96 Mathematics Semester 3 STPM Chapter 2 Probability 2 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 52 = 10 13 and P(B) = 40 52 = 10 13 . Since P(B | A) = P(B), events A and B are independent. 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.
97 Mathematics Semester 3 STPM Chapter 2 Probability 2 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 way to check whether events are independent. Example 32 A couple has five children. Assume that the probability of getting a boy or a girl is 1 2 . Find the probability that (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 2 × 1 2 × 1 2 × 1 2 × 1 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 2 × 1 2 × 1 2 × 1 2 × 1 2 = 1 32 The number of possible combinations to position three girls in the birth order of five children is 5 C3 = 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 .
98 Mathematics Semester 3 STPM Chapter 2 Probability 2 Example 33 Two events C and D are such that P(C) = 2 5 and P(D) = 1 3 . If (a) P(C D) = 11 15 , (b) P(C D) = 9 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 15 = 2 5 + 1 3 – P(C D) P(C D) = 0 \ C and D are mutually exclusive. P(C) × P(D) = 2 5 × 1 3 = 2 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 15 = 2 5 + 1 3 – P(C D) 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. 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. fault First printer 0.2 0.2 0.8 0.2 0.8 0.8 fault Second printer no fault no fault fault no fault
99 Mathematics Semester 3 STPM Chapter 2 Probability 2 Let F1 and F2 be the events that the first and second printers developing fault respectively. We understand that events F1 and F2 are independent. P(F1 ) = P(F2 ) = 0.2, P(F1 ) = P(F2 ) = 1 – 0.2 = 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 will develop a fault. The probability that the first printer develops a fault, P(F1 F2 ) = P(F1 ) × P(F2 ) = 0.2 × 0.8 = 0.16 And the probability that the second printer develops a fault, P(F1 F2 ) = P(F1 ) × P(F2 ) = 0.8 × 0.2 = 0.16 Thus the probability that only one printer develops a fault = P(F1 F2 ) + P(F1 F2 ) = 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. accident Accident or no accident no accident accident no accident rain Weather no rain 0.08 0.25 0.75 0.03 0.97 0.92
100 Mathematics Semester 3 STPM Chapter 2 Probability 2 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 The probability of a driver will not have an accident on not-raining tomorrow, P(A R) = P(R) × P(A | R) = 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 exclusive events, A1 , A2 and A3 . By definition, the three events do not overlap and they occupy the entire sample space. The Venn diagram on the right displays the events A1 , A2 and A3 and any event B. From the diagram, the event B is composed of three mutually exclusive events A1 B, A2 B and A3 B. So, P(B) = P(A1 B) + P(A2 B) + P(A3 B) By applying the conditional probability formula to each term on the right hand side of this equation, we obtain P(B) = P(A1 ) × P(B | A1 ) + P(A2 ) × P(B | A2 ) + P(A3 ) × P(B | A3 ) 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 A1 , A2 , …, Ak be such that 1. S = A1 A2 … Ak 2. Ai Aj = 0 if i ≠ j Then the collection of sets {A1 , A2 , …, Ak } is said to be a ‘partition’ of S. Note: If B is any subset of S, and{A1 , A2 , …, Ak } is a partition of S, B can be decomposed as follows: B = (A1 B) (A2 B) … (Ak B) Thus, the rule of total probability states that: if {A1 , A2 , …, Ak } is a partition of S such that P(Ai ) . 0 for i = 1, 2, …, k, then for any event B P(B) = k ∑ i=1 P(Ai ) P(B | Ai ) A1 A2 A3 B S Law of Total Probability INFO
101 Mathematics Semester 3 STPM Chapter 2 Probability 2 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 A1 represents a morning call, A2 represents an afternoon call, B represents a sale of a product, Events A1 and A2 are mutually exclusive. P(A1 ) = 60 100 P(A2 ) = 40 100 = 0.6, = 0.4, P(B | A1 ) = 0.25 and P(B | A2 ) = 0.16 By the rule of total probabilities, we have P(B) = P(A1 ) × P(B | A1 ) + P(A2 ) × P(B | A2 ) = 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 S1 , S2 and S3 represent the events that the item unit are from suppliers 1, 2 and 3 respectively and D be the event that the selected unit is defective. defective Defective or good good defective good defective good 1 Supplier 2 3 0.02 0.6 0.24 0.16 0.98 0.03 0.97 0.05 0.95
102 Mathematics Semester 3 STPM Chapter 2 Probability 2 We are interested in finding P(S3 | D), the probability of a unit from supplier 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(S3 D) = 0.16 × 0.05 = 0.008 P(S3 | D) = product of branch probabilities leading to D through S3 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.
103 Mathematics Semester 3 STPM Chapter 2 Probability 2 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 C : No head is observed Define the following events, and find their respective probabilities. (a) A (b) B (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.
104 Mathematics Semester 3 STPM Chapter 2 Probability 2 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. (b) List the simple events. (c) Define the sample space. (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 E1 , E2 and E3 are defined as follows: E1 = The number shown uppermost on the black dice exceeds that on the white dice. E2 = The number shown uppermost on the black dice exceeds 2. E3 = The total number shown uppermost on both dice is less than 9. (a) Verify that P(E1 ) + P(E2 ) = P(E1 E2 ) + P(E1 E2 ) and P(E1 ) + P(E3 ) = P(E1 E3 ) + P(E1 E3 ) (b) Identify a pair of events that are exhaustive. (c) What is the relation between E2 ' and E3 '. (d) Find P(E2 E3 ) and P(E2 ' E3 '). 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 5 and P(D) = 3 10 . Find (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?
105 Mathematics Semester 3 STPM Chapter 2 Probability 2 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 warranty Poor service under warranty Total Brand A water purifier dealers 65 15 80 Brand B water purifier dealers 45 75 120 Total 110 90 200 (a) If one of these water purifier dealers is randomly selected, find the probability of choosing a 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.
106 Mathematics Semester 3 STPM Chapter 2 Probability 2 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? 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. 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 C1 and C2 . C1 has eight blue balls and two red balls, while C2 has four 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 C1 C2 R B R 0.5 0.8 0.2 0.4 0.6 0.5 Find the probability that a selected red ball come from C1 .
