Vedanta Excel in Mathematcs Book -10 Final (2078)

Trigonometry

Solution:

Here, area of 'BCD = 1 BC u BD sin30q
2
1 1
= 2 u 10 cm u 6 cm u 2 = 15 cm2

Median BD bisects 'ABC.

? Area of 'ABC = 2 u Area of 'BCD

= 2 u 15 cm2 = 30 cm2

Example 3: In the given figure, AB = 6 cm, BC = 8 cm, A D
‘ABC = 60q, find the area of ABCD. C
6 cm
Solution: 60°

Diagonal AC is drawn in the ABCD. B 8 cm

Here, area of 'ABC = 1 AB u BC sin60q
2
1 3
= 2 u 6 cm u 8 cm u 2 = 12 3 cm2

Diagonal AC bisects the parallelogram ABCD.

?Area of ABCD = 2 u Area of 'ABC

= 2 u 12 3 cm2 = 24 3 cm2

Example 4: Calculate the area of the given parallelogram ABCD.

Solution: A 26 cm D
Here, ‘A = 180q – (80q + 40q) = 180q – 120q = 60q 80°

Also, AB = DC = 43 cm. 43 cm

Now, area of 'ABD = 1 AD u AB sin60q 40°
2 B
1 3
= 2 u 26 cm u 43 cm u 2 = 484.094 cm2 C

Diagonal BD bisects the parallelogram ABCD.

?Area of ABCD = 2 u Area of 'ABD = 2 u 484.094 cm2 = 968.19 cm2

Example 5: The area of the given parallelogram PQRS is P S
14 sq. cm. If SR = 4 cm, QR = 7 cm, find the Q R
value of the ‘QRS.

Solution:

Diagonal QS is drawn.

Here, area of 'QRS = 1 u Area of PQRS
2
1
= 2 u 14 sq. cm. = 7 sq. cm.

Also, area of 'QRS = 1 u QR u SR sinR
2

? 1 u QR u SR sinR = 7
2

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Trigonometry

or, 7 u 4 sinR = 14
14 1
or, sinR = 28 = 2

or, sinR = sin30q

or, R = 30q.

So, the value of the required angle is 30°.

A D
C
Example 6: In the given figure, ABCD is a rhombus and
'DPC is an equilateral triangle. Find the
area of the trapezium ABPD.

Solution: 60° P
In the figure, ABCD is a rhombus and 'DPC is an equilateral B 6 cm

triangle.

So, AB = BC = CD = DP = PC = 6 cm.
Also, ‘ABC = ‘DCP = 60q

Now, area of the rhombus ABCD = AB × BC × sin 60q

= 6 cm × 6 cm × 3 = 18 3 cm2.
2
1
Similarly, area of equilateral triangle DPC = 2 CD × CP × sin 60q

= 1 × 6 cm × 6 cm × 3 =9 3 cm2
2 2

? Area of the trapezium ABPD = 18 3 cm2 + 9 3 cm2 = 27 3 cm2.

Example 7: In the adjoining figure, PQRS is a trapezium. P S
T
If ‘PQR = 60q, ST = SR, PQTS is a rhombus
and the area of the trapezium is 108 3 sq. cm,

find QR.

Solution: 60° R

In the figure, PQTS is a rhombus, ‘PQT = 60q and ST = SR Q

So, ‘PQT = ‘STR = ‘SRT = 60q

Also, ‘TSR = 180q - (60q + 60q) = 60q

Thus, 'STR is an equilateral triangle. = 3 × QT2
Now, area of the rhombus PQTS = PQ × QT × sin 60q 2

And, the area of equilateral 'STR = 1 ST × TR × sin 60q = 3 QT2
2 4

Again, the area of trapezium PQRS = 3 × QT2 + 3 × QT2
2 4

or, 108 3 = 2 3 QT2 + 3 QT2
4
or, QT2 = 144

or, QT = 144 = 12 cm.

? QR = 2 × QT = 2 × 12 cm = 24 cm.

