose 2 players from a group of 5 players and arrange them in order. It was found to be P = 5!
5
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2 3!
= 20 ways. Assume the players are designated by A, B, C, D or E. For a group of 2 players A and B, we
have two arrangements, AB and BA. Note that the doubles consisting of AB is the same doubles formed by 2
BA. This implies that within the 20 ordered arrangements, the order of the groups of 2 players should be
disregarded. Consequently, we divide the ordered arrangements of 5 players taken 2 at a time, P , by the
5
2
number of ways of 2 players arranged between them, i.e. 2!.
5 P
This gives 2 = 5!
2! 3! 2!
5 × 4
=
2
= 10 doubles
Generally, we could extend the above analysis by choosing r elements from a set of n elements without regard
r 1 2
to order. This type of selections is classified as combination and is denoted by C or n . The number of
n
r
possible combinations, C , is calculated by the second permutations formula divided by r!. Thus
n
r
n P
C = =
n r n!
r r! (n – r)! r! Choosing without
Return –
A School Election
VIDEO
Example 9
How many different ways can you choose 4 kings from a standard deck of 52 playing cards?
Solution: This problem involves combinations, because the order of the cards does not
matter. We select 4 kings from 4 available kings. So,
4!
4 C =
4 (4 – 4)! 4!
= 1
There is only 1 way you could select 4 kings.
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02 STPM Math(T) T3.indd 77 28/10/2021 10:21 AM
Mathematics Semester 3 STPM Chapter 2 Probability
Example 10
From 5 men and 3 women, find the number of committees of 5 that can be formed with 3 men and 2
women.
Solution: The number of ways of selecting 3 men from 5 is
5 C = 5!
3 (5 – 3)! 3!
= 10
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The number of ways of selecting 2 women from 3 is
2 3 C = 3!
2 (3 – 2)! 2!
= 3
Thus, the required number of committees that can be formed = 10 × 3
= 30
Note: We apply the multiplication principle in this example.
Note: A commonly raised question is, “When do we use permutation and when do we use combination?”
To answer this we must first realise that objects are drawn from the lot without replacement. Thus,
the total number of ways is P if the order in which the objects are drawn is important, and C if
n
n
r
r
it is not important.
Example 11
Consider the collection of objects consisting of the six letters a, b, c, d, e, f. Write the answers in factorial
notations.
(a) Find the number of three-letter permutations.
(b) Find the number of three-letter combinations.
(c) Which is greater, the number of permutations or the number of combinations?
(d) Find an equation relating parts (a) and (b).
6!
Solution: (a) The number of three-letter permutations is, P =
6
3 (6 – 3)!
= 6!
3! 6!
(b) The number of three-letter combinations is, C =
6
3 (6 – 3)! 3!
6!
=
3! 3!
(c) From parts (a) and (b), it is obvious that the number of permutations is
greater than the number of combinations
6! 6 P
6
(d) The number of three-letter combinations is, C = = 3
3 3! 3! 3!
6 6 .
⇒ P = C 3!
3 3
.
6 P = C 3! is the required equation relating parts (a) and (b).
6
3 3
.
n
n
In general, we can deduce that P = C r!.
r r
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02 STPM Math(T) T3.indd 78 28/10/2021 10:21 AM
Mathematics Semester 3 STPM Chapter 2 Probability
Some useful tips for counting techniques
When you use counting techniques, you must always keep these in mind: “Do I count everything?” and “Do
I count anything more than once?” to avoid mistakes. It is not easy to answer the question which counting
methods you should use. You may find some of the suggestions below helpful.
1. If the set of elements breaks up into disjoint subsets, then the addition rule can be applied.
2. If the elements are from different sets or the elements to be counted are collected through multiple
steps, then the multiplication rule is used.
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3. Permutations involve choosing a specific number of elements from a set of elements and placing them
in order.
4. Combinations involve choosing a specific number of elements from a set of elements without regard 2
to order.
Exercise 2.1
1. There are two routes from a student’s home to a bus station and three routes from the bus station to
the school. Use the multiplication principle to find the number of ways a journey from the student’s
home to the school via the bus station may be completed.
2. If a Mathematics department schedules 4 lecture sections and 12 tutorial groups for a course in
Introductory Statistics, in how many different ways can a student choose a lecture section and a tutorial
group?
3. In a practical class, students are asked to classify the specimens according to their colours: red, blue,
green, yellow, white and also according to their sizes: small, medium, large and lastly to their gender:
male, female. In how many different ways can a specimen be classified according to colour, size and
gender?
4. Jenny has 9 different blouses and 6 different skirts. Find the number of possible different blouse-skirt
outfits she can form. If she also has 7 pairs of different stockings, how many different blouse-skirt-
and-stocking outfits are possible?
5. The license plates of some cars consist of two letters followed by three digits. Find the number of
possible different license plates if
(a) there is no restrictions,
(b) no letter or digit is repeated.
6. How many lunches are possible consisting of soup, a sandwich, desserts, and a drink if one can select
from 3 soups, 4 types of sandwiches, 4 desserts, and 5 drinks?
7. Find the number of ways for the letters of the word MONDAY to be arranged in a row. How many
different ways can three of these letters be chosen and written in a row if
(a) there is no restriction?
(b) the first letter must be D?
8. How many ways can a football team schedule 3 exhibition games with 3 teams if they are all available
on any of 5 possible dates?
9. How many ways can 8 people be assigned to 2 triple and 1 double rooms?
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Mathematics Semester 3 STPM Chapter 2 Probability
10. A club has 30 members. They are to select 3 office holders consisting of the president, secretary, and
treasurer for the following year. They always select these office holders by drawing 3 names randomly
from names of all members. The first person selected becomes the president, the second is the secretary,
and the third takes over as treasurer. Thus, the order in which 3 names are selected is important. Find
the total arrangements of 3 names from 30.
11. Find the number of permutations that can be formed from the letters of the word POPULAR. How
many of these permutations,
(a) begin and end with P?
(b) have the two P’s separated?
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(c) have the vowels together?
2 12. (a) In how many different ways can a student answers 8 true-false questions?
(b) In how many ways may the test be completed if a student is imposed for each incorrect answer,
so that the student may leave some questions unanswered?
13. An ice cream parlour has seven flavours of ice cream. Nathan wants to buy two flavours of ice cream.
If he randomly selects two flavours out of seven, how many possible combinations are there?
14. Consider the collection of objects consisting of the five letters c, d, e, f, g.
(a) List all possible combinations of three letters from this collection of five letters.
(b) Use part (a) to determine the number of possible combinations of three letters that can be formed
from this collection of five letters.
(c) Use combination formula to confirm the answer obtained in part (b).
15. To recruit new members, a compact-disc (CD) club advertises a special introductory offer:
A new member agrees to buy one CD at regular club prices and receives free any four CDs of the
member’s choice from a collection of 50 CDs. Find the number of possibilities a new member has for
the selection of the four free CDs.
16. In how many different ways can a coach choose two badminton players from among seven students
and three table tennis players from among nine students?
17. Evaluate
(a) 5 P 5 (b) P 3 (c) 6 P 4 (d) 12 P 0
8
(e) 7 C 4 (f) 10 C 10 (g) 9 C 5 (h) 15 C 0
18. Out of 5 class representatives and 7 prefects, a committee consisting of 2 class representatives and 3
prefects is to be formed. In how many ways can this be done if,
(a) any class representative and any prefect can be included,
(b) one particular prefect must be in the committee, and
(c) two particular class representatives cannot be in the committee.
19. Santi has 6 flowers, each of a different variety. How many different bouquets can she form?
20. There are 25 paintings in a collection. How many ways can the following groups be selected?
Write the answers in factorial notations.
(a) 5 paintings,
(b) 20 paintings
What conclusion can you deduce from parts (a) and (b)?
21. A developer of a residential subdivision has four house styles and wants to build on eight adjacent
lots. How many distinguishable arrangements are possible if the developer decides to build two houses
of each style?
22. A school has 4 Mathematics teachers, 3 Chemistry teachers and 2 Biology teachers. Find the number
of different sets of teachers a student could have for these 3 subjects.
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02 STPM Math(T) T3.indd 80 28/10/2021 10:21 AM
Mathematics Semester 3 STPM Chapter 2 Probability
23. A company manager has to visit four of the ten subsidiaries that the company owns.
(a) In how many different ways can the manager plan his itinerary in visiting four of the twelve
subsidiaries?
(b) How many sets of four companies are there from which the manager can pick one set to visit?
24. A company has 10 construction workers. The manager plans to assign 3 to job site A, 4 to job site B,
and 5 to job site C. In how many different ways can the manager make this assignment?
2.2 Probability
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Sample spaces 2
The scientists perform experiments to produce observations or measurements that will assist them in arriving
at conclusions. In statistics the word experiment is used to describe any process that generates raw data or
outcome. Thus, an outcome is a result of some activity. For example:
Rolling a die has six outcomes: 1, 2, 3, 4, 5, 6
A sample space is a set of all possible outcomes for an activity and is represented by S. For example: The
sample space for rolling a die is: S = {1, 2, 3, 4, 5, 6}
The sample space for tossing a coin consists of the two outcomes, head H and tail T is: S = {H, T}
An event is the collection of outcomes of particular interest in an experiment. For the experiment of tossing
a coin, suppose we are interested only in getting a head, then this event consists of a single element, H.
Thus, in this example of tossing a coin;
The event that a head occurs is: E = {H},
If a coin is tossed 3 times in succession;
The event that the number of heads is appears more than once is: E = {HHH, HTH, THH, HHT},
Obviously the event E is the subset of the sample space S.
For another example in rolling a die;
The event of obtaining a 3 is: E = {3},
The event of obtaining an odd number is: E = {1, 3, 5}
Example 12
Consider the experiment of a team’s result in a football match. Determine the sample space, S.