107 Mathematics Semester 3 STPM Chapter 2 Probability 2 Summary 1. Addition principle of counting: Let A1 , A2 , … Ak be disjoint events with n1 , n2 , … nk possible outcomes, respectively. Then the total number of outcomes for the event “A1 or A2 or … or Ak ” is n1 + n2 + … + nk . 2. Multiplication principle of counting: Let A1 , A2 , …, Ak be events with n1 , n2 , …, nk possible outcomes, respectively. Then the total number of outcomes for the sequence of these k events is n1 × n2 × … × nk 3. First permutation rule The number of permutation of n distinct elements is n!. 4. Second permutation rule The number of permutations of n distinct elements taken r at a time is, n Pr = n! (n – r)!. 5. The number of distinct permutations of n elements of which n1 are of one kind, n2 of a second kind, …, nk of a kth kind is given by the formula n! n1 ! n2 ! … nk ! . 6. The number of possible combinations of choosing r elements from a set of n elements without regard to order is, n Cr = n! (n – 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 P(E) = frequency of the event occured total number of observations = f n
108 Mathematics Semester 3 STPM Chapter 2 Probability 2 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). 20. If two events A and B are mutually exclusive and exhaustive, then 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: P(A B) = P(A) + P(B) – P(A B) 23. Conditional probability The probability of event A happening given that event B has happened, P(A | B) = P(A B) P(B) , provided that P(B) ≠ 0. 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 events. 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 {A1 , A2 , …, Ak } is a partition of S such that P(Ai ) . 0 for i = 1, 2, …, k, then for any event B, P(B) = k ∑ i=1 P(Ai )P(B | Ai ) A B S
109 Mathematics Semester 3 STPM Chapter 2 Probability 2 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? 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 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 P2 (b) 10P6 (c) 4 P4 (d) 20P1 (e) 8 C3 (f) 12C12 (g) 7 C6 (h) 13C0 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. (a) Spin a number 1. (b) Spin a number 3. (c) Spin an even number. (d) Spin a number larger than 4. 3 1 1 41 2 2 1
110 Mathematics Semester 3 STPM Chapter 2 Probability 2 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). The cards J, Q, K, and A are called face cards. (a) State the event that the chosen card is a spade. (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 (♥).
111 Mathematics Semester 3 STPM Chapter 2 Probability 2 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 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. (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.
112 Mathematics Semester 3 STPM Chapter 2 Probability 2 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, (a) find the probability that the item is (i) defective, (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.
113 Mathematics Semester 3 STPM Chapter 2 Probability 2 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%. 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 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(A9 > 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.
Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 CHAPTER Subtopic Learning Outcome 3.1 Discrete random variables (a) Identify discrete random variables. (b) Construct a probability distribution table for a discrete random variable. (c) Use the probability function and cumulative distribution function of a discrete random variable. (d) Calculate the mean and variance of a discrete random variable. 3.2 Continuous random variables (a) Identify continuous random variables. (b) Relate the probability density function and cumulative distribution function of a continuous random variable. (c) Use the probability density function and cumulative distribution function of a continuous random variable. (d) Calculate the mean and variance of a continuous random variable. 3.3 Binomial distribution (a) Use the probability function of a binomial distribution, and find its mean and variance. (b) Use the binomial distribution as a model for solving problems related to science and technology. 3.4 Poisson distribution (a) Use the probability function of a Poisson distribution, and identify its mean and variance. (b) Use the Poisson distribution as a model for solving problems related to science and technology. 3.5 Normal distribution (a) Identify the general features of a normal distribution, in relation to its mean and standard deviation. (b) Standardise a normal random variable, and use the normal distribution tables. (c) Use the normal distribution as a model for solving problems related to science and technology. (d) Use the normal distribution, with continuity correction, as an approximation to the binomial distribution, where appropriate. binomial distribution – taburan binomial continuous random variable – pembolehubah rawak selanjar cumulative distribution function – fungsi taburan longgokan discrete random variable – pembolehubah rawak diskret mean – min normal distribution – taburan normal Poisson distribution – taburan Poisson probability density function – fungsi ketumpatan kebarangkalian probabilty distribution – taburan kebarangkalian random variable – pembolehubah rawak variance – varians PROBABILITY 3 DISTRIBUTIONS Bilingual Keywords
115 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 3.1 Discrete Random Variables Consider an experiment where a coin is thrown twice. Let T represent the event ‘coin shows tail’ and H represent the event the ‘coin shows head’. The sample space is {TT, TH, HT, HH}. Sometimes, we need to know the number of tails obtained. In this case, the number set to consider is {0, 1, 2}. We map {TT, TH, HT, HH} to the set {0, 1, 2} as shown in Figure 3.1. A number is associated with each outcome in the sample space by what we call a random variable. A random variable is a variable whose value is determined by the outcome of an experiment. Random variables are represented by capital alphabets such as X and Y, while their values are represented by lower case alphabets, such as x and y. Example 1 In order to test the understanding of a student on the concept of momentum, the student is required to answer two questions. For a correct answer, he is given 2 marks and 1 mark is subtracted for a wrong answer. If the random variable X represents the total marks obtained, find the possible values of X. Solution: Suppose C represents the event the answer is correct and W represents the event the answer is wrong. The sample space = {CC, CW, WC, WW}. For the outcome CC, the marks obtained is 2 + 2 = 4. For the outcome CW or WC, the mark obtained is 2 – 1 = 1. For the outcome WW, the marks obtained is –1 – 1 = –2. Hence, the values of X are 4, 1, –2. The mapping of set {CC, CW, WC, WW} to the set {4, 1, –2} is shown in the diagram below. –2 C C C W W C W W 4 1 Sample space For the above example, the values of the random variable are exact values. Random variables which have exact values are known as discrete random variables. Probability distributions If X is a discrete random variable which takes values x1 , x2 , ... , xn with probabilities P(X = x1 ), P(X = x2 ), ... , P(X = xn ) respectively, then all probabilities are non-negative and n ∑ i = 1P(X = xi ) = 1. Note that the number of values that X can take may be finite or infinite. T T T H H T H H 2 Sample space 1 0 Figure 3.1
116 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 A probability distribution of a discrete random variable lists all the possible values of a discrete random variable with their corresponding probabilities. It is usually in the form of a table or a formula (called probability distribution function or probability function). Example 2 Three fair coins are tossed. The discrete random variable X represents the number of tails obtained. (a) Find P(X = x), for x = 0, 1, 2, 3, and show that 3 ∑ i = 1P(X = xi ) = 1. (b) Tabulate all values of X and their corresponding probabilities. Solution: Suppose T is the event of obtaining a ‘tail’ and H is the event of obtaining a ‘head’. The sample space = {TTT, HTT, THT, TTH, HHT, HTH, THH, HHH}. Hence, the random variable X takes the values 0, 1, 2 and 3. (a) P(X = 0) = P(HHH) = 1 2 3 = 1 8 P(X = 1) = P(THH or HTH or HHT) = P(THH) + P(HTH) + P(HHT) = 1 2 3 + 1 2 3 + 1 2 3 = 3 8 P(X = 2) = P(TTH or THT or HTT) = P(TTH) + P(THT) + P(HTT) = 1 2 3 + 1 2 3 + 1 2 3 = 3 8 P(X = 3) = P(TTT) = 1 2 3 = 1 8 ∑ all x P(X = x) = 1 8 + 3 8 + 3 8 + 1 8 = 1 (b) We can summarise P(X = x) in the following table. x 0 1 2 3 P(X = x) 1 8 3 8 3 8 1 8 This table is known as the probability distribution table for X.