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Trigonometry

Example 8: Find the area of the given figure ABCD. A D10 cm
55° 51° 99°
Solution:
In the figure, ABCD is a quadrilateral in which, 14 cm C
‘ABC = 65q, ‘BAC = 55q, ‘CAD = 51q, 65°

‘ADC = 99q, BC = 10 3 cm, AC = 14 cm and CD = 10 cm B 10 3 cm
Now, in 'ABC, ‘ACB = 180q - (65q + 55q) = 60q

? Area of 'ABC = 1 AC × BC × sin 60q = 1 × 14 × 10 3 × 3 = 105 cm2
2 2 2

Also, in 'ACD, ‘ACD = 180q - (99q + 51q) = 30q

? Area of 'ACD = 1 AC × CD × sin 30q = 1 × 14 × 10 × 1 = 35 cm2
2 2 2

Now, the area of the quadrilateral ABCD = 105 cm2 + 35 cm2 = 140 cm2

Example 9: In the given figure, AB = 12 cm, BC = 16 3 cm, A

AD = 18 cm, ‘ABC = 60q and ‘ADC = 30q. If 18 cm

'ACD = 1 'ABC, find the length of CD. 12 cm 30° D
4 C
60°
Solution:
B 16 3 cm
In the figure, AB = 12 cm, BC = 16 3 cm, AD = 18 cm,
‘ABC = 60q and ‘ADC = 30q

Now, area of 'ABC = 1 AB × BC × sin 60q = 1 × 12 × 16 3 × 3 = 144 cm2
2 2 2

Also, area of 'ACD = 1 AD × CD × sin 30q = 1 × 18 × CD × 1 = 9 CD
2 2 2 2

According to the question,

'ACD = 1 'ABC
4

or, 9 CD = 1 × 144
2 4

or, CD = 8 cm.

EXERCISE 16.2

General section

1. Find the area of the following triangles:

a) b) C 12 cm c) A d) P

P RA 45° BB 10 cm 60° 110° 4 cm
8 cm 15 2 cm
30° CQ 40° R
Q 12 cm
7 cm

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e) f) A g) A h) P

20 cm L 60° D 30° 8 cm
K 35° 100° 8 cm 6 cm 6 cm
12 cm 12 cm 30° QR

MB 48 cm C B 10 cm C d) D S
C
2. Find the area of the following parallelograms: 30°

a) A D b) A D c) S 10 cm R

6 cm 120° 10 cm
4 cm 8 cm

60° C 30° C P Q A 8 cm B
B 8 cm G B 6 cm
g) D
e) H 25 cm f) P Q 25° C h) 35° N
20 cm
60° K

42° 10 cm 8 cm

35° 115°
A 16 cm
78° F S 8 cm R B LM
E

3. Find the area of the following kites:

a) P b) A c) A 3 cm d) P 10 cm
30°
Q 30 cm S B 45° DB 5 cm D 6 cm S
30° 16 cm 30° Q 60°
27 cm
CR
8cm C
C
R

4. a) In the given figure, AD is a median of ' ABC. If ‘BAD = 60°, D
AD = 12 cm and AB = 10 m, find the area of ' ABC.
12 cm

60° B
A 10 cm

A

b) In the given figure, AB = 10.2 cm, AC = 14 cm, AD A BC 10.2 cm 14 cm
and AD = DC. Calculate the area of ' ABC.
C
75° C
BD

A

5. a) In the figure given alongside, if AB = 6 cm, BC = 8 cm 6 cm

and the area of ' ABC is 12 3 sq. cm, find the value

of ‘ ABC. B 8 cm

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Trigonometry

D

b) In the given triangle DEF, DF = 7 cm, EF = 10 cm, area of 7 cm

' DEF = 35 3 cm2 and ‘ DEF = 50°, find the value of 50°
2 10 cm
‘ EDF.. E F
P

c) In the given triangle, PQ = 10 cm, PR = 18 cm and area of 10 cm 18 cm
' PQR = 45 3 cm2. Find the value of ‘ QPR. B
R
Q
30° 12 cm
6. a) In the given ' ABC, BC = 12 cm, ‘ ABC = 30°, and the area of
' ABC is 27 cm2. Find the length of AB.

b) In the given figure, AB = 6 2 cm, ‘ABC = 45° and the area A C
of ' ABC = 24 cm2. Find the measurement of side BC. A C

B 6 2 cm
45°

A

c) In the adjoining figure, area of ' ABC is 10 sq.cm, 5 cm 80°
70°
AB = 5 cm, ‘ A = 80°, and ‘C = 70°. Find the C

measurement of BC. B

A 12 cm B

7. a) The adjoining figure is a rhombus ABCD. If AB = 12 cm and
the area of the rhombus is 72 cm2, find the measurement of
‘ BAD.