Solution: The possible outcome of the team’s result in a football match is win, lose or draw.
So, the sample space is S = {win, lose, draw}.
Example 13
For the experiment in Example 12, write the following events.
(a) Event A: The team wins the match
(b) Event B: The team does not lose the match
Solution: (a) Event A has only one possible outcome, win, so the event A is: A = {win}
(b) Two outcomes fulfil this given condition: win or draw. So the event B is:
B = {win, draw}
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Example 14
A number is randomly picked from a set of integers, {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.
(a) Find the sample space of the above experiment.
(b) List the outcomes in the following events for the above experiment.
(i) The number is divisible by 3.
(ii) The number is an even number.
(iii) The number is 11.
Note: By picking an integer at random, we mean that each integers has equally likely chance of
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being picked.
2 Solution: (a) There are ten possible outcomes, so the sample space is S = {1, 2, 3, 4, 5,
6, 7, 8, 9, 10}.
(b) (i) The event of the number divisible by 3 has three possible outcomes,
3, 6 or 9.
E = {3, 6, 9}
(ii) The event of obtaining an even number has five possible outcomes,
2, 4, 6, 8, 10.
E = {2, 4, 6, 8, 10}
(iii) The set has ten integers and none of these integers is 11. So,
picking a number 11 is impossible. We say it is an impossible event
and E = {φ}
Probability of an event
If you are planning to visit your friend who is staying nearby, then you look at the sky and you are not sure
whether it is going to rain. So, you are hesitating to bring along an umbrella with you. This is one of the
non-deterministic situations that we encounter often. In the real world, many problems cannot be predicted
with accuracy. Probabilities are thus introduced to deal with situations involving randomness or uncertainty
about the outcome.
Probability is a measure of how likely an event is to happen. In an experiment, if all the outcomes are
equally likely to occur, then the probability of an event to occur is the number of ways that an event can
occur divided by the total number of outcomes in the sample space.
number of ways that an event can occur
P(Event) =
total number of possible outcomes
The probability of an event is a number between 0 and 1 that indicates the likelihood the event will occur
as shown below:
P(E) = 0 P(E) = 0.5 P(E) = 1
Note: 1. P(E) = 0 means that the event will not occur.
2. P(E) = 1 means that the event is certain to occur.
3. P(E) = 0.5 means that the event is equally likely to occur or not occur.
4. The closer the probability of a given event is to 1, the more likely it is to occur.
5. Probability can be expressed as decimal, fraction, ratio or percentage.
6. For example: P(E) = 0.5, 1 or 50%.
2
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Mathematics Semester 3 STPM Chapter 2 Probability
Definition:
1. The probability that an event will happen is between 0 and 1 inclusive, i.e. 0 < P(E) < 1.
2. P(φ) = 0; φ is 0 collections.
3. P(S) = 1
Example 15
A fair die is tossed. If the event of interest is obtaining a number less than 3, find the probability of the
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event happening.
Solution: There are six different possible outcomes in the sample space for tossing a die,
so S = {1, 2, 3, 4, 5, 6}. Let E be the event of obtaining a number less than 3. 2
Event E has two outcomes satisfying the requirement, we thus have E = {1, 2}.
We use a Venn diagram to show the relationship between S and E.
S
1 2 E
3 4 5 6
The probability of the event E occurring is
P(E) = 2 = 1
6 3
Thus, the probability of getting a number less than 3 is 1 .
3
Definition: If an experiment can result in any one of N different equally likely outcomes, and if exactly n
of these outcomes correspond to event E, then the probability of event E is
P(E) = n
N
Example 16
Two fair coins are tossed.
(a) Use a tree diagram to find the sample space of the above experiment.
(b) List the simple events in the sample space and their corresponding probabilities.
(c) Find the probability of observing exactly one head in the two tosses.
Solution: (a) Let H and T denote head or tail respectively.
The tree diagram is as follows:
First coin Second coin Outcome
H HH
H
T HT
H TH
T
T TT
The sample space of the above experiment is {HH, HT, TH, TT}
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(b) The simple events in the sample space and their corresponding probabilities
are listed in the following table:
Event First coin Second coin P(E)
E 1 H H 1
4
E H T 1
2 4
E T H 1
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3 4
E T T 1
2 4 4
(c) Let F = event of observing exactly one head in the two tosses.
The probability of observing exactly one head in the two tosses is,
F = {HT, TH}
P(F) = 2
4
= 1
2
Example 17
3 students are chosen from a group of 10 students consisting of 7 boys and 3 girls to represent a school
in chess competition. If the selection is merely based on random picking, find the probability that the
representatives are formed by
(a) 2 boys and 1 girl,
(b) at least 2 girls.
10
Solution: (a) There are C = 10! = 10 × 9 × 8 = 120 combinations by taking
3 (10 – 3)! 3! 3 × 2 × 1
3 from 10 students.
There are C × C = 7 × 6 × 3 = 63 combinations by exactly taking 2
7
3
2 1 2 × 1 1
boys and 1 girl from the group.
Hence, the probability that the representatives are formed by exactly 2 boys
and 1 girl is 63 = 21 .
120 40
(b) The selection could be either 2 girls and 1 boy or 3 girls and no boys.
So, the number of combinations of 2 girls and 1 boy or 3 girls and no boys
= C × C + C × C
7
3
3
7
2 1 3 0
= 3 × 7 + 1 × 1
= 22
The probability that the representatives are formed by at least 2 girls is
22 = 11 .
120 60
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Mathematics Semester 3 STPM Chapter 2 Probability
In summary, if the number of equally probable outcomes in the sample space S is denoted by n(S) and the
number of equally probable outcomes in an event E is written as n(E), then the probability of an event to
occur can be expressed as P(E) = n(E) .
n(S)
As E is a subset of S, we have 0 < n(E) < n(S).
Dividing all by n(S), 0 < n(E) < n(S)
n(S) n(S) n(S)
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Thus, we get 0 < P(E) < 1.
2
Another approach in determining the probability is based on relative frequency. If an experiment is repeated
n times under the identical condition and an event is observed to happen f times, the probability of the
event happening is then estimated to be
P(E) = frequency of the event occured = f
total number of observations n
Example 18
Data are collected on the gender of customers who enter a supermarket on a particular day. It is found
that out of 360 customers, 249 are females. Find the probability that a customer who visits the supermarket
on that day is a female customer.
Solution: Let A be the event that a female customer visits the supermarket.
P(A) = 249 = 83
360 120
Complementary
Events
Complementary events INFO
Two events are said to be complementary, if one event happens then the other event cannot happen at the
same time. Both events contain all the experimental outcomes in the sample space. Let E (read as E prime)
denotes the event E does not happen where E is called the complement of E. If n(S) is the size of the sample
space, n(E) is the number of outcomes in event E, then n(E) = n(S) – n(E). Hence, in terms of probability
P(E) + P(E) = n(E) + n(E) = n(E) + n(S) – n(E) = n(S) = 1
n(S) n(S) n(S) n(S)
This rule for complementary events states that if two events are complementary, then the sum of their
probabilities equal to 1. Hence,
P(event E happens) + P(event E does not happen) = 1.
Rearranging this equation we obtain the complement rule as follows:
P(E) = 1 – P(E)
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Example 19
A die is tossed.
(a) Calculate the probability of,
(i) obtaining an odd number,
(ii) getting an even number.
(b) Show the validity of the rule for complementary events.
Solution: (a) The sample space for tossing a die is S = {1, 2, 3, 4, 5, 6}.
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(i) Let A be the event of obtaining an odd number.
Event A = {1, 3, 5}
2 P(A) = 3
6
= 1
2
(ii) Let B be the event of getting an even number.
Event B = {2, 4, 6}.
P(B) = 3
6
= 1
2
(b) P(A) + P(B) = 1 + 1
2 2
= 1
We have thus illustrated the rule of complementary events.
Example 20
A letter is randomly selected from the alphabet. Find the probability of not getting a vowel.
Solution: Let A be the event of getting a vowel. There are 5 vowels {a, e, i, o, u} and a
total of 26 letters in the alphabet. Thus,
P(A) = n(A) = 5 .
n(S) 26
The event of not getting a vowel is the complement of the event A. By the rule
of complementary events, we have
P(A) = 1 – P(A)
5
= 1 – 26
= 21
26
Alternatively, we can find the probability directly by counting the letters that are
not vowels. So, n(A) = 21 and P(A) = 21 .
26
Note: Sometimes using complementary events can make the probability calculation easier.
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Exhaustive events
Two events are said to be exhaustive if it is certain that at least one of them occurs. If the events A and B
are exhaustive, they together form the whole sample space. In set language, A B = S.
For example, when tossing a die the events “getting an even number” and “getting an odd number” are
exhaustive, because they include all possible outcomes.
Example 21
In a group of 10 students, 5 are form-one boys, 3 are form-one girls and the remaining 2 are form-two
girls. A student is randomly chosen from the group. The events A, B, C and D are defined as follows. 2
A : The selected student is a form-one student,
B : The selected student is a girl,
C : The selected student is a boy,
D : The selected student is a form-two girl,
Identify which pairs of the events are exhaustive.
Solution: The following pairs of events are exhaustive.
• A and B, because it is certain that at least one of both events occur.
• A and D, because it is certain that at least one of both events occur.
• B and C, because it is certain that at least one of both events occur.
If the events A and B are exhaustive, then A B = S.
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So, P(A B) = 1.
Demonstrating
Mutually exclusive events Exclusive Events
VIDEO
We are often interested in finding the probability of events whose outcomes are described by two or more
other events. For example, a secondary school data shows that 28% of students age 13 years and 16%
age 17 years. If a student from the school is selected at random, what is the probability that the student
ages 13 years or 17 years? In the following section we expand the probability calculation to include two or
more events.