117 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 Example 3 The discrete random variable U has probability function u 2k for u = 2, 3, 5, P (U = u) = u 5k for u = 7, 8, 10, 0 otherwise. Find the value of k. Hence, (a) construct a probability distribution table for U, (b) find P(4 U 8). Solution: ∑ P(U = u) = P(U = 2) + P(U = 3) + P(U = 5) + P(U = 7) + P(U = 8) + P(U = 10) = 2 2k + 3 2k + 5 2k + 7 5k + 8 5k + 10 5k = 10 2k + 25 5k = 5 k + 5 k = 10 k Since U is a discrete random variable, ∑ all x P(U = u) = 1 10 k = 1 k = 10 (a) The probability distribution table for U is u 2 3 5 7 8 10 P(U = u) 0.10 0.15 0.25 0.14 0.16 0.20 (b) P(4 U 8) = P(U = 5) + P(U = 7) + P(U = 8) = 0.25 + 0.14 + 0.16 = 0.55 Exercise 3.1 1. A random variable X has probability distribution. x 0 2 4 6 P(X = x) 0.20 0.35 0.30 0.15 Find (a) P(X 3), (b) P(0 X 6).
118 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 2. A random variable Y has probability function 0.1 for y = 1, 3, 5, 7, P(Y = y) = 0.2 for y = 2, 4, 6, 0 otherwise. Find (a) P(Y 4), (b) P(3 Y 6). 3. W is a discrete random variable with probabilities a for w = 0, 1, P(W = w) = a + b for w = 2, b for w = 3, 4. If P(W 2) = 5 9 , find the values of a and b. Construct a probability distribution table for W. Draw a graph to illustrate this probability distribution. 4. The probability distribution of a discrete random variable X is given by P(X = x) = m (x + 1)2 for x = 1, 2, 3, 4. (a) Determine the value of m. (b) Construct a probability distribution table for X. (c) Find P(X 2). (d) Find P(X 2). 5. The probability function of a random variable X is x m for x = 1, 3, 5, 7, P(X = x) = x2 m for x = 2, 4, 6. (a) Determine the value of m. (b) Find P(|X – 4| 1). (c) Find P(X 3.5). 6. Two tetrahedral dice with faces marked 1, 2, 3 and 4 are thrown. The score obtained is the sum of the numbers on the bottom face. Construct a probability distribution table for the score obtained. 7. A box contains two yellow balls and five red balls. Balls are drawn at random from the box, one after the other without replacement, until a red ball is obtained. By using X to represent the number of balls drawn from the box, show that P(X = 2) = 5 21 . Construct a probability distribution table for X. 8. A bag contains two small dolls, two medium dolls and two large dolls. Dolls are drawn at random from the bag, one by one without replacement until two dolls of the same size are obtained. If X represents the number of dolls drawn from the bag, construct a probability distribution table for X. Hence, find P(2 X 3). 9. A fair dice is thrown continuously until the total score is 3 or more. The random variable S represents the number of times the dice is thrown. (a) Construct a probability distribution table for S. (b) Draw a graph for the probability distribution of S.
119 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 10. A random variable X has probability function P(X = x) = 2 5 3 5 x , for x = 0, 1, 2, ... . Verify that P(X = x) > 0 for x = 0, 1, 2, … and ∞ ∑ i = 1P(X = x) = 1. 11. The discrete random variable X has a probability distribution P(X = x) = k 3 4 x for x = 0, 1, 2, ... (a) Find the constant k. (b) Determine P(2 X 5). 12. In a game, a player is required to throw tennis balls into a basket at a fixed distance. Every player is given three tries. The probability that Zheyi succeeds in throwing the tennis ball into the basket is 0.4. If X represents the number of times the tennis balls enter the basket, show that X is a discrete random variable and find its probability distribution. 13. For each game of chess, a player who wins is given 1 point, 1 2 point if there is a draw and 0 point if he loses. Ming Kuen played three games of chess. Construct a sample space by using W for a win, D for a draw and L for a loss. Hence, by using T to represent the total points obtained by Ming Kuen, show that T is a discrete random variable and obtain its probability distribution table. 14. Based on the points discipline system in Sekolah Menengah Memir, 2 points are given for good behaviour and –1 point is given for bad behaviour every week. The assessment is done at random in a week. If + indicates good conduct and – indicates bad conduct, construct a sample space for a student’s behaviour after three weeks. If X represents the points obtained by a student in three weeks, show that X is a discrete random variable. 15. During a unity quiz, a participant tries to answer questions in order to win prizes. If a participant gives the correct answer, he will get a prize and leave the stage. If his answer is wrong, he will be given another question to answer. Every participant is given four chances to win a prize. Based on past records, the probability that a participant will be able to answer a question correctly is 2 3 . Let X represent the number of tries before a participant leaves the stage (either by winning or not). Show that X is a discrete random variable and obtain its probability distribution. What is the probability that a participant wins a prize after one or two tries only? Cumulative distribution function If X is a discrete random variable, the corresponding cumulative probabilities are obtained by summing all the probabilities up to a particular value. If X is a discrete random variable with probability function P(X = x) for x = x1 , x2 , …, xn , then the cumulative distribution function is given by F(t) = P(X < t) = t x Σ = x1 P(X = x), t = x1 , x2 , …, xn