DC

A D
B
b) Given figure is a parallelogram ABCD of area 96 2 sq. cm.
If AD = 12 cm and DC = 16 cm, find the measurement of

‘ ABC.

C

c) In the given figure, PQRS is a parallelogram in which QR S P
Q
is produced to the point T. If the area of the parallelogram

is 27 3 sq.cm, the sides QP = 6 cm and PS = 9 cm, find

the measurement of ‘ SRT. TR

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Trigonometry

P 5 3cm MQ
S 8 cm R
d) In the given figure, parallelogram PQRS of area 60 sq.cm
and the triangle MRS are standing on the same base SR. If
SM = 5 3 cm and SR = 8 cm, evaluate ‘ MSR.

8. a) In the adjoining figure, quadrilateral ABCD is a rhombus. D

If the area of ' ABD = 9 cm2 and ‘ BAD = 30°, find the C

measurement of AB. 30°
A

B

S 12 cm P
Q
b) Adjoining figure is a parallelogram PQRS of area 60 3 cm2. 60°
If ‘SRQ = 60° and PS = 12 cm, calculate the length of PQ. R

PS

c) The area of the given parallelogram is 48 cm2. If 8 cm
‘ PQR = 30° and PQ = 8 cm, find the length of QR.
30°
Q R

P O
N
d) In the given figure, MNOP is a rhombus. If the area of the

rhombus is 25 2 cm2 and ‘ PMN = 45°, find the measurement

of ON. 45°
M

E DC

9. a) In the given figure, ABCD is a rhombus and AED 120°
is an equilateral triangle. Find the area of the
trapezium ABCE.

A 8 cm B

PQ

b) In the given figure, PQRS is parallelogram and 4 3cm
QRT is an equilateral triangle. If SP = 4 3 cm,
SR = 8 cm, and ‘ PSR = 60°, calculate the area of S 8 cm R T
the trapezium PSTQ.

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Trigonometry

c) The adjoining figure is a quadrilateral ABCD. Find 12 2 cm D
65° 35°
its area. A 115° C
70° 8 cm
B10 cm
D

d) In the given figure, AB // DC, DC = 12 cm, ‘BDC = 45°, 45°

BD = 8 2 cm and 3AB = DC. Find the area of the A 8 2 cm 12 cm

trapezium.

BC

e) Find the area of the given quadrilateral ABCD in which A 20 cm D
AB = 30 cm, BC = 40 cm, CD = 24 cm, DA = 20 cm, 120°

‘ BAD = 120°, and ‘ BCD = 60°. (sin 120° = 3 ) 30 cm 24 cm
2

60° C

B 40 cm

A D
55°
f) From the in formation given in the figure, find 8 cm
the area of the trapezium ABCD. 65°
E 6 3 cm C
B 6 3 cm

B 4 2 cm A

10. a) In the adjoining figure, area of the quadrilateral ABCD is 7 cm 30° 8 cm

28 cm2. Find the measurement of ‘ BAC.

C D
P
DC

b) In the given figure, BPC is an equilateral triangle and B

ABCD is a rhombus. If the area of APCD is 27 3 cm2,
find the length of AB.

A

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Trigonometry

A D
120° C 6 3 cm E

c) In the adjoining figure, the area of trapezium ABED
is 132 cm2. Find the measurement of AB.

B 8 3 cm

S P
QR
d) In the figure alongside, PQTS is a parallelogram and

PQR is an equilateral triangle. If TQ = 3 ST and the

area of the trapezium PSTR is 175 3 sq.cm, find the

length of ST. T

11. Project work

a) Prepare different models of triangles and parallelograms by cutting chart papers.
Find the lengths of adjacent sides and the angles between them. Then, calculate the
area of each model.

b) Make a model of each of quadrilateral and triangle by using wires and juice pipe.
Trace the diagrams of these models on chart paper and find their area.

Objective Questions
Tick the correct alternatives.

1. In a triangle, if T be the angle contained by the sides with lengths x and , what is the

area of the triangle?