Two or more events are mutually exclusive or disjoint if the events cannot occur at the same time when the
experiment is performed. Mutually exclusive events can be shown by using a Venn diagram as follows:
S
A B
If A and B are two mutually exclusive events, then
(a) they do not have any outcomes in common or cannot both occur at the same time,
i.e. A B = φ, the intersection of A and B is the empty set.
(b) P(A and B) = 0, i.e. P(A B) = 0
(c) P(A or B) = P(A) + P(B), i.e. P(A B) = P(A) + P(B).
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Note: If events A and B are not only exhaustive but they are also mutually exclusive, then exactly one of
the events can happen. For instance, an event and its complement are always exhaustive and mutually
exclusive:
P(A A) = 1 as well as P(A A) = 0.
Example 22
A fair die is tossed, let event A = obtain a one,
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event B = obtain a four,
event C = obtain an even number,
2 event D = obtain an odd number.
Which of the following groups of events are mutually exclusive?
(a) A and B (b) B and C (c) B and D
Solution: (a) Events A and B are mutually exclusive because a single die can only land
one way. Obtaining both “one” and “four” at the same time is impossible.
(b) Events B and C are not mutually exclusive because they have the common
outcome “four”. Both events occur if the number “four” is obtained.
(c) Events B and D are mutually exclusive. As shown in the Venn diagram,
there is no overlapping between the two events. It is impossible to get a
number that is both “four” and an odd number in a single toss.
A
B
4 2 1 5
6 3
C D
Example 23
A coin is tossed.
(a) List the possible outcomes.
(b) Define the sample space.
(c) List the simple events.
(d) Are the events mutually exclusive?
(e) Are the events exhaustive?
(f) Assuming that there is equal probability for the coin to land with any of its faces up and that it will
not stand on its edge, find the probability of each event.
Solution: (a) The possible outcomes are head and tail.
Let H represents head and T represents tail.
(b) The sample space, S = {H, T}.
(c) The simple events are {H} and {T}.
Let E = {H} and E = {T}.
2
1
(d) Since on any toss, either H or T may turn up, but not both; the events are
mutually exclusive.
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(e) Since the events E and E make up the whole of the possibility space, hence,
2
1
E and E are exhaustive events.
1
2
(f) For exhaustive events E and E , P(E E ) = 1
2
1
1
2
For equally probable outcomes P(E ) = P(E ),
1
2
P(E ) + P(E ) = 1
1 2
2P(E ) = 1
1
P(E ) = 1 = P(E )
1 2 2
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Probability of the union of events 2
In Example 22 we conducted an experiment by tossing a die. Event B was “obtaining a four” and event D
was “obtaining an odd number”. The probability of event B, P(B), is 1 and the probability of event D, P(D),
6
is 3 or 1 . Suppose we want to find the probability of tossing an odd number or four, which is denoted
2
6
by P(B D).
We notice that events B and D are mutually exclusive. Hence,
P(B D) = n(B D) = n(B) + n(D)
n(S) n(S)
= n(B) + n(D)
n(S) n(S)
= P(B) + P(D)
= 1 + 1 = 2
6 2 3
Example 24
It is found that out of 100 students of a school, 65 students go to school by bus, 15 students walk to
school and the remaining 20 students go to school by other transportations. If a student is randomly
selected from this group of 100 students, what is the probability that the student selected either goes to
school by bus or walks to school?
Solution: Let event B be “the student selected goes to school by bus” and
event W be “the student selected walks to school”.
The two events are mutually exclusive as the selected student cannot go to school
by two different means at the same time.
Thus, we have P(B W) = P(B) + P(W)
= 65 + 15
100 100
= 4
5
If two events A and B are mutually exclusive, the probability of A or B occurring is
P(A B) = P(A) + P(B)
This is the addition rule for mutually exclusive events:
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The events linked to an experiment may not be mutually exclusive. Consider the following case:
Suppose 10 students were asked what sports they have participated recently, their answers showed that 5
students played football and 7 students played badminton. What is the probability that one of the students
selected has played football or badminton? The probability of selecting a student who has played football is
5 or 0.5 and the probability of a student playing badminton is 7 or 0.7. If the addition law for mutually
10 10
exclusive events is used, the sum of these two probabilities is 1.2. We know that the value of probability
cannot exceed 1. So, let us check the counting. To count all the students playing either football or badminton,
we need to count all the students playing football, all the students playing badminton and subtract from this
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the number of students who were counted twice because they were playing both football and badminton.
2 To help us visualise the above reasoning, the Venn diagram is used. Assume that 3 students played both
football and badminton in the above example. The diagram indicates clearly why 3 outcomes in the overlap
area of the event A B are being counted twice – once in A and once in B. Where event A represents “a
selected student playing football” and event B represents “a selected student playing badminton”.
A B
For events that are not mutually exclusive, the addition law is modified to take into account of double
counting. If the number of outcomes in event A is n(A) and the number of outcomes in event B is n(B),
then to find the number of outcomes in event A B, we must count the outcomes in A B. So, to get
the correct total without counting the outcomes in the overlap A B twice, we must subtract the number
of outcomes in A B. Thus
n(A B) = n(A) + n(B) – n(A B)
Let S be a sample space with n(S) possible outcomes. The probability of event A B is given by
P(A B) = n(A B)
n(S) S
A B
= n(A) + n(B) – n(A B)
n(S)
= n(A) + n(B) – n(A B)
n(S) n(S) n(S)
= P(A) + P(B) – P(A B)
This is called addition rule of probability:
P(A B) = P(A) + P(B) – P(A B)
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Example 25
Consider the experiment of selecting one card at random from a standard deck of 52 playing cards. Find
the probability of drawing either a king or a diamond card.
Solution: Let event A = a king card is drawn,
event B = a diamond card is drawn.
As there are 4 king cards in the deck, P(A) = 4 = 1 ,
52 13
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and the deck has 13 diamond cards, so P(B) = 13 = 1 .
52 4
Since there is 1 card corresponding the king of diamond, P(A B) = 1 . 2
52
By applying the formula, P(A B) = P(A) + P(B) – P(A B), we have
P(A B) = 1 + 1 – 1
13 4 52
= 4
13
The probability of drawing either a king or a diamond card is 4 .
13
Relative frequency data for two or more events is often summarised in a table called a contingency table.
We can easily determine probabilities from this table.
Example 26
A survey of 150 students on their reading habit over the weekend is presented in the contingency table.
Newspaper Total
Yes No
Yes 16 21 37
Magazine
No 75 38 113
Total 91 59 150
If a student under the survey is selected at random, find the probability that the student reads newspaper
or reads a magazine.
Solution: Let A be the event that the student selected reads newspaper,
B be the event that the student selected reads a magazine.
From the table, there are 91 students out of a total of 150 students reading
newspapers,
hence P(A) = 91 .
150
There are 37 students out of a total of 150 students reading magazine, hence
37
P(B) = 150 .
The probability that the selected student reads both newspaper and a magazine,
16
P(A B) = 150 .
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Using the formula, P(A B) = P(A) + P(B) – P(A B), we get
P(A B) = 91 + 37 – 16
150 150 150
= 56
75
The probability that the selected student reads newspaper or a magazine is 56 .
75
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2 Conditional probability
A box contains 6 red and 4 green balls. Suppose two balls are chosen at random from this box. Let us
define two events, A = the first ball is red and B = the second ball is red. Consider the situation that the
ball is chosen without replacement. The calculation of the probability that the first ball chosen is red P(A)
is straight forward and it is found to be 6 = 3 . What about the probability of choosing the second red
10 5
ball P(B)? To compute P(B) we need to know whether the event that the first ball drawn is red or did not
happen. This example introduces an important concept called conditional probability.
In many circumstances the probability of an event is affected by the occurrence of another event. Conditional
probability is defined as the probability of event A happening given that event B has happened. It is denoted
by P(A | B).
Note: Conditional events:
If A and B are two events such that A will occur given that B has already occurred, then A given
B or A | B is the event A conditional on B happening.
The vertical bar “ | ” is read as “given that”.
Consider the following Venn diagram. Suppose that event B is chosen and we wish to find the probability
that event A is being picked. Now the sample space is reduced from the original sample space S to B with
this additional information.
S
A B
Thus, to compute the conditional probability of the event A given that event B occurred, denoted by P(A |
B), we need to count the numbers of elements in event A B as well as in event B.
We could write P(A | B) = n(A B) . Divide both numerator and denominator by the total number of
n(B)
elements in the sample space n(S) to get
n(A B)
P(A | B) = n(S)
n(B)
n(S)
= P(A B) .
P(B)
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Hence, the conditional probability can be determined using the following formula.
P(A | B) = P(A B) , provided that P(B) ≠ 0
P(B)
Rearranging the above equation gives
P(A B) = P(A | B) × P(B)
Note: 1. P(B A) = P(B | A) × P(A),
since P(A B) = P(B A)
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we deduce that, P(A | B) × P(B) = P(B | A) × P(A)
2. If A and B are mutually exclusive events, i.e. A B = φ, then P(A B) = 0. Thus, 2
P(A | B) × P(B) = P(B | A) × P(A) = 0.
3. For mutually exclusive and exhaustive events A and A, A A = φ, A A = S,
P(A A | B) = P(A | B) + P(A | B)
i.e. P(S | B) = P(A | B) + P(A | B).
P(S | B) = P(S B)
P(B)
= P(B)
P(B)
= 1
Hence, P(A | B) + P(A | B) = 1 or P(A | B) = 1 – P(A | B)
4. Consider the following Venn diagram:
S
A B
A B
A B
A = (A B) (A B)
P(A) = P(A B) + P(A B) (A B) and (A B) are mutually exclusive.
.
.