120 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 Example 4 Find the cumulative distribution function of the random variable X where its probability distribution is given by the table below. X = x 1 2 3 4 5 P(X = x) 1 5 1 5 1 5 1 5 1 5 Solution: F(1) = P(X < 1) = 1 5 F(2) = P(X < 2) = 1 5 + 1 5 = 2 5 F(3) = P(X < 3) = 1 5 + 1 5 + 1 5 = 3 5 F(4) = P(X < 4) = 1 5 + 1 5 + 1 5 + 1 5 = 4 5 F(5) = P(X < 5) = 1 5 + 1 5 + 1 5 + 1 5 + 1 5 =1 Therefore, F(t) = t 5 , t = 1, 2, …, 5 Usually, we write the cumulative distribution function in terms of x, F(x). Hence, F(x) = x 5 , x = 1, 2, …, 5 Example 5 A discrete random variable X has cumulative distribution function F(x) as follows. x 0.2 0.3 0.4 0.5 0.6 0.7 F(x) 0.1 0.2 0.4 0.6 0.9 1 Find (a) P(X = 0.5), (b) P(X . 0.4). Solution: (a) F(0.5) = 0.6 P(X < 0.5) = 0.6 P(X = 0.2) + P(X = 0.3) + P(X = 0.4) + P(X = 0.5) = 0.6 ................ a F(0.4) = 0.4 P(X < 0.4) = 0.4 P(X = 0.2) + P(X = 0.3) + P(X = 0.4) = 0.4 ...........................................b a – b: P(X = 0.5) = 0.6 – 0.4 = 0.2 (b) P(X . 0.4) = 1 – P(X < 0.4) = 1 – F(0.4) = 1 – 0.4 = 0.6 Alternative Method P(X = 0.5) = F(0.5) – F(0.4) = 0.6 – 0.4 = 0.2
121 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 Exercise 3.2 1. Construct a cumulative distribution table for the following discrete random variables: (a) the number of heads when two fair coins are tossed, (b) the number of ones obtained when three fair dice tossed. 2. The probability distribution table for the discrete random variable X is given by the table below. X = x 4 5 6 7 8 P(X = x) 0.1 0.1 0.3 0.4 0.1 Construct the cumulative distribution table. 3. The cumulative distribution function F(x) of a discrete random variable X is shown below. x 5 6 7 8 9 F(x) 0.05 0.16 0.42 0.76 1 Construct the probability distribution of the random variable X. 4. For a discrete random variable Y, the cumulative distribution function F(y) is shown below. y 1 2 3 4 5 F(y) 0.15 0.50 0.72 0.84 1 Find (a) P(Y = 2), (b) P(Y , 3), (c) P(Y > 4). 5. For a discrete random variable X, the cumulative distribution function is given by F(x) = kx², x = 2, 3, 4. Find (a) the value of the constant k, (b) P(X < 3), (c) the probability distribution of X. Mean and variance of a discrete random variable Mean Two coins are tossed and the number of heads that occurred is recorded. What will be the mean number of heads obtained? For this experiment, the sample space is {HH, HT, TH, TT}. If X is a discrete random variable representing ‘the number of heads obtained’, then P(X = 0) = P(TT) = 1 4 , P(X = 1) = P(HT, TH) = 2 4 = 1 2 , P(X = 2) = P(HH) = 1 4 . Summarising, we get the following probability distribution table for X. X = x 0 1 2 P(X = x) 1 4 1 2 1 4
122 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 We can calculate the expected number of heads by multiplying each value of x with its corresponding probabilities. Hence, the expected number of heads, E(X) = ∑x P(X = x) = 0 1 4 + 1 1 2 + 2 1 4 = 1 Mean number of heads = 1 The expected value of a random variable, X, is written briefly as E(X) and is given by E(X) = Σx P(X = x) E(X) is also called the mean value of X. Note: If a probability distribution for a random variable X is symmetrical about x = k, then the mean of X = E(X) = k. Example 6 X is a discrete random variable with probability distribution given by P(X = x) = k, x = 2, 4, 6, 2kx, x = 1, 3, 5. Find (a) the value of k, (b) E(X), (c) P[X E(X)]. Solution: (a) ∑ P(X = x) = 2k + k + 6k + k + 10k + k = 21k Since ∑P(X = x) = 1, then 21k = 1 k = 1 21 (b) E(X) = ∑ x P(X = x) = 1(2k) + 2(k) + 3(6k) + 4(k) + 5(10k) + 6(k) = 2k + 2k + 18k + 4k + 50k + 6k = 82k = 82 1 21 Substitute k = — 1 21 = 82 21 (c) P[X E(X)] = P X 82 21 = P(X = 4) + P(X = 5) + P(X = 6) = k + 10k + k = 12k = 12 21 Substitute k = — 1 21 = 4 7
123 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 Example 7 In a game, Jemang throws 3 fair coins. He will receive RM8 if all the three coins show ‘heads’, RMx if two ‘heads’ are obtained, RM3 if one ‘head’ is obtained and RM1 if no ‘heads’ appear. State, in terms of x, the expected gain for Jemang in each game. If Jemang pays RM3.75 to participate in each game, find (a) the value of x so that each game is fair, (b) the expected gain or loss for Jemang after he plays 100 games if x = 3.5. Solution: Suppose H = The event ‘head appears’ when a coin is thrown. T = The event ‘tail appears’ when a coin is thrown. P(3 heads) = P(HHH) = 1 8 P(2 heads) = P(HHT) + P(HTH) + P(THH) = 3 8 P(1 head) = P(HTT) + P(THT) + P(TTH) = 3 8 P(0 head) = P(TTT) = 1 8 Let the random variable Y represent ‘Jemang’s gain for each game’. Therefore, the probability distribution for Y is y RM8 RMx RM3 RM1 P(Y = y) 1 8 3 8 3 8 1 8 E(Y) = ∑ y P(Y = y) = RM8 1 8 + RMx 3 8 + RM3 3 8 + RM1 1 8 = RM 18 + 3x 8 = RM 3(6 + x) 8 Expected gain for Jemang from each game = RM3(6 + x) 8 (a) If Jemang pays RM3.75 for each game, expected gain = RM 18 + 3x 8 – RM3.75. If the game is fair, expected gain = RM0. Therefore RM 18 + 3x 8 – RM3.75 = RM0 18 + 3x – 30 = 0 3x = 12 x = 4 (b) If x = 3.5 expected gain for one game = RM3 18 + 3(3.5) 8 4 – RM3.75 = – RM0.1875 expected gain for 100 games = 100(– RM0.1875) = – RM18.75 (negative sign means loss) Therefore, the expected loss after 100 games = RM18.75.