(A) ∆ = 1 xy sinT (B) ∆= 1 xy cosT (C) ∆= 1 xy tanT (D) ∆ = xy sinT
2 2 2
A

2. In ∆ABC, ‘ABC = 300, AB = 9 cm, and BC = 12 cm. The area of ∆ABC is 6 cm 9 cm

(A) 108 cm2 (B) 54cm2 (C) 27cm2 (D) 18cm2 B 30° C
12 cm

3. In ∆PQR, QR= 8 cm, ‘PRQ= 300 and ‘PQR = 750. What is the area of ∆PQR?

(A) 64cm2 (B) 32cm2 (C) 16 cm2 (D) 8cm2 A

4. What is the value of x0 if the area of ∆ABC is 12 3 cm2?

(A) 300 (B) 450 (C) 600 (D) 750 x°
B 8 cm C

5. In ∆ABC ‘ABC = 45°, AB = 6 2 cm, and area of ∆ABC = 24 cm2. Find the length of BC.

(A) 6cm (B) 8cm (C) 4 2 cm (D) 12cm

6. ABCD is a rhombus in which AB = 6 cm and ‘ABC = 60°. A D

8cm
The area of rhombus ABCD is 60°
B C

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(A) 32 2 cm2 (B) 32 3 cm2 (C) 8 3 cm2 (D) 32cm2

PS

7. In the given parallelogram ABCD if QR = 8cm, ‘PQR = 300, and the
area is 48 cm2, what is the length of PQ?
30°
(A) 8cm (B) 10cm (C) 12cm (D) 16cm Q 8 cm R

8. A tree of the height 18 3 m is situated on the edge of a river. If the angle of elevation
of the tree observed from the opposite edge of the river is found to be 60°, what is the
breadth of the river?

(A) 3 m (B) 9 3 m (C) 18 m (D) 18 ft

9. The angle of elevation of the top of the tree from a point on the ground, which is 24m
away from the foot of the tree, is 30°. What is the height of the tree?

(A) 8 3 m (B) 24 3 m (C) 24 m (D) 12m

10. A circus artist is climbing a 30m long rope, which is tightly stretched and tied to the top
of a vertical pole from a point on the ground. If the angle made by the rope with ground
is 600, then what is the height of the pole?

(A) 3 m (B) 15 3 m (C) 30 3 m (D) 60 3 m

11. What is the altitude of the sun when the length of shadow of a vertical pole is equal to
its height?

(A) 300 (B) 450 (C) 600 (D) none of these

12. A 5 ft tall person observed the top of a tower of 55 ft high from a point 50 ft away from
the bottom of the tower on the horizontal level. Then, the angle of elevation of the tower
is

(A) 450 (B) 300 (C) 600 (D) 750

13. If a 3 ft tall child observes the angle of elevation of the top of a building 47 ft high which
is in front of him and finds to be 450, the distance between the child and the building
is...

(A) 44 ft (B) 54 ft (C) 44 3 ft (D) 44 m

14. If the angle of depression of a the top of a temple as observed from the roof of a house 30
ft high is found to be 300 and the distance between the temple and the house is 10 3 ft,
what is the height of the temple?

(A) 10 ft (B) 20 ft (C) 30 ft (D) 60 ft

15. A tree 18m high is broken by the wind so that its top touches the ground. If the length of
broken part of the tree is 12m, the angle made by the top of the tree with the ground is...

(A) 450 (B) 300 (C) 600 (D) 750

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Unit Statistics

17

17.1 Statistics - review

Statistics is the branch of mathematics in which facts and information are collected,
sorted, displayed, and analysed. Statistics are used to make decision and prediction
about the future plans and policies.
The word ‘statistics’ comes from the word ‘state’ largely because it was the job of the
state to keep records and make decisions based on census results.

17.2 Measures of central tendency

The measure of central tendency gives a single central value that represents the
characteristics of the entire data. A single central value is the best representative of
the given data towards which the values of all other data are approaching.

Average of the given data is the measure of central tendency. There are three types of
averages which are commonly used as the measure of central tendency.

They are: mean, median, and mode.
There are mainly three types of presentation of statistical data.
(i) Individual series (ii) Discrete series (iii) Grouped and continuous series
The process of finding averages of the given data is based on the way of presentation
of the data.