⇒ P(A) = P(A | B) P(B) + P(A | B) P(B)
Example 27
A bag contains 8 yellow and 4 green marbles. Two marbles are taken randomly from the bag. Determine
the probability that only one of them is yellow.
Solution: Let event A = first marble selected is yellow,
event B = second marble selected is yellow.
Then, P(A) = 8
12
= 2 .
3
When the second marble is selected, there are only 11 marbles left and 4 green
marbles remain untouched. So the conditional probability of B given that A has
happened is:
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P(B | A) = 4 . By applying the conditional probability formula,
11
P(yellow first and green second) = P(A B) = P(B | A) × P(A)
= 4 × 2
11 3
8
= 33
Similarly, P(A) = 4 = 1 .
12 3
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When the second marble is selected, there are only 11 marbles left and 8 yellow
marbles are still there. The conditional probability of B given that A has occurred
8
2 is thus determined and P(B | A) is equal to 11 .
Using the conditional probability formula,
P(green first and yellow second) = P(A B) = P(B | A) × P(A)
= 8 × 1
11 3
= 8 .
33
P(only one yellow) = P(first yellow followed by second green or first green
followed by second yellow) = P(A B) + P(A B)
= 8 + 8
33 33
= 16 .
33
The probability that only one of the two marbles taken is yellow is 16 .
33
Note: The result whether the two marbles are taken from the bag at the same time or one after the other
is the same.
Example 28
A and B are two events in a sample space S such that:
P(A | B) = 0.5, P(A) = 0.65 and P(B) = 0.7. Calculate
(a) P(B | A), (b) P(B | A).
Solution: (a) P(B) = 1 – P(B)
= 1 – 0.7
= 0.3
P(A | B) = P(A B)
P(B)
0.5 = P(A B)
0.3
P(A B) = 0.15
P(B | A) = P(A B)
P(A)
= 0.15
0.65
= 3
13
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Mathematics Semester 3 STPM Chapter 2 Probability
(b) P(A) = 1 – P(A)
= 1 – 0.65
= 0.35
P(A B) = P(B) – P(A B)
= 0.3 – 0.15
= 0.15 S
P(B | A) = P(A B) A B
P(A)
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= 0.15 A B
0.35
= 3 2
7
Example 29
In a factory, three machines A, B and C are operated to make certain parts. The percentages of the parts
manufactured by the machines A, B and C are 35%, 50% and 15% respectively. It is known that 8%, 5%
and 16% of the parts produced by the machines A, B and C respectively are defective. If a finished part
is randomly picked, calculate the probability that the part is from the machine A given that it is defective.
Solution: Let event J = a part is made by machine A,
event K = a part is defective.
Assume that the total parts manufactured by the three machines are n. The
numbers of parts produced by machines A, B and C are 0.35n, 0.5n and 0.15n
respectively. Thus, the defective parts produced by machines A, B and C are
0.08 × 0.35n, 0.05 × 0.5n and 0.16 × 0.15n respectively.
Thus, P(K) = 0.08 × 0.35n + 0.05 × 0.5n + 0.16 × 0.15n
n
= 0.077
0.08 × 0.35n
and P(J K) = n
= 0.028
Applying the conditional probability formula, the probability of the part made
by machine A given that the part is defective,
P(J | K) = P(J K)
P(K)
= 0.028
0.077
= 0.364
Independent events
Suppose a fair coin is tossed and a ‘head’ is shown face up. What would the coin land on for the next toss?
The probability of getting a ‘head’ or a ‘tail’ is still 0.5. The outcome of the second toss is not affected by
the previous result. When the knowledge that an event has happened provides no information about the
occurrence of another event, the two events are said to be independent. Thus, if the outcome of event A
does not affect the outcome of event B, then A and B are independent events,
i.e. P(A | B) = P(A) or, equivalently P(B | A) = P(B)
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Mathematics Semester 3 STPM Chapter 2 Probability
Example 30
A card is drawn randomly from a standard deck of 52 cards with replacement.
Determine whether the events “getting a spade” and “getting a numeric card” are independent.
Solution: Let event A = a spade is chosen,
event B = a numeric card is chosen.
P(B | A) = 40 = 10 and P(B) = 40 = 10 .
52 13 52 13
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Since P(B | A) = P(B), events A and B are independent.
2
Example 31
The table below shows 80 students registered for a programming course.
Basic Advanced
Girls 11 24
Boys 16 29
Determine whether the events “a girl is selected” and “a student register advanced programming is
selected” are independent.
Solution: Let event A be a girl is selected,
event B be a student register advanced programming is selected.
From the table, P(A) = 11 + 24
80
= 35
80
= 7
16
= 0.4375
and P(A | B) = P(A B)
P(B)
24
80
=
24 + 29
80
= 24
53
= 0.4528
Since P(A | B) ≠ P(A), the two events are dependent.
Probability of the intersection of events
Based on the definition of the conditional probability we have
P(A B) = P(A | B) × P(B)
This is called the multiplication rule of probability.
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Mathematics Semester 3 STPM Chapter 2 Probability
If events A and B are independent, then P(A | B) = P(A).
Hence, P(A B) = P(A) × P(B)
This is the multiplication rule for independent events.
In words, two events are independent if and only if the probability that both events will happen is found
by multiplying their individual probabilities.
The above relationship is the multiplication law of probability for independent events. It provides a simple
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way to check whether events are independent.
Example 32 2
A couple has five children. Assume that the probability of getting a boy or a girl is 1 . Find the probability
that 2
(a) all five are boys, (b) three are girls and two are boys.
Solution: Let event A = a child is a boy,
event B = the couple has three girls and two boys.
(a) Knowing that the child is a boy has no influence on the next birth, so the
five births are independent of one another.
Given P(A) = 1 ,
2
we have P(all five children are boys) = 1 × 1 × 1 × 1 × 1
2 2 2 2 2
= 1
32
Hence, the probability that all five are boys is 1 .
32
(b) One of the possible birth orders to get three girls and two boys is GGGBB.
The probability of this combination is
= 1 × 1 × 1 × 1 × 1
2 2 2 2 2
= 1
32
The number of possible combinations to position three girls in the birth
5
order of five children is C
3
= 5 × 4 × 3
3 × 2 × 1
= 10
Each combination would have the same probability of 1 .
32
Thus, P(B) = 10 × 1
32
= 5
16
Hence, the probability of getting three girls and two boys is 5 .
16
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Mathematics Semester 3 STPM Chapter 2 Probability
Example 33
Two events C and D are such that P(C) = 2 and P(D) = 1 . If
5 3
(a) P(C D) = 11 , (b) P(C D) = 9 ,
15 15
determine if C and D are mutually exclusive and find also if C and D are independent.
Solution: (a) By using P(C D) = P(C) + P(D) – P(C D)
11 = 2 + 1 – P(C D)
15 5 3
2 P(C D) = 0
\ C and D are mutually exclusive.
P(C) × P(D) = 2 × 1 = 2
5 3 15
Since P(C D) ≠ P(C) × P(D), C and D are not independent.
(b) By using P(C D) = P(C) + P(D) – P(C D)
9 = 2 + 1 – P(C D)
15 5 3
P(C D) = 2
15
Since P(C D) ≠ 0, C and D are not mutually exclusive.
As P(C D) = P(C) × P(D), C and D are independent.
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Example 34
The probability that a new released model of printer will develop a fault within a year is 0.2. If two new
printers are selected at random from a store, determine the probability that only one printer will develop
a fault.
Solution: We start with a tree diagram showing all the possible combined outcomes of the
two experiments, the happening of the first and second printers.
The first set of branches of the tree shows what could happen to the first printer
and the second set of branches indicates what could happen to the second printer.
Since a printer has two possible outcomes, fault or no fault and each of these
may lead to two other possible outcomes of the second printer, we have a total
of 2 × 2 = 4 possible outcomes.
First printer Second printer
fault
0.2
fault
0.2
0.8 no fault
0.2 fault
0.8
no fault
0.8
no fault
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Let F and F be the events that the first and second printers developing fault
1
2
respectively.
We understand that events F and F are independent.
1
2
P(F ) = P(F ) = 0.2,
1
2
P(F ) = P(F ) = 1 – 0.2
2
1
= 0.8
Probability associated with each branch on the tree is written down.
From the tree diagram, two paths give the outcomes that one of the two printers
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will develop a fault.
The probability that the first printer develops a fault, 2
P(F F ) = P(F ) × P(F )
1
1
2
2
= 0.2 × 0.8
= 0.16
And the probability that the second printer develops a fault,
P(F F ) = P(F ) × P(F )
1
2
1
2
= 0.8 × 0.2
= 0.16
Thus the probability that only one printer develops a fault
= P(F F ) + P(F F )
1
2
1
2
= 0.16 + 0.16
= 0.32.
Note: To use the tree diagram, multiply the probabilities along the branches of a path and add the
probabilities when more than one path fulfilling the requirements.
Example 35
A past record in a town provides the following information:
On a rainy day the probability of a driver involved in an accident is 0.08, whereas the probability of a
driver involved in an accident is 0.03 if there is no rain. The probability of rain in the town is forecasted
to be 0.25 in these few days. Find the probability that a driver will not involve in an accident tomorrow.
Solution: The tree diagram is displayed below where,
Event R: it is raining,
Event A: a driver involved in an accident.
Weather Accident or no accident
accident
0.08
rain
0.25
0.92 no accident
accident
0.03
0.75
no rain
0.97 no accident
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Mathematics Semester 3 STPM Chapter 2 Probability
P(R) = 0.25,
P(R) = 1 – 0.25
= 0.75
P(A | R) = 0.08, P(A | R) = 0.03.