124 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 Variance For a discrete random variable X, where the mean of X = E(X) = µ, variance of X is written as Var(X) and is defined by Var(X) = E(X – µ) 2 Therefore, Var(X) = Σ(X – µ) 2 P(X = x) This formula involves the term (x – µ) 2 and because of this, the calculation of variance becomes complicated. To overcome this problem, we write the above formula in a different form to simplify the calculation of the variance as shown below. Var(X) = ∑(x – µ) 2 P(X = x) = ∑ (x2 – 2µx + µ 2) P(X = x) = ∑ x2 P(X = x) – 2µ ∑ x P(X = x) + µ2 ∑ P(X = x) ∑ P(X = x) = 1 = ∑ x2 P(X = x) – 2µ2 + μ2 = ∑ x2 P(X = x) – µ2 E(X) = µ Var (X) = E(X2 ) – [E(X)]2 Example 8 A discrete random variable X has the following probability distribution. x 0 1 2 3 4 P(X = x) 0.30 0.15 0.10 0.15 0.30 Find the (a) expected value of X, (b) variance of X. Solution: (a) E(X) = ∑ x P(X = x) = 0(0.30) + 1(0.15) + 2(0.10) + 3(0.15) + 4(0.30) = 2 Hence, expected value of X = 2. (b) E(X2 ) = ∑ x2 P(X = x) = 0(0.30) + 1(0.15) + 4(0.10) + 9(0.15) + 16(0.30) = 6.7 Var(X) = E(X2 ) – [E(X)]2 = 6.7 – 22 = 2.7 Hence, variance of X = 2.7. ∑ x P(X = x) = µ
125 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 Example 9 In a bag, there are 4 red towels and 3 yellow towels. Towels are drawn at random from the bag, one after the other without replacement, until a red towel is obtained. If X is the total number of towels drawn from the bag, obtain the probability distribution of X. Hence, find (a) E(X), (b) Var(X). Solution: The random variable X takes the values 1, 2, 3 and 4. Let R represent the event ‘a red towel is drawn’ and Y represent the event ‘a yellow towel is drawn’. P(X = 1) = P(R) = 4 7 P(X = 2) = P(YR) = 3 7 × 4 6 = 2 7 P(X = 3) = P(YYR) = 3 7 × 2 6 × 4 5 = 4 35 P(X = 4) = P(YYYR) = 3 7 × 2 6 × 1 5 × 4 4 = 1 35 Hence, the probability distribution of X is x 1 2 3 4 P(X = x) 4 7 2 7 4 35 1 35 (a) E(X) = ∑ x P(X = x) = 1 4 7 + 2 2 7 + 3 4 35 + 4 1 35 = 4 7 + 4 7 + 12 35 + 4 35 = 8 5 (b) E(X2 ) = ∑ x2 P(X = x) = 12 4 7 + 22 2 7 + 32 4 35 + 42 1 35 = 4 7 + 8 7 + 36 35 + 16 35 = 16 5 Var (X) = E(X2 ) – [E(X)]2 = 16 5 – 8 5 2 = 16 25
126 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 Exercise 3.3 1. The probability distribution for a random variable X is shown in the following table. x 0 1 2 3 4 P(X = x) 1 12 1 6 1 3 1 6 1 4 Find E(X). 2. The probability distribution of a discrete random variable Y is given in the table below. y 14 15 16 17 18 19 P(Y = y) 0.1 0.2 p 0.3 0.1 0.1 Find (a) the value of p, (b) E(Y). 3. In a game, a fair dice is tossed. If the number 6 appears, a person receives RM10. If an odd number appears, that person receives RM5 and if other numbers appear, he will be fined RM6. Determine the expected gain if the person pays RM3 to play once. 4. A bag contains 4 red coloured beads and 6 green coloured beads. Two beads are drawn, one after the other, without replacement. Suppose X is a random variable which represents the number of red beads drawn, find E(X). 5. 2 5 4 1 3 5 1 2 3 4 3 4 1 2 5 1 3 2 5 4 In a game, a player is blindfolded and he is required to place two RM1 coins at random on the squares labelled, as shown in the above diagram. Both coins can be placed on the same square and assume that each square has an equal chance of being filled by the coins. The player’s score is the sum of the numbers on the squares filled by the coins. Construct a probability distribution table for all the possible scores and hence find the player’s expected score after playing 10 times. 6. Copy and complete the following table. x P(X = x) x – 3 (x – 3)2 (x – 3)2 P(X = x) 1 0.1 2 0.2 3 0.3 4 0.4 Hence, find Var(X).
127 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 7. A discrete random variable X has the following probability distribution. 0.1, x = 1, 5, 6 0.25, x = 2, 4, P(X = x) = 0.2, x = 3, 0, otherwise. Find (a) E(X), (b) E(X2 ), (c) Var(X). 8. A discrete random variable V has the following probability distribution. 1 v , v = 2, 4, 8, 16, P(V = v) = 1 16 , v = 20, 0, otherwise. Find (a) the expectation of V, (b) the variance of V. 9. A discrete random variable X takes the values 0, 1, 2 only. The probability distribution of Y is shown in the following table, where p is a constant (0 p 1 3 ). x 0 1 2 P(X = x) 1 – 3p 2p p If Var(X) = 1 2 , find two possible values for E(X). 10. A tetrahedral dice has faces marked with the numbers 1, 2, 3 and 4 respectively. In a game, this dice is thrown on a table. If the bottom face of the dice is the number 1, 10 marks are subtracted from a player. If the bottom face of the dice shows the number 2 or 4, the player is given 5 marks and if the bottom face is the number 3, then 3 marks are given. Find the expected total marks when a player throws the dice once. 11. A student receives a set of three keys from his form teacher to open a drawer. If the student tries to open the drawer by choosing a key at random one by one without repetition, draw a tree diagram to show the different selections until he gets the correct key. Use your tree diagram to calculate the expected number of keys the student has to try before he opens the drawer. 12. A bag contains 3 red balls and 2 black balls. Balls are drawn at random from the bag, one by one without replacement, until a black ball is obtained. The number of draws required to obtain a black ball is represented by a random variable X. Copy and complete the table below to show the probability distribution of X. x 1 2 3 4 P(X = x) 3 10 Show that E(X) = 2 and find Var(X).
128 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 3.2 Continuous Random Variables In the previous section, we know that discrete random variables can only take exact values. Continuous random variables do not take exact values but are defined on an interval instead. Probability density function A probability density function, when plotted, is called a probability density curve which may be viewed as the limiting form of a relative frequency density histogram. The probability density function, f of a discrete random variable X has the properties that f(x) 0 for all values of x in an interval [a, b] and ∫ b a f(x) dx = 1, that is the area under the curve y = f(x) between x = a and x = b is 1. If a x1 x2 b, then P(x1 X x2 ) = ∫x1 x2 f(x) dx. P(x1 X x2 ) is the area under the curve y = f(x) between x = x1 and x = x2 . Note: If X is a continuous random variable, then (a) P(X = x1 ) = ∫x1 x1 f(x) dx = 0 (b) P(x1 X x2 ) = P(x1 X x2 ) = P(x1 X x2 ) = P(x1 X x2 ) If X is a continuous random variable, its probability density function, f satisfies the conditions below. (a) f(x) 0 for all values of x (b) ∫–∞ ∞ f(x) dx = 1 Example 10 A continuous random variable Y has probability density function hy2 , –2 y 2, g(y) = 0, otherwise. (a) Find the value of h. (b) If P(|Y| m) = 0.5, find m. Solution: (a) g(y) is the probability density function of Y, thus ∫ 2 –2 g(y) dy = 1 ∫ 2 –2 hy2 dy = 1 3 hy3 3 4 –2 2 = 1 16h 3 = 1 h = 3 16 Figure 3.2 f(x) x y = f(x) a b Figure 3.3 f(x) x y = f(x) a x1 x2 b
129 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 (b) Because P(| Y | m) = P(–m Y m) = 0.5, thus ∫ m –m g(y) dy = 0.5 ∫ m –m hy2 dy = 0.5 3 hy3 3 4 –m m = 0.5 h = — 3 16 3 y3 16 4 –m m = 0.5 m3 8 = 0.5 m3 = 4 m = 3 4 = 1.587 Example 11 The random variable X has probability density function x 2 , 0 x 1, f(x) = 1 – x 6 , 3 x 6, 0, otherwise. (a) Sketch the graph of y = f(x), (b) Find P(0 x 4). Solution: (a) f(x) x 0 1 2 3 4 5 6 1 – 2 (b) P(0 X 4) = ∫ 4 0 f(x) dx = ∫ 1 0 x 2 dx + ∫ 3 1 0 dx + ∫ 4 3 1 – x 6 2 dx = 3 x2 4 4 0 1 + 0 + 3x – x2 12 4 3 4 = 1 4 – 02 + 4 – 4 3 – 3 + 3 4 2 = 1 4 + 5 12 = 2 3
130 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 Exercise 3.4 1. A probability density function f is defined by f(x) = 4 15 (x – 1)3 , 2 x 3, 0, otherwise. Verify that f(x) > 0 for 2 < x < 3 and that ∫ 3 2 f(x) dx = 1. 2. A function g is defined as follows: g(y) = 3 44 (5 – y2 ), –2 y 2, 0, otherwise. (a) Sketch the graph g. (b) Show that g is the probability density function of a continuous random variable. (c) Find (i) P(Y 1), (ii) P(| Y | 1). 3. Y is a continuous random variable with probability density function g(y) = 3y2 16 , –2 , y , 2, 0, otherwise. Calculate (a) P(Y 1), (b) P(| Y | 3 2 ). 4. The continuous random variable X has probability density function 1 3 x, 0 x 2, f(x) = 2 3 (3 – x), 2 x 3, 0, otherwise. Sketch the graph of y = f(x) and verify that f(x) satisfies the conditions for a probability density function. Calculate (a) P(X 1), (b) P(X 0.5), (c) P(| X – 2 | 0.5). 5. The continuous random variable X has probability density function m(5 – x), 1 x 4, f(x) = 0, otherwise. Find the value of m. Sketch the graph y = f(x). Find also (a) P(X 3), (b) P(3 X 5).