17.3 Mean of grouped and continuous data

We have already discussed the methods of finding mean of individual and discrete
series in the earlier classes. So, in this chapter, we are discussing the methods of
finding mean of grouped and continuous series.

In case of grouped and continuous data, we should find the mid-value (m) of each
class interval and it is written in the second column of the frequency table. The mid-
value of each class interval is obtained by using the following rule:

Mid-value (m) = lower limit + upper limit
2

Then, each mid - value is multiplied by the corresponding frequency and the product
fm is written in the fourth column.

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Statistics

Thus, if the corresponding frequency of each of mid-value m1, m2, m3 ... mn be f1, f2,
f3, .... fn, respectively, then

Mean (X) = f1 m1 + f2 m2 + f3 m3 + ... fn mn = ¦fm
f1 + f2 + f3 + ... fn N

Worked-out examples

Example 1: Calculate the mean from the table given below:

Marks 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50
No. of students
3 8 12 7 2

Solution:

Calculation of mean

Marks (X) mid–value (m) No. of students (f) fm
0 – 10 5 3 15
10 – 20 15 8 120
20 – 30 25 12 300
30 – 40 35 7 245
40 – 50 45 2 90

Total N = 32 6fm = 770

Now, mean marks (X) = ¦fm = 770 = 24.06
N 32

Alternative method (I) (Deviation method)

Alternatively, a number is taken as mean called assumed mean and it is denoted by
the letter ‘A’. The difference between each mid–value and assumed mean is then
calculated. The difference is called the deviation and it is denoted by ‘d’. Each deviation
is multiplied by the corresponding frequency. Then, from the sum of products of fd
column, the mean is calculated by using the following formula.

Mean (X ) = A + ¦fd
N
Let’s calculate the mean from the data of example 1.

Suppose, the assumed mean be 25.

Marks (X) mid–value (m) No. of students (f) d = m – 25 fd
0 – 10 5 3 – 20 – 60
10 – 20 15 8 – 10 – 80
20 – 30 25 12 0
30 – 40 35 7 10 0
40 – 50 45 2 20 70
Total 40
N = 32 6fd = – 30

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Now, mean (X ) =A+ ¦fd
N

= 25 – 30 = 25 – 0.9375 = 24.06
32

Alternative method (II) (Step deviation method)

In this method another deviation (d') is found by dividing the deviation (d) by

the length of class interval. Then the mean is calculated by using the following

formula: ¦fd'
N
Mean (X ) = A + uc

where, A = the assumed mean, c = the length of class internal.

This method is very much useful in the case of large values of x and f.

Let’s calculate the mean from the data of example 1.

Marks mid–value No. of students d = m – 25 d' = d fd'
(X) (m) (f) – 20 10 –6
5 3
0 – 10 –2

10 – 20 15 8 – 10 –1 – 8

20 – 30 25 12 0 00

30 – 40 35 7 10 1 7

40 – 50 45 2 20 2 4

Total N = 32 6fd' = – 3

Now, mean (X ) =A+ ¦fd' uc
N

= 25 – 3 u 10 = 25 – 0.9375 = 24.06
32

Example 2: If the arithmetic mean of the data given below is 31, calculate the value of P.

X 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60
f 4 6 10 p 6 4

Solution:

X mid-value (m) f fm

0 – 10 5 4 20
10 – 20 15 6 90
20 – 30 25 10 250
30 – 40 35 p 35p
40 – 50 45 6 270
50 – 60 55 4 220

Total N = 30 + p ¦fm = 850 + 35p

Now, mean ( X )= ¦fm
N

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31 = 850 + 35p
30 + p

or, 850 + 35p = 930 + 31p

or, p = 20

So, the required value of p is 20.

EXERCISE 17.1

General section

1. a) If ¦fx = 450 and N = 15, find X .
b) If ¦fx = 960 and n = 24, find X .

2. a) If X = 35 and ¦fx = 455, find the number of terms (N).
b) If the mean of a grouped data having ¦fm = 320 is 16, find the value of N.

3. a) If X = 25, N = 10 and ¦fx = p, find the value p.
b) If X = 32, ¦f = 15 and ¦fm = k, find the value k.