The probability of a driver will not involve in an accident on raining tomorrow,
P(A R) = P(R) × P(A | R)
= 0.25 × 0.92
= 0.23
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The probability of a driver will not have an accident on not-raining tomorrow,
P(A R) = P(R) × P(A | R)
2 = 0.75 × 0.97
= 0.728
The probability that a driver will not have an accident tomorrow,
= P(A R) + P(A R)
= 0.23 + 0.728
= 0.958
Rule of total probability
Suppose that a sample space consists of three exhaustive and mutually S A 1 A 2 A 3
exclusive events, A , A and A . By definition, the three events do not
2
1
3
overlap and they occupy the entire sample space. The Venn diagram on
the right displays the events A , A and A and any event B.
1 2 3
B
From the diagram, the event B is composed of three mutually exclusive events A B, A B and A B.
3
1
2
So, P(B) = P(A B) + P(A B) + P(A B)
1 2 3
By applying the conditional probability formula to each term on the right hand side of this equation, we obtain
P(B) = P(A ) × P(B | A ) + P(A ) × P(B | A ) + P(A ) × P(B | A )
1 1 2 2 3 3
This formula is known as the rule of total probability. This rule states that the whole is the sum of its parts.
In general, for some positive integer k, let A , A , …, A be such that
1 2 k Law of
1. S = A A … A k Total
1
2
2. A A = 0 if i ≠ j INFO Probability
j
i
Then the collection of sets {A , A , …, A } is said to be a ‘partition’ of S.
1 2 k
Note: If B is any subset of S, and{A , A , …, A } is a partition of S, B can be decomposed as follows:
1 2 k
B = (A B) (A B) … (A B)
k
2
1
Thus, the rule of total probability states that: if {A , A , …, A } is a partition of S such that
1
2
k
P(A) . 0 for i = 1, 2, …, k, then for any event B
i
k
P(B) = ∑ P(A) P(B | A)
i = 1 i i
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Mathematics Semester 3 STPM Chapter 2 Probability
Example 36
A telemarketing company makes a total of 100 phone calls to its customers, 60 calls in the morning and
40 calls in the afternoon. The successful rates of selling its products in the morning and in the afternoon
are 0.25 and 0.16 respectively. Find the total successful rate of the company in selling its products.
Solution: Let A represents a morning call,
1
A represents an afternoon call,
2
B represents a sale of a product,
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Events A and A are mutually exclusive.
1 2
40
60
P(A ) = 100 P(A ) = 100
2
1
= 0.6, = 0.4, 2
P(B | A ) = 0.25 and P(B | A ) = 0.16
1
2
By the rule of total probabilities, we have
P(B) = P(A ) × P(B | A ) + P(A ) × P(B | A )
2
1
2
1
= 0.6 × 0.25 + 0.4 × 0.16
= 0.214
Hence, the total successful rate of the company in selling its products is 0.214.
Example 37
A shop ordered certain item from three different suppliers. The table below shows the distribution of the
current stock of the item and the corresponding percentages of defective units.
Supplier 1 Supplier 2 Supplier 3
Stock 150 60 40
Defective units 2% 3% 5%
If a unit is selected at random from the item stock and is found to be defective, find the probability that
the selected unit came from supplier 3.
Solution: We form a tree diagram and place appropriate probability on each branch as
shown.
Let S , S and S represent the events that the item unit are from suppliers 1, 2
3
2
1
and 3 respectively and D be the event that the selected unit is defective.
Supplier Defective or good
defective
0.02
1
0.98 good
0.6
0.03 defective
0.24
2
0.97 good
0.16 defective
0.05
3
0.95 good
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Mathematics Semester 3 STPM Chapter 2 Probability
We are interested in finding P(S | D), the probability of a unit from supplier 3
3
given that the unit is defective.
From the tree diagram, the probability of selecting a defective unit is
P(D) = 0.6 × 0.02 + 0.24 × 0.03 + 0.16 × 0.05
= 0.0272
and the probability that the selected unit is from supplier 3 and is defective is
P(S D) = 0.16 × 0.05
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3
= 0.008
2 P(S | D) = product of branch probabilities leading to D through S 3
3 sum of all branch probability products leading to D
= 0.16 × 0.05
0.6 × 0.02 + 0.24 × 0.03 + 0.16 × 0.05
= 0.008
0.0272
= 0.2941
The probability of a unit from supplier 3 given that the unit is defective is 0.2941.
Exercise 2.2
1. Consider the experiment about a team’s result in a football match.
(a) Determine the sample space.
(b) Write the following events.
(i) The team wins the match
(ii) The team does not lose the match
2. A number is randomly picked from a set of integers, {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}.
(a) Find the sample space of the above experiment.
(b) List the outcomes in the following events for the above experiment.
(i) The number is divisible by 4.
(ii) The number is an odd number.
(iii) The number is 12.
3. An experiment involves flipping a fair coin twice and recording the resulting sequence of happenings.
(a) Describe the sample space of this experiment.
(a) Determine the event that at least one tail appears.
4. A fair die is thrown once. If the event of interest is obtaining a number less than 3, find the probability
of the event happening.
5. A jar contains 2 red marbles, 4 yellow marbles and 3 green marbles. A marble is drawn at random.
(a) Describe the sample space S of this experiment.
(b) Find the probability that
(i) the marble is yellow,
(ii) the marble is not green.
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Mathematics Semester 3 STPM Chapter 2 Probability
6. A bag contains one blue and two green balls. Two balls are randomly chosen, one at a time. Their
colours are recorded
(a) Use a tree diagram to find the sample space.
(b) Find the probability that both balls are green.
7. Two fair coins are tossed, and the outcome is recorded. The events of interest are listed as
follows:
A : At least one head is observed
B : At least one tail is observed
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C : No head is observed
Define the following events, and find their respective probabilities.
(a) A (b) B 2
(c) C (d) A B
(e) A B (f) A
8. A bowl contains three balls, one blue, one green, and one red. A student draws two balls at random,
one at a time with replacement. What is the probability that at least one ball is red?
9. A box contains six balls, two blue and four green. Two balls are drawn at random in succession without
replacement. Find the probability that exactly one is green.
10. An experiment consists of casting a die and recording whether the number of up face is even or odd.
If the number is odd, then a coin is tossed once and the up face is recorded. If the number is even,
then the coin is tossed twice and the up face is recorded for each toss. Use a tree diagram to determine
the sample space of the experiment and find the probability that,
(a) at least one head appears,
(b) no heads appear, and
(c) no tails appear.
What is the relation between the events for parts (a) and (b)?
11. An ordinary deck of cards contains 52 cards divided into four suits:
The red suits are diamonds (◆) and hearts (♥) and
the black suits are clubs (♣) and spades (♠).
Each suit contains 13 cards of the following denominations:
2, 3, 4, 5, 6, 7, 8, 9, 10, J(jack), Q(queen), K(king), and A(ace).
The cards J, Q, K, and A are called face cards.
(a) List the sample space of the outcomes.
(b) State the event that the chosen card is a red face card.
(c) Find the probability that the chosen card is a red face card.
12. A die is one of a pair of dice. It is a cube with six sides, each containing from one to six dots,
called pips. Suppose a black die and a green die are tossed together, and the numbers of dots that
occur face up on each are recorded.
(a) Give a collectively exhaustive list of the possible outcomes.
(b) Use set notation to write the event E that the numbers showing face up have a sum of 7.
(c) Find the probability that the numbers showing face up have a sum of 7.
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Mathematics Semester 3 STPM Chapter 2 Probability
13. When drawing a card from a deck of playing cards, determine whether the following events are
mutually exclusive:
(a) the events “ace” and “king”,
(b) the events “ace” and “spade”.
Hence, find in a single draw, the probability of drawing,
(c) either an “ace” or a “king”,
(d) an “ace” or “spade” or both.
14. A die is cast.
(a) List the possible outcomes.
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(b) List the simple events.
(c) Define the sample space.
2 (d) Are the events mutually exclusive?
(e) Are the events exhaustive?
(f) Assuming that there is equal probability for the die to land with any of its faces up and that it
will not stand on its edge, find the probability of each event.
15. Consider the experiment of casting two fair dice, one black and one white and the separate numbers
shown uppermost are observed. List the outcomes as ordered pairs, (b, w) on a table where b and w
represent the numbers shown uppermost on black and white respectively.
Events E , E and E are defined as follows:
2
3
1
E = The number shown uppermost on the black dice exceeds that on the white dice.
1
E = The number shown uppermost on the black dice exceeds 2.
2
E = The total number shown uppermost on both dice is less than 9.
3
(a) Verify that P(E ) + P(E ) = P(E E ) + P(E E )
1 2 1 2 1 2
and P(E ) + P(E ) = P(E E ) + P(E E )
1 3 1 3 1 3
(b) Identify a pair of events that are exhaustive.
(c) What is the relation between E ' and E '.
2
3
(d) Find P(E E ) and P(E ' E ').
2
3
3
2
16. It is given that, for events A and B,
P(A) = 0.5, P(A B) = 0.9 and P(A B) = 0.2. Find
(a) P(B) (b) P(A B' )
(c) P(A' B) (d) P(A' B' )
17. In a swimming competition in which there are no dead heats, the probability that swimmer A wins is
0.4, the probability that swimmer B wins is 0.3 and the probability that swimmer C wins is 0.2. Find
the probability that
(a) swimmer A or B wins,
(b) swimmer A or B or C wins,
(c) someone else wins.
18. Events C and D are mutually exclusive such that, P(C) = 2 and P(D) = 3 . Find
5 10
(a) P(C) (b) P(D) (c) P(C D)
(d) P(C D) (e) P(C D) (f) P(C D)
19. The probabilities that it will rain in a town on a day in mid-September, that there will be a thunderstorm
on that day, and that there will be rain as well as a thunderstorm are 0.28, 0.23 and 0.16 respectively.
What is the probability that there will be rain and/or a thunderstorm in the town on such a day?