131 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 6. The continuous random variable X has probability density function k(x + 2)2 , –2 x 0, f(x) = 4k, 0 x 1 1 3 , 0, otherwise. Determine the value of the constant k. Find (a) P(–1 X 1), (b) P(X 1). 7. The continuous random variable X has probability density function kx3 , 0 x c, f(x) = 0, otherwise. If P(X 1 2 ) = 1 16 , find the value of the constants c and k. 8. The probability density function g of a continuous random variable W is defined as follows: aw + b , 0 w 4, g(w) = bw + a , 4 w 5. If P(2 W 4) = 23 51 , find the value of the constants a and b. Hence, find (a) P(W 3), (b) the value v if P(W v) = 39 68 . 9. The probability density function f of the continuous random variable X is given as 10 – x 42 , 0 x 6, f(x) = 0, otherwise. (a) If P(X u) = 9 84 , find the value u. (b) Find the value q such that P(X q) = 1 4 . 10. The continuous random variable X has probability density function f defined as follows. 1 8 , 0 x 3, f(x) = 1 16 , 5 x 15, 0, otherwise. (a) Find the value m if P(X m) = 1 2 . Hence, state the median of X. (b) Find the first quartile and third quartile of X. Cumulative distribution function Just as the cumulative frequency for data in Chapter 1 (Data Description), we can also accumulate the probabilities for random variables. The function produced is known as the cumulative distribution function. The method of accumulating the probabilities is different for discrete random variables and continuous random variables. The probability for continuous random variables can be calculated by integration. This can also be done for cumulative distribution functions.
132 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 The cumulative distribution function, F of a continuous random variable X with its probability density function f defined on an interval [a, b], is defined by F(x) = P(X x) = ∫ x a f(t) dt The cumulative distribution function of a continuous random variable is the area between the curve which represents the probability density function and the x-axis from a to x. The cumulative distribution function is also known as the distribution function. Note: If f(t) is defined for –∞ , t , ∞, then F(x) = ∫ x –∞ f(t) dt. We can use the cumulative distribution function to find P(x1 X x2 ). Refer to the figures below. t a x2 b F(x2) f(t) t a x1 b F(x1) f(t) t a x2 x1 b F(x2) – F(x1) f(t) P(a X x2 ) = F(x2 ) P(a X x1 ) = F(x1 ) P(x1 X x2 ) = F(x2 ) – F(x1 ) Figure 3.6(a) Figure 3.6(b) Figure 3.6(c) P(x1 X x2 ) = P(a X x2 ) – P(a X x1 ) P(x1 X x2 ) = F(x2 ) – F(x1 ) The median, first quartile (lower quartile) and third quartile (upper quartile) can also be determined by using the cumulative distribution function. Figure 3.4 a 0 x b f(t) t F(x) Figure 3.5 F(x) 0 x t f(t)
133 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 The median, m, divides the area under the curve y = f(x) into two equal sections. Hence, ∫ m a f(x) dx = 1 2 = ∫ b m f(x) dx F(m) = 1 2 The first quartile, q1 , satisfies the equation F(q1 ) = 1 4 and the third quartile, q3 , satisfies the equation F(q3 ) = 3 4 . Example 12 The continuous random variable X has probability density function f(x) = 3 – x 2 , 1 x 3, 0, otherwise. (a) Find the cumulative distribution function F of X and P(X 2). (b) Sketch the graphs f and F. Solution: (a) When x 1, ∫ x –∞ f(t) dt = ∫ x –∞ 0 dt = 0 When 1 x 3, ∫ x –∞ f(t) dt = ∫ 1 –∞ 0 dt + ∫ x 1 (3 – t) 2 dt = 0 + 3 3t 2 – t 2 4 4 1 x = 6x – x2 4 – 6 – 1 4 = 6x – x2 – 5 4 When x 3, ∫ x –∞ f(t) dt = ∫ 1 –∞ 0 dt + ∫ 3 1 (3 – t) 2 dt + ∫ x 3 0 dt = 0 + 3 3t 2 – t 2 4 4 1 3 + 0 = 9 4 – 5 4 = 1 Hence, F(x) is defined by 0, x 1, F(x) = 6x – x2 – 5 4 , 1 x 3, 1, x 3. Figure 3.7 x a m b Area = —1 2 f(x)
134 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 P(x 2) = F(2) = 6(2) – 22 – 5 4 = 3 4 (b) Graphs of f and F are as follows. f(x) x 0 1 1 2 3 4 F(x) x 0 1 1 2 3 4 Example 13 The continuous random variable X has probability density function 1 12 , 0 x 4, f(x) = 1 6 , 10 x 14, 0, otherwise. Find the cumulative distribution function of X. Hence, find the (a) median, m, (b) first quartile, q1 , (c) third quartile, q3 . Solution: When x 0, ∫ x –∞ f(t) dt = ∫ x –∞ 0 dt = 0 When 0 x 4, ∫ x –∞ f(t) dt = ∫ 0 –∞ 0 dt + ∫ x 0 1 12 dt = 3 t 124 0 x = x 12 When 4 x 10, ∫ x –∞ f(t) dt = ∫ 0 –∞ 0 dt + ∫ 4 0 1 12 dt + ∫ x 4 0 dt = 3 t 124 0 4 = 1 3
135 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 When 10 x 14, ∫ x –∞ f(t) dt = ∫ 0 –∞ 0 dt + ∫ 4 0 1 12 dt + ∫ 10 4 0 dt + ∫ x 10 1 6 dt = 1 3 + 3 t 6 4 10 x = x – 8 6 When x 14, ∫ x –∞ f(t) dt = ∫ 0 –∞ 0 dt + ∫ 4 0 1 12 dt + ∫ 10 4 0 dt + ∫ 14 10 1 6 dt + ∫ x 14 0 dt = 1 Hence, x 12 , 0 x 4, 1 3 , 4 x 10, F(x) = x – 8 6 , 10 x 14, 1, x 14. (a) P(X m) = 1 2 or F(m) = 1 2 Since F(10) = 1 3 and F(14) = 1, hence 10 m 14. Therefore, m – 8 6 = 1 2 m = 11 (b) P(X q1 ) = 1 4 or F(q1 ) = 1 4 Since F(4) = 1 3 , hence q1 4. Therefore, q1 12 = 1 4 q1 = 3 (c) P(X q3 ) = 3 4 or F(q3 ) = 3 4 Since F(10) = 1 3 and F(14) = 1, hence 10 q3 14. Therefore, q3 – 8 6 = 3 4 q3 – 8 = 9 2 q3 = 12.5