4. a) In a series, if ¦fm = 150 + 5a, X = 10 and ¦f = 10 + a, find the value of a.
b) If ¦fm = 72 + 8k, X = 6 and ¦f = 16 + k, find the value of k.
c) If mean ( X ) = 12, ¦fm = 70 + 10a and the number of frequency (N) = 5 + a, find
the value of a.
d) In a series if X = 20, ¦fm = 800 + 15p and ¦f = 10 + p, find the value of p.
e) If X = 15, ¦fm = 350 + 13p and number of terms (N) = 32, find the value of p.

Creative section
5. Find the arithmetic mean (or average) from the data given in the following tables:

a) X 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50
f5 7 8 6 4

b) Marks 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70

Frequency 12 7 8 3 10

c) Wages (Rs) 50 – 60 60 – 70 70 – 80 80 – 90 90 – 100

No. of workers 6 8 12 8 6

d) Age (in years) 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70

No. of workers 50 70 100 180 150

e) Marks 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80

No. of students 4 5 2 4 3 2

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Statistics

6. a) Find the mean by constructing a frequency table of class interval of 10 from the
data given below.

7, 47, 36, 39, 31, 19, 41, 49, 9, 51, 29, 22,
59, 17, 49, 21, 24, 12, 31, 8, 36, 18, 32, 16, 23

b) Construct a frequency table of class interval of 10 from the given data and find the
mean.
23, 5, 17, 28, 39, 52, 16, 22, 69, 75
41, 33, 9, 49, 34, 59, 72, 46, 65, 58
60, 48, 64, 32, 50, 73, 57, 51, 63, 36

7. a) The mean of the data given below is 36. Determine the value of p.

Age (in years) 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60

No. of teachers 3 8 15 p 4

b) The mean of the given data below is 28. Find the value of k.

Class interval 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60
6
Frequency 12 18 27 k 17

c) The mean of the data given below is 45. Find the value of a.

Wages (in Rs) 10 – 20 20 – 30 30 – 40 40 – 50 50 –60 60 – 70 70 – 80
No. of workers 4 7 9 18 a 7 3

17.4 Median – review

Median is an average that divides the given set of data into two equal parts. For
example,

10, 16, 22, 28 34 40, 46

Here, the data are arranged in ascending order. The item 28 has 3 items before it and
3 items after it. So, 28 is the middle item that divides the series into two parts. Thus,
28 is the median of the given series of data.

17.5 Median of grouped and continuous series

In the case of grouped and continuous series, the position of the median is obtained

by using the following formula.

Position of the median = N th class.
2
Then, the actual median within the median class is obtained by using the following

formula.

N – c.f.
2
Median =L+ uc
Where L f
= the lower limit of the class

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Statistics

c.f. = c.f. of the class just before the N th class
2

f = frequency of N th class (i.e. median class)
2

c = length of class interval

Worked-out examples

Example 1: The table given below shows the daily wages of 56 workers in a factory.
Compute the median wage.

Wages (in Rs) 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 80 – 90

No. of workers 5 12 18 10 6 3 2

Solution: Cumulative frequency table

Wages No. of workers (f) c.f.
20 – 30 5 5
30 – 40 12 17
40 – 50 18 35
50 – 60 10 45
60 – 70 6 51
70 – 80 3 54
80 – 90 2 56
Total
N = 56

Now, the position of the median class = N th class = 56 th class = 28th class
2 2

In c.f. column, the c.f. just greater than 28 is 35 and its corresponding class is 40 – 50.

? Median class is 40 – 50.

Here, L = 40, N = 28, c.f. = 17, f = 18 and c = 10
2

N – c.f. 28 – 17
2 18
? Median = L + u c = 40 + u 10 = 46.1
f

So, the median wage is Rs 46.1.

Example 2: Find the median from the following data.

Wages (Rs) 15 – 25 15 – 35 15 – 45 15 – 55 15 – 65

No. of workers 4 10 22 27 30

Solutions:

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Statistics

Cumulative frequency table

Wages No. of workers (f) c.f
15 – 25 4 4
25 – 35 10
35 – 45 10 – 4 = 6 22
45 – 55 22 – 10 = 12 27
55 – 65 27 – 22 = 5 30
Total 30 – 27 = 3

N = 30

Now, the position of the median class = N th class = 30 th
2 2

= 15th class

In c.f. column, the c.f. just greater than 15 is 22 and its corresponding class is 35 - 45.