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Mathematics Semester 3 STPM Chapter 2 Probability
20. A consumer research organisation studies the service under warranty provided by the 200 water purifier
dealers in a city. The information is summarised in the following contingency table:
Good service under Poor service Total
warranty under warranty
Brand A water purifier dealers 65 15 80
Brand B water purifier dealers 45 75 120
Total 110 90 200
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(a) If one of these water purifier dealers is randomly selected, find the probability of choosing a 2
Brand A dealer who provides good service under warranty.
(b) Find the probability that the water purifier dealer provides good service under warranty given
that,
(i) he is from Brand A, (ii) he is from Brand B.
21. There are 48% boys and 52% girls in a classroom. The proportions of wearing spectacles for boys and
girls are shown in the following probability table
Boys Girls Total
Wearing spectacles 0.36 0.42 0.78
Not wearing spectacles 0.12 0.10 0.22
Total 0.48 0.52 1.00
If a student is selected at random from the classroom and is found
(a) to be a boy, find the probability that he wears spectacles.
(b) to be a girl, find the probability that she wears spectacles.
22. Consider the experiment of tossing two coins. The H and T events are defined as follows:
H : Head on the first coin
T : Tail on the second coin
Determine whether the events H and T are independent.
23. In a telephone survey of 900 adults, respondents are asked about the expense of owning a car and
the relative necessity of some form of financial assistance. The respondents are classified according
to whether they currently own a car and whether they think the loan burden for most car owners is
too high, moderate or low. The proportions responding in each category are shown in the following
probability table.
Too high (C) Moderate (D) Low (E)
Owns a car (A) 0.38 0.09 0.01
Does not own a car (B) 0.29 0.13 0.10
If one respondent is randomly chosen, find the probability that the respondent
(a) owns a car,
(b) does not own a car,
(c) owns a car or thinks that the loan burden is too high.
Are the events A and C independent? Explain.
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Mathematics Semester 3 STPM Chapter 2 Probability
24. Consider the experiment of tossing a fair dice. The events E and F are defined as follows:
E : Observe an even number
F : Observe a number less than or equal to 4 is obtained
Determine whether the events E and F are independent.
25. The probability that John will like a new sport is 0.70 and the probability that Peter, his brother, will
like it is 0.60. If the probability is 0.28 that he will like it and Peter will dislike it, what is the probability
that he will like it given that Peter is not going to like it?
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26. Two thousand randomly selected adults are asked whether or not they have never shopped on the
internet. The following table gives a two-way classification of the responses obtained.
2
Have shopped (H) Have never shopped (N)
Male (M) 554 386
Female (F) 612 448
Suppose one adult is selected at random from these 2000 adults. Find the probability that
(a) the adult has never shopped on the internet.
(b) the adult has shopped on the Internet or is a female.
(c) the adult is a male given that the adult has shopped on the internet.
27. A production process uses two machines in its daily production. A random sample of 500 items
produced are inspected and the following contingency table is obtained.
Defective Non-defective
Machine X 15 285
Machine Y 6 194
If an item is selected randomly, what is the probability that the item is
(a) defective,
(b) produced by machine Y and defective,
(c) produced by machine X or non-defective,
(d) defective given that it is produced by machine X.
28. Suppose there are two containers C and C . C has eight blue balls and two red balls, while C has four
2
2
1
1
blue balls and six red balls. If a container is selected randomly, and a ball is then selected randomly
from that container, the sequential process and probabilities is shown by the following tree diagram.
B
0.8
C 1
0.5
0.2
R
B
0.4
0.5
C
2
0.6
R
Find the probability that a selected red ball come from C .
1
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Mathematics Semester 3 STPM Chapter 2 Probability
Summary
1. Addition principle of counting:
Let A , A , … A be disjoint events with n , n , … n possible outcomes, respectively.
2
1
2
k
k
1
Then the total number of outcomes for the event “A or A or … or A ” is n + n + … + n .
1 2 k 1 2 k
2. Multiplication principle of counting:
Let A , A , …, A be events with n , n , …, n possible outcomes, respectively.
2
2
1
k
k
1
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Then the total number of outcomes for the sequence of these k events is
n × n × … × n
1 2 k
3. First permutation rule 2
The number of permutation of n distinct elements is n!.
4. Second permutation rule
n!
The number of permutations of n distinct elements taken r at a time is, P = (n – r)! .
n
r
5. The number of distinct permutations of n elements of which n are of one kind, n of a second kind,
2
1
n!
…, n of a kth kind is given by the formula n ! n ! … n ! .
k
2
k
1
6. The number of possible combinations of choosing r elements from a set of n elements without regard
n!
n
to order is, C = (n – r)! r! .
r
7. An outcome is a result of some activity.
For example: Rolling a dice has six outcomes: 1, 2, 3, 4, 5, 6
8. In statistics the word experiment is used to describe any process that generates raw data or outcome.
9. A sample space is a set of all possible outcomes for an activity and is represented by S.
For example: The sample space for rolling a dice is, S = {1, 2, 3, 4, 5, 6}.
10. An event is the collection of outcomes of particular interest in an experiment.
11. An event is the subset of the sample space S.
12. Probability is a measure of how likely an event is to happen.
P(Event) = number of ways that an event can occur
total number of possible outcomes
13. (a) The probability that an event will happen is between 0 and 1 inclusive, i.e.
0 < P(E) < 1
(b) P(φ) = 0; φ is 0 collections.
(c) P(S) = 1
14. If an experiment can result in any one of N different equally likely outcomes, and if exactly n of these
outcomes correspond to event E, then the probability of event E is P(E) = n .
N
15. If an experiment is repeated n times under the identical condition and an event is observed to happen
f times, the probability of the event happening is then estimated to be
frequency of the event occured f
P(E) = =
total number of observations n
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Mathematics Semester 3 STPM Chapter 2 Probability
16. The complement rule: P(E) = 1 – P(E)
17. Two events are said to be exhaustive if it is certain that at least one of them occurs. If the events A
and B are exhaustive, then A B = S.
18. Two or more events are mutually exclusive or disjoint if the events cannot occur at the same time.
19. If A and B are two mutually exclusive events, then
(a) A B = φ, (b) P(A B) = 0,
(c) P(A B) = P(A) + P(B).
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20. If two events A and B are mutually exclusive and exhaustive, then
2 P(A B) = 0 and P(A) + P(B) = 1
21. An event and its complement are always exhaustive and mutually exclusive:
P(A A) = 1 as well as P(A A) = 0
22. Addition rule of probability: S
P(A B) = P(A) + P(B) – P(A B) A B
23. Conditional probability
The probability of event A happening given that event B has happened,
P(A | B) = P(A B) , provided that P(B) ≠ 0.
P(B)
24. Multiplication rule of probability: P(A B) = P(A | B) × P(B)
25. P(A | B) × P(B) = P(B | A) × P(A)
26. If A and B are mutually exclusive events, i.e. A B = φ, then P(A B) = 0. Thus,
P(A | B) × P(B) = P(B | A) × P(A) = 0
27. For mutually exclusive and exhaustive events A and A, A A = φ, A A = S,
(a) P(S | B) = P(A | B) + P(A | B)
(b) P(S | B) = 1
(c) P(A | B) + P(A | B) = 1
.
.
28. P(A) = P(A | B) P(B) + P(A | B) P(B)
29. If the outcome of event A does not affect the outcome of event B, then A and B are independent
30. If events A and B are independent, then
(a) P(A | B) = P(A) or, equivalently P(B | A) = P(B).
(b) P(A B) = P(A) × P(B)
31. The rule of total probability states that: if {A , A , …, A } is a partition of S such that
1 2 k
k
P(A) . 0 for i = 1, 2, …, k, then for any event B, P(B) = ∑ P(A)P(B | A)
i i i
i = 1
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Mathematics Semester 3 STPM Chapter 2 Probability
STPM PRACTICE 2
1. Three sets A, B and C are given by
A = {k, d, m, a}
B = {l, o, w}
C = {h, b, v, f, y}
If one letter is chosen from among the sets A, B or C, how many ways are there?
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2. There are 100 customers in a store, 20 of these customers have both coffee and tea, 60 have coffee and
35 have tea. How many customers have coffee or tea?
3. A password consists of three symbols. If the first symbol is a character, the second and the third 2
symbols are digits, how many three-symbol passwords could be formed?
4. APEC stands for Asia Pacific Economic Cooperation. How many different permutations are possible
for the set of letters {A, P, E, C}?
5. There is a picture taking session in a birthday party. 6 people line up for a picture. How many different
ways can they be arranged from the left to the right?
6. Evaluate each of the following permutations.
(a) 7 P 2 (b) 10 P 6 (c) 4 P 4 (d) 20 P 1
8
(e) C 3 (f) 12 C 12 (g) 7 C 6 (h) 13 C 0
7. There are first, second and third prizes to be awarded to 12 participants in a singing contest. How
many ways can the prizes distributed among the participants?
8. A badminton team has 5 players. How many ways can a coach select the first and second singles?
9. In how many different ways can the 25 committee members of a sports club choose a president, a
vice-president, a secretary, and a treasurer?
10. Find the number of permutations from arranging 2 yellow cards, 3 red cards and 1 green card.
11. A set has 8 elements. How many subsets of the set can be formed such that each subset consists of 5
elements?
12. A teacher prepares a list of 9 questions. A test paper consists of 4 questions from the list. How many
test papers can be set?
13. How many different ways can 4 kings be chosen from a standard deck of 52 playing cards?
14. A fair dice is rolled. If the events of interest are:
(i) Event A of getting a number three
(ii) Event B of getting a prime number
(a) Describe the sample space of this experiment.
(b) Determine the probabilities of P(A) and P(B).