136 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 Exercise 3.5 1. The probability density function of a continuous random variable X is given by 1 10, 1 x 11, f(x) = 0, otherwise. (a) Find the cumulative distribution function of X and sketch its graph. (b) Find P(3 X 9). 2. The probability density function of a continuous random variable Y is given by y 10 , 0 y 4, g(y) = 1 15 , 4 y 7, 0, otherwise. (a) Find the cumulative distribution function of Y and sketch its graph. (b) Find P(2 Y 5). 3. The probability density function of a continuous random variable X is given by 2x, 0 x a, f(x) = 6 – 6x, a x 1, 0, otherwise. (a) Show that a = 1 2 . (b) Find the cumulative distribution function of X. (c) Find P(X 2 3 ). 4. X is a continuous random variable with probability density function x2 , 0 x 1, f(x) = 5 – x 12 , 1 x 5. Find the cumulative distribution function of X. Calculate (a) P(X 1 2 ), (b) P( 1 2 X 2). 5. A continuous random variable X takes values in the interval 0 x 3 and has probability density function ax , 0 x 1, f(x) = a 2 (3 – x), 1 x 3, 0, otherwise. (a) Determine the value of the constant a. (b) Find the cumulative distribution function of X. (c) Find P(| X – 1 | 1 2 ).
137 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 6. The continuous random variable Y has probability density function 1 10 , 3 y 7, g(y) = 1 5 , 10 y 13, 0, otherwise. Find the cumulative distribution function of Y. Hence, find (P| Y – 3 | 1). 7. The probability density function of a continuous random variable W is given by w 6 , 0 w 2, h(w) = w 15 , 4 w 6, 0, otherwise. Find the cumulative distribution function of W. Find P(| W – 4 | 1). 8. X is a continuous random variable with probability density function 3x2 , 0 x 1, f(x) = 0, otherwise. Find (a) the cumulative distribution function of X, (b) the median for X. 9. The continuous random variable X has probability density function e–x , x 0, f(x) = 0, otherwise. Find the cumulative distribution function. Hence, find the median of X. 10. X is a continuous random variable with probability density function x 12 , 1 x 5, f(x) = 0, otherwise. (a) Find the distribution function of X. (b) Determine the lower quartile, upper quartile and the interquartile range. 11. The continuous random variable X has the probability density function kx, 0 x 2, f(x) = k, 2 x 3, 0, otherwise. Determine the constant value k. Hence, find (a) the cumulative distribution function of X, (b) the median of X.
138 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 We know that a cumulative distribution function, F, is given by F(x) = ∫ x a f(t) dt for a t b. Thus, f(x) = dF(x) dx Differentiation is the inverse of integration. that is, f(x) = F(x) . Example 14 X is a continuous random variable with cumulative distribution function 0 , x – 1, 1 5 x + a, –1 x 0, F(x) = 1 5 + 1 5 x + bx2 , 0 x 1, 1, x 1. (a) Find the values a and b. (b) Find the probability density function, f, of X. (c) Sketch the graph of f. Solution: (a) F is a continuous function. To determine the values a and b, we use the boundary values for each interval. When x –1, F(–1) = 0. When –1 x 0, F(–1) = – 1 5 + a Therefore, – 1 5 + a = 0 a = 1 5 When 0 x 1, F(1) = 1 5 + 1 5 + b = 2 5 + b When x 1, F(1) = 1 Therefore, 2 5 + b = 1 b = 3 5 Hence, 0, x –1, 1 5 x + 1 5 , –1 x 0, F(x) = 1 5 + 1 5 x + 3 5 x2 , 0 x 1, 1, x 1.
139 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 (b) f(x) = d dx [F(x)] When x –1, d dx [F(x)] = 0 When –1 x 0, d dx [F(x)] = d dx 1 5 x + 1 5 2 = 1 5 When 0 x 1, d dx [F(x)] = d dx 1 5 + 1 5 x + 3 5 x2 2 = 1 5 + 6 5 x When x 1, d dx [F(x)] = 0 Hence, the probability density function of X is 1 5 , –1 x 0, f(x) = 1 5 + 6 5 x , 0 x 1, 0 , otherwise. (c) f(x) x 1 1 –1 0 7 – 5 3 – 5 1 – 5 Exercise 3.6 1. The cumulative distribution function of a continuous random variable X is given as follows. 0, x 0, x 8 , 0 x 2, F(x) = x – 1 4 , 2 x 5, 1, x 5. Find the probability density function of X. Calculate P(X 4) and P(1 X 3).
140 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 2. The continuous random variable X has cumulative distribution function 0, x –2, F(x) = (x + 2)2 9 , –2 x 1, 1, x 1. Find (a) the probability density function of X, (b) P(X 0), (c) the value k if P(X k) = 8 9 . 3. X is the continuous random variable with cumulative distribution function 0, x 0, kx , 0 x 4, F(x) = h(x – 4)2 + 2 5 , 4 x 7, 1, x 7 with h and k as constants. Determine the values h and k, and find the probability density function of X. 4. The continuous random variable X takes values in the interval [0, 3]. If P(X x) = a + bx3 for 0 x 3, find (a) the value of the constants a and b, (b) the cumulative distribution function of X, (c) the probability density function of X. 5. The continuous random variable X has cumulative distribution function as follows. 0, x m, F(x) = x2 + 5mx – 6m2 , m x 2m, 1, x 2m. (a) Verify that the value m is 1 22 . (b) Find the probability density function of X. (c) Calculate P(X 1 2 ). 6. The cumulative distribution function of a continuous random variable X is as follows. 0, x 1, F(x) = (x – 1)2 , 1 x 2, 1, x 2. Find the probability density function of X. Calculate (a) P(2X 3), (b) P(4 4X – 1 5), (c) P(X 2 3 2 ).