15. You spin once on the spinner as shown on the right:
Find the probability of each of the following event. 1 3
(a) Spin a number 1. 2 1
(b) Spin a number 3. 2 1
(c) Spin an even number. 1 4
(d) Spin a number larger than 4.
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Mathematics Semester 3 STPM Chapter 2 Probability
16. The probability that an ambulance from Red Crescent reaches a patient in less than half an hour is
0.83. If the Red Crescent service receives 7500 calls in one year, find the number of patients reached
within half an hour.
17. An ordinary deck of cards contains 52 cards divided into four suits:
The red suits are diamonds (◆) and hearts (♥) and
the black suits are clubs (♣) and spades (♠).
Each suit contains 13 cards of the following denominations:
2, 3, 4, 5, 6, 7, 8, 9, 10, J(jack), Q(queen), K(king), and A(ace).
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The cards J, Q, K, and A are called face cards.
(a) State the event that the chosen card is a spade.
2 (b) Find the probability that the chosen card is a spade.
18. A typical personal identification number, PIN, is a sequence of any three symbols chosen from the
26 letters in the alphabet and the 10 digits.
(a) Find the number of different possible PINs, if
(i) repetition is allowed,
(ii) repetition is not allowed.
(b) If all PINs are equally likely, find the probability that a randomly chosen PIN
(i) contains no repeated symbols,
(ii) contains a repeated symbol.
19. (a) Find the number of ways the letters in the word COMPUTER can be arranged in a row if,
(i) there is no restrictions,
(ii) the letters ER must remain next to each other in order as a unit.
(b) Find the probability that the letters ER appear together in order if the letters of the word
COMPUTER are randomly arranged in a row.
20. Two letters are selected at random from the word REJECT. Find the probability that the selection
(a) does not contain the letter E,
(b) contains one letter E,
(c) contains both the letters E.
21. It is forecasted that there is a 40% chance of raining on Tuesday and a 60% chance of raining on
Wednesday. It is expected that the chance of raining on both days is 35%. Determine the probability
that it will rain on either Tuesday or Wednesday.
22. 3 students are chosen from a group of 10 students consisting of 7 boys and 3 girls to represent a school
in chess competition. If the selection is merely based on random picking, find the probability that the
representatives are formed by
(a) 2 boys and 1 girl,
(b) at least 2 girls.
23. Data was collected on the gender of customers entered a supermarket on a particular day. It was found
that out of 360 customers, 249 were female. Estimate the probability that a person who visited the
supermarket on that day is female.
24. Consider the events of drawing a diamond (◆) or drawing a heart (♥) out of a deck of cards.
(a) Determine whether the events are mutually exclusive.
(b) Find the probability of drawing either a diamond (◆) or a heart (♥).
25. Consider the events of drawing a 5 or drawing a heart (♥) out of a deck of cards.
(a) Determine whether the events are mutually exclusive.
(b) Find the probability of drawing either a 5 or a heart (♥).
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02 STPM Math(T) T3.indd 110 28/10/2021 10:21 AM
Mathematics Semester 3 STPM Chapter 2 Probability
26. The two way classification table below shows the responses of 100 employees of a company regarding
paying high salaries to directors of companies.
In Favour (I) Against (A) Total
Male (M) 15 45 60
Female (F) 4 36 40
Total 19 81 100
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Suppose one employee is selected at random. Find the probability that the employee selected is,
(a) in favour of paying high salaries to directors.
(b) a male who is in favour of paying high salaries to directors. 2
(c) against of paying high salaries to directors given that this employee is a female. Draw a tree
diagram to illustrate this.
(d) a male given that this employee is in favour of paying high salaries to directors. Draw a tree
diagram to illustrate this.
27. In a cannery, assembly lines A, B and C account for 50%, 30% and 20% of the total output. The
corresponding percentages of the cans from assembly lines that are improperly sealed are listed in the
following table:
Assembly line A Assembly line B Assembly line C
Output 50% 30% 20%
Improperly sealed 0.4% 0.5% 1.1%
Find the probability that an improperly sealed can be discovered at the final inspection of outgoing
products comes from assembly line A.
28. Eighty job applicants are assessed as either good or poor for their aptitude test and communication
skill. The results are recorded in the following two-way classification table.
Good aptitude Poor aptitude
Good communication 24 18
Poor communication 28 10
Calculate the probability that an applicant has
(a) good aptitude test or communication skill,
(b) good aptitude test and communication skill.
29. In a survey on the number of electrical fans in a house, the following probability table is obtained.
Number of fans 0 1 2 3 or more
Probability 0.10 0.28 0.36 0.26
Calculate the probability of a house having
(a) more than one electrical fans,
(b) one or fewer electrical fans.
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Mathematics Semester 3 STPM Chapter 2 Probability
30. A production process uses two machines in its daily production. A random sample of 500 items
produced were inspected and listed in the table below.
Defective Non-defective
Machine A 15 285
Machine B 6 194
If an item is selected randomly,
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(a) find the probability that the item is
(i) defective,
2 (ii) produced by Machine A or non-defective,
(iii) defective given that it is produced by Machine A,
(b) determine whether
(i) the events “Machine A” and “defective” are independent,
(ii) the events “Machine B” and “non-defective” are mutually exclusive.
31. In a school, 45% of the students are males and 25% of the students play badminton. 11.25% of the
students are male students who play badminton. Events A and B are defined as follows:
A : A male student of the school is selected.
B : A student of the school, who play badminton, is selected.
(a) Find P(A), P(B), and P(A B).
(b) Determine whether
(i) A and B are mutually exclusive,
(ii) A and B are independent.
(c) Find P(A | B). What can you conclude?
32. Of the applications to a certain course, 70% are eligible to enter and 30% are not. To aid in the selection
process, an admissions test is conducted that is designed so that an eligible candidate will pass 80%
of the time, while an ineligible student will pass only 20% of the time.
(a) Find the probability that a student will pass the admissions test.
(b) If a student passes the admissions test, what is the probability that the student is eligible?
33. A bag contains 10 table tennis balls, of which 4 are dented. All table tennis balls look alike and have
equal probability of being chosen. Three table tennis balls are selected and placed in a bag. Find the
probability that
(a) all 3 are dented,
(b) exactly 2 are dented,
(c) at least 2 are dented.
34. A pharmaceutical company had developed a new diabetes treatment which is being tested on 1000
volunteers. In the test, 600 volunteers received the treatment and some a placebo (a harmless neutral
substance). It is found that 250 showed some improvement. It is also found that 450 received treatments
showed no improvement.
(a) Construct a two way classification table based on the above information.
(b) Find the probability that a random chosen volunteer
(i) showed some improvement after receiving a placebo,
(ii) received treatment or showed no improvement.
(c) Determine whether the events “a volunteer received treatment” and “a volunteer showed some
improvement” are independent. Explain.
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Mathematics Semester 3 STPM Chapter 2 Probability
35. There are nine wrapping bags, three of which contain 2 GB pen drives and the rest 1 GB pen drives.
There are three girls and five boys randomly select a bag each. Find the probability that
(a) the girls select more bags which contain 2 GB pen drives than the boys,
(b) none of the girls has a bag which contain 2 GB pen drives.
36. The result of a diagnostic test for a certain infection may be negative or positive, but the test is not
completely reliable. If a person has the infection, the probability that the result will be negative is 0.02.
If a person does not have the infection, the probability that the result will be negative is 0.94. In a
certain population, the percentage affected by the infection is 7%.
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A person is chosen at random and tested.
(a) Find the probability that the result of the test is negative.
(b) If the result of the test is negative, show that the probability of the individual not infected is 2
0.9974, by giving your answer correct to four decimal places.
37. Ahmad and Suhaimi frequently play each other in a series of games of tennis. Records of the outcomes
of these games indicate that Ahmad has the probability 0.6 of winning the first game and that in every
subsequent game in the series, his probability of winning the game is 0.7 if he won the preceeding
game but only 0.5 if he lost the preceding game. A game cannot end in a draw. Find the probability
that Ahmad will win the third game in the next series he plays with Suhaimi.
38. A man goes to work by bus, taxi or MRT. The probability that he travels by MRT is 0.56 and he is
equally likely to take a bus or a taxi. The probability that he is late for work if he goes by bus, taxi or
MRT is 0.3, 0.2 or 0.1 respectively. Calculate the probability that
(a) he is late for work on a randomly chosen working day,
(b) he goes to work by bus if he is late for work,
(c) he is not late for work if he goes to work by bus or MRT.
39. A box contains 2 blue balls, 3 red balls and 4 green balls. Five balls are taken out at random from the
box without replacement. Calculate the probability that
(a) at least one ball is green,
(b) exactly three are red balls, given that at least one ball is green.
40. Let E and F be two events such that P(E) = 0.3, P(F) = 0.5, and P(E F) = 0.7.
(a) Find P(E' F' ).
(b) Determine whether the events E' and F' are
(i) mutually exclusive,
(ii) independent.
41. Two events, A and B, are such that P(A) = 0.4, P(B | A) = 0.5 and P(B | A' ) = 0.3.
(a) Find
(i) P(A B)
(ii) P(A B)
(iii) P(B)
(iv) P(A B)
(b) Determine whether
(i) A and B are mutually exclusive. Explain.
(ii) A and B are independent. Explain.
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Mathematics Semester 3 STPM Chapter 5 Hypothesis Testing
CHAPTER
5 HYPOTHESIS TESTING
Learning Outcome
(a) Explain the meaning of a null hypothesis and an alternative hypothesis.
(b) Explain the meaning of the significance level of a test.
(c) Carry out a hypothesis test concerning the population mean for a normally distributed population with
known variance.
(d) Carry out a hypothesis test concerning the population mean in the case where a large sample is used.
(e) Carry out a hypothesis test concerning the population proportion by direct evaluation of binomial
probabilities.
Carry out a hypothesis test concerning the population proportion using a normal approximation.
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Bilingual Keywords
alternative hypothesis – hipotesis alternatif
critical region – rantau genting
critical value – nilai genting
5
hypothesis testing – pengujian hipotesis
null hypothesis – hipotesis nol
one-tailed test – ujian satu hujung
rejection region – rantau penolakan
significance level – aras keertian
test statistic – statistik ujian
two-tailed test – ujian dua hujung
Type I error – ralat jenis I
Type II error – ralat jenis II
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Mathematics Semester 3 STPM Chapter 5 Hypothesis Testing
5.1 Hypothesis Tests
In the last chapter we have discussed an important statistical inference: estimation of parameters, where
our objective is to estimate the unknown true value of a parameter. In this chapter we introduce another
important type of statistical inference: testing of statistical hypothesis, where we shall be interested to examine
whether the data from a random sample support or refute a conjecture about the true value of a parameter.
As an example, the manufacturer of a certain water filter may claim that the filtered water, has an average,
a pH value of 6.4. As another example, an insurance company may claim that the percentage of population
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who buy health insurance this year has increased compared to last year’s 12%. In the first example, it is
to test a hypothesis about the population mean. In the second example, it is to test a hypothesis about the
population proportion.
The truth or falsity of a statistical hypothesis is never known to us with absolute certainly unless we examine
the entire population. This, of course, would be impractical in most situations. Instead, we examine a random
sample from the population to produce evidence that either supports or refutes the hypothesis. The evidence
from the sample that is inconsistent with the stated hypothesis leads to the rejection of the hypothesis. A
hypothesis test or significance test is a method of using sample data as evidence to test a statistical hypothesis
about a population parameter.
The null and alternative hypothesis
The structure of hypothesis testing will be formulated with a null hypothesis denoted by H and an alternative
0
hypothesis denoted by H . Usually H specifies a particular value for a population parameter, and H specifies
0
1
1
a range of values. In general, the null hypothesis H represents there is no difference between the claim and
0
reality whereas the alternative hypothesis H represents there is statistically significant difference between
1
the claim and reality. In a hypothesis test, H is assumed to be true and information obtained for a sample
0
statistic is used to determine whether there is strong evidence to reject H .
0
Example 1
A hypothesis test is performed to determine whether the mean value of filtered water from a certain
type of water filter is 6.4.
State the null hypothesis and alternative hypothesis for the hypothesis test.
Solution: Let µ denote the mean pH value of filtered water.
The null hypothesis is that the mean pH value is 6.4, i.e. H : µ = 6.4
0
5 The alternative hypothesis is H : µ ≠ 6.4
1
Example 2
A hypothesis test is performed to determine whether the percentage of population who buy health
insurance this year has increased compared to last year’s 12%.
State the null hypothesis and alternative hypothesis for the hypothesis test.
Solution: Let p denote the proportion of population who buy health insurance this year.
The null hypothesis is that this year’s percentage remains the same as last year.
H : p = 0.12
0
The alternative hypothesis is that this year’s percentage has increased.
H : p . 0.12
1
Note: The null hypothesis H is usually stated using the equality sign.
0
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05 STPM Math(T) T3.indd 238 28/10/2021 10:24 AM
Mathematics Semester 3 STPM Chapter 5 Hypothesis Testing
Test statistics
A test statistic is a random variable whose value is used to determine whether a null hypothesis is rejected
in a hypothesis test.
The choice of a test statistic depends on the assumed probability distribution and the hypothesis under
question.
Consider the following example. A study claims that 20% drivers in a city involve running red lights. We
choose, at random, 20 drivers from the city. If more than 7 drivers admit to running red lights, it indicates
a higher percentage. In this case, we are essentially testing the null hypothesis that 20% drivers involve
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running red lights against the alternative hypothesis that the percentage is higher.
This can be written as follows:
H : p = 0.2,
0
H : p . 0.2.
1
The test statistic on which we base our decision is random variable X, the number of drivers in our test. The
possible values of X, from 0 to 20, are divided into two groups: those numbers less than or equal to 7 and
those greater than 7. All possible values obtained greater than 7 constitute what is called the critical region.
The set of values that leads to the rejection of H in favour of H is called a critical region or rejection region.
0
1
Thus, if x . 7, we reject H in favour of the alternative hypothesis H . If x < 7, we fail to reject H .
0
0
1
This decision criterion is illustrated in the figure below.
Critical region
Do not reject H 0 Reject H 0
(p = 0.2) (p > 0.2)
x
0 20
Critical value, 7
Figure 5.1
Type I and Type II errors
The decision procedure described in test statistics above could lead to either of two wrong conclusions. We
may reject H when in fact H is true, that is, the percentage of running of red lights by drivers does not
0
0
increase. This may occur because we happen to choose this particular selected group of drivers who have
such rude behaviour. Or, alternatively, we may not reject H when in fact H is false, that is, running red
0
0
lights is getting worse. 5
A Type I error occurs when a true H is rejected; a Type II error occurs when a false H is not rejected.
0 0
It is obvious that we cannot completely avoid making these errors. Our goal is try to keep the probability
of making these errors relatively small.
In testing any statistical hypothesis, there are four possible outcomes that determine whether our decision
is correct or in error. These four possibilities are listed in the following table.
H is true H is false
0 0
Do not reject H 0 Correct decision Type II error
Reject H Type I error Correct decision
0
Table 5.1
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Mathematics Semester 3 STPM Chapter 5 Hypothesis Testing
The probability of making a Type I errror is called the significance level and is denoted by the Greek letter
a. In our example, a type I error will occur when more than 7 drivers rush through red lights that is actually
an odd sample taken. Hence, if X is the number of drivers who involve running red lights,
20 20
x
P(Type I error) = P(X . 7 when p = 0.2) = ∑ 1 2 (0.2) (1 – 0.2) 20 – x ,
x = 8 x
7 20
x
1 – ∑ 1 2 (0.2) (1 – 0.2) 20 – x = 1 – 0.9679 = 0.0321.
x = 0 x
We say that the null hypothesis is being tested at a significance level of 3.21%.
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The probability of making Type II error is denoted by the Greek letter b. It is impossible to calculate this
probability unless a specific value is stated in the alternative hypothesis. We shall not discuss the determination
of b.
It can be shown that, for a fixed sample size, a decrease in the probability of one error will usually result in
an increase in the probability of the other error. However, we can reduce both types of errors by increasing
the sample size.
Example 3
A food company produces a box of 250 g corn flakes. It periodically conducts a statistical test to decide
whether the mean net mass of all boxes is 250 g. The null and alternative hypotheses are stated below.
H : µ = 250,
0
H : µ ≠ 250.
1
The results of carrying out the hypothesis test lead to no rejection of the null hypothesis. Comment on
the conclusion by error type or as a right decision to make if
(a) µ is in fact 250 g,
(b) µ is in fact not 250 g.
Solution: (a) If in fact µ = 250 g, the null hypothesis is true. Thus, by not rejecting the
null hypothesis, we have made a right decision. This interprets that the
package machine is functioning properly.
(b) If in fact µ ≠ 250 g, the null hypothesis is false. Thus, by not rejecting the
null hypothesis, we have committed a Type II error. This interprets that the
package machine is out of control even though the inspected output sample
indicates a satisfactory position.
5
One-tailed and two-tailed tests
Consider the null hypothesis that the mean weight of new born babies in a certain city is 3 kg. We test
H : µ = 3
0
against H : µ ≠ 3 or H : <3 or H : >3
1
1
1
Only one of these alternative hypotheses can be used at a time. We examine each case in turn.
Two-tailed test (H : µ ≠ 3)
1
A random sample of size n = 64 new born babies is taken. Assume that the standard deviation of the
–
population to be 1.60 kg. From the central limit theorem, we know that the sampling distribution of X is a
approximate normal distribution with standard deviation s – = s n = 1.6 = 0.2.
8
X
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05 STPM Math(T) T3.indd 240 28/10/2021 10:24 AM
Mathematics Semester 3 STPM Chapter 5 Hypothesis Testing
A sample mean that falls close to the hypothesised value of 3 kg would be considered evidence in favour of
H . Conversely, a sample mean that is significantly less than or more than 3 kg would be evidence favouring
–
0
H . A critical region, indicated by the shaded areas in the figure 5.2, is arbitrarily chosen to be X , 2.7 and
–
–
1
X . 3.3. If the sample mean X falls inside the critical region, H is rejected; otherwise H is not rejected.
0 0
–
–
2 2
–
x
2.7 = 3.0 3.3
Critical region Nonrejection region Critical region
Figure 5.2
The significance level of the test is equal to the total of the areas shaded in each tail of the normal distribution.
We have, – –
a = P(X , 2.7) + P(X . 3.3).
–
–
The z values corresponding to x = 2.7 and x = 3.3 are
1 2
z = 2.7 – 3.0 = –1.5
1 0.2
z = 3.3 – 3.0 = 1.5
2 0.2
Hence,
a = P(Z , –1.5) + P(Z . 1.5)
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= 2P(Z , –1.5)
= 0.1336
That is to say 13.36% of all samples of size 64 would lead to the rejection of H when it is true.
0
One-tailed test (H : µ < 3 or µ > 3)
1
The critical region for the alternative hypothesis H : µ , 3 lies entirely in the left tail of the normal
1
distribution, while the alternative hypothesis µ . 3 lies entirely in the right tail as shown in Figure 5.3.
5
– –
x
x
Critical region = 3.0 = 3.0 Critical region
Figure 5.3 (a) Figure 5.3(b)
In testing hypothesis about a continuous population, it is common to choose the value of a to be 1%, 5%
and 10%.
What is
Hypothesis
Testing?
INFO
241
05 STPM Math(T) T3.indd 241 28/10/2021 10:24 AM
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