141 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 7. The continuous random variable X has cumulative distribution function 0, x 0, 1 8 x , 0 x 4, F(x) = x – 3 4 , 4 x 7, 1, x 7. Find the probability density function of X. 8. The continuous random variable X has cumulative distribution function as follows. 0, x 1, F(x) = (x – 1)2 12 , 1 x 3, 14x – x2 – 25 24 , 3 x 7, 1, x 7. (a) Find the probability density function of X, and sketch its graph. (b) Find P(2.8 X 5.2). Mean and variance of a continuous random variable The mean of a continuous random variable X is the expected value of X, defined by E(X) = ∫ ∞ –∞ xf(x) dx The variance X is defined as the expected value of (X – µ) 2 , that is, Var(X) = E[(X – µ) 2 ] E[(X – µ) 2 ] = ∫ ∞ –∞ (x – µ) 2 f(x) dx = ∫ ∞ –∞ (x2 – 2µx + µ2 )f(x) dx = ∫ ∞ –∞ x2 f(x) dx – ∫ ∞ –∞ 2µxf(x) dx + ∫ ∞ –∞ µ2 f(x) dx = ∫ ∞ –∞ x2 f(x) dx – 2µ∫ ∞ –∞ xf(x) dx + µ2 ∫ ∞ –∞ f(x) dx = E(X 2 ) – 2µ 2 + µ 2 = E(X 2 ) – µ 2 Hence, Var(X) = E(X 2 ) – µ 2 or Var(X) = E(X 2 ) – [E(X)]2 µ = E(X) If X is a continuous random variable with probability density function f(x), then Var(X) = ∫ ∞ –∞ x2 f(x) dx – 3∫ ∞ –∞ xf(x) dx4 2
142 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 Example 15 In the meteorological station at a town, observations were taken on the rate, X, for the sky to be covered by clouds every half day. Past analysis of the records shows that X has probability density function given below. 4 3 (1 – x), 0 x 1 2 , f(x) = 4 3 x, 1 2 x 1. Find the expectation and variance of X. Solution: E(X) = ∫ ∞ –∞ xf(x) dx = ∫ —1 2 0 x 4 3 (1 – x) dx + ∫ 1 —1 2 4 3 x2 dx = 3 4 3 1 x2 2 – x3 3 24 —1 2 0 + 3 4 3 1 x3 3 24 1 —1 2 = 4 3 1 1 8 – 1 24 2 + 4 3 1 1 3 – 1 24 2 = 1 2 E(X2 ) = ∫ ∞ –∞ x2 f(x) dx = ∫ —1 2 0 x2 4 3 (1 – x) dx + ∫ 1 —1 2 4 3 x3 dx = 3 4 3 1 x3 3 – x4 4 24 —1 2 0 + 3 x4 3 4 1 —1 2 = 4 3 1 1 24 – 1 64 2 + 1 1 3 – 1 48 2 = 25 72 Var(X) = E(X2 ) – [E(X)]2 = 25 72 – 1 1 2 2 2 = 7 72 Example 16 The continuous random variable X has a uniform distribution with p.d.f. as follows. f(x) = 1 b – a , a x b, X is uniformly distributed. 0, otherwise. (a) Sketch the graph of f. (b) Determine the mean and variance of X.
143 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 Solution: (a) Since a and b are constants, 1 b – a is also a constant. The graph of f is a straight line parallel to the x-axis as shown in the figure on the right. (b) Graph f(x) = 1 b – a is symmetrical about the line x = 1 2 (a + b). Hence, mean of X = 1 2 (a + b) E(X 2 ) = ∫ b a x2 f(x) dx = ∫ b a x2 1 1 b – a 2 dx = 3 x3 3(b – a) 4 b a = b3 – a3 3(b – a) = 1 3 (b2 + ab + a2 ) Var(X) = E(X2 ) – [E(X)]2 = 1 3 (b2 + ab + a2 ) – [ 1 2 (a + b)]2 = 1 3 (b2 + ab + a2 ) – 1 4 (a2 + 2ab + b2 ) = 1 12 (4b2 + 4ab + 4a2 – 3a2 – 6ab – 3b2 ) = 1 12 (b2 – 2ab + a2 ) Variance of X = 1 12 (b – a) 2 Example 17 The lifespan of a light bulb has a negative exponential distribution with a mean lifespan of 500 hours. (a) Find the probability that the bulb functions after 600 hours. (b) Given that the bulb still functions after 600 hours, find the probability that the bulb will function after 700 hours. Solution: Assume that X is a random variable which represents the lifespan of a light bulb in hours. λe–λx , x 0, Hence, f(x) = 0, otherwise. f(x) x 0 a b – 1 b – a
144 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 E(X) = 1 λ 500 = 1 λ λ = 0.002 Therefore, f(x) = 0.002e–0.002x (a) P(X 600) = ∫ ∞ 600 0.002e–0.002x dx = 3–e–0.002x 4 ∞ 600 = 0 + e–0.002(600) = e–1.2 = 0.3012 (b) P(X 700 | X 600) = P(X 700) P(X 600) P(X 600) = P(X 700) P(X 600) = e–0.002(700) e–0.002(600) = e–1.4 e–1.2 = e–0.2 = 0.8187 Exercise 3.7 1. The continuous random variable X has probability density function 1 8 , 2 x 10, f(x) = 0, otherwise. Find (a) E(X), (b) Var(X), (c) Var(5X), (d) Var(5X + 2). 2. X is a continuous random variable with probability density function 1 4 (x – 1), 1 x 3, g(x) = 1 4 (7 – x), 5 x 7, 0, otherwise. Find (a) E(X), (b) E(X 2 ), (c) Var(X), (d) Var ( 1 4 X), (e) Var(3 – X).
145 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 3. The probability density function of a continuous random variable Y is defined by 3 16 (y + 2), –2 y 0, f(y) = 3 16 (2 – y), 0 y 2, 0, otherwise. Determine the expectation and variance of Y. 4. The continuous random variable X has probability density function ax, 0 x 1, f(x) = b(3 – x), 1 x 3, 0, otherwise where a and b are positive constants. If E(X) = 4 3 , find the values of a and b. Find also the variance of X. 5. The probability density function of a continuous random variable X is given by 1 a , a x 2a, f(x) = 0, otherwise. If Var(X) = 3, find the value of a. Sketch the graph of f and deduce the value of E(X). 6. The output from a machine in a factory is a continuous random variable with probability density function given by ax, 0 x 10, f(x) = a(20 – x), 10 x 20, 0, otherwise. Determine the value of the constant a. Find the expectation and variance of the output of the machine. 7. The continuous random variable X is defined by its cumulative distribution function 0, x 0, F(x) = x2 , 0 x 1, 1, x 1. Find (a) the probability density function (b) E(X), (c) Var(X), (d) Var(6X + 1). 8. A continuous random variable X has cumulative distribution function 0, x –1, F(x) = αx + α, –1 x 0, 2αx + α, 0 x 1, 3α, x 1.