Median class is 35 – 45.

Here, L = 35, N = 15, c. f. = 10, f = 12 and c = 10
2

Now, median =L+ c N – c.f.
f 2

= 35 + 10 (15 – 10) = 39.17
12

So, the required median is 39.17

Example 3: If the median of the given data is 24, find the value of x.

Marks 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50

No. of students 5 25 x 18 7

Solutions:

Cumulative frequency table

Marks No. of Students c.f
0 – 10 5 5
10 – 20 25 30
20 – 30 x 30 + x
30 – 40 18 48 + x
40 – 50 7 55 + x
Total
N = 55 + x

Since the median of the data is given as 24, it lies in the class 20 – 30. So, the median class
is 20 – 30.

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Here, L = 20, N = 55 + x , c.f. = 30, f = x, c = 10
2 2

N – c.f.
2
Now, Median = L + f ×c

55 + x – 30
24 = 20 + 2 x
or, × 10

or, 24 – 20 = x–5 ×5
x

or, 5x – 25 = 4x
or, x = 25
So, the required value of x is 25.

17.6 Quartiles

Quartiles are the values that divide the arrayed data into four equal parts. A
distribution is divided into four equal parts by three quartiles.

(i) The first or lower quartile (Q1) is the point below which 25% of the items lie and
above which 75 % of the items lie.

(ii) The second quartile (Q2) is the point below which 50% of the items lie and
above which 50% of the items lie. So, the second quartile is the median of a
distribution.

(iii) The third or upper quartile (Q3) is the point below which 75% of the items lie
and above which 25% of the items lie.

25% Q1 25% Q2 25% Q3 25%

17.7 Quartiles of grouped and continuous series

In this case, we obtain the position of the first quartile and the third quartile by
using the following formulae.

The position of the first quartile, Q1 = N th class.
4

The position of the third quartile, Q3 = 3N th class
4

Then, the actual first and third quartiles are obtained by using the following

formulae. N – c.f.
The first quartile, Q1 = L + 4
uc and
f

3N – c.f.
4
the third quartile, Q3 =L+ uc
f

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Example 4: Compute the lower quartile (Q1) and the upper quartile (Q3) from the
following data.

Marks 20 – 40 40 – 60 60 – 80 80 – 100

No. of students 8 12 10 6

Solution:

Cumulative frequency table

Marks No. of students (f) c.f.

20 – 40 8 8
40 – 60 12 20
60 – 80 10 30
80 – 100 6 36

Total N = 36

Now, the position of the lower quartile, Q1 = N th class = 36 th class = 9th class
4 4

In c.f. column, the c.f. just greater than 9 is 20 and its corresponding class is 40 – 60.

Here, L = 40, N = 9, c.f. = 8, f = 12 and c = 20
4
N
4 – c.f. 9–8
12
? The lower quartile, Q1 = L + f u c = 40 + u 20 = 41.67

Again, the position of the upper quartile, Q3 = 3N th class = (3 u 9)th class = 27th class
4

If c.f. column, the c.f. just greater than 27 is 30 and its corresponding class is 60 – 80.

Here, L = 60, 3N = 27, c.f. = 20, f = 10 and c = 20
4
3N
4 – c.f. 27 – 20
10
? The upper quartile, Q3 = L+ f uc = 60 + × 20 = 74

So, the required lower quartile (Q1) and the upper quartile (Q3) are 41.67 and 74
respectively.

EXERCISE 17.2

General section

1. a) The median of a series lies in the class interval of 20 – 30. If the necessary variables
N
in computing median are given as 2 = 15, c.f. = 12, f = 5 and c = 10, compute the

median.

b) The median of the given data lies in class interval of 40 – 50. If N = 18, c.f. = 16,
2
f = 8 and c = 10, find the value of median.

c) If the lower quartile class of a series is 40 – 60, N = 9, c.f. = 8, f = 12 and c = 20,
compute the first quartile. 4

d) If the Q3 of a series lies in 30 – 40, 3N = 25, c.f. = 20, f = 5 and c = 10, find Q3.
4

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2. a) Find the median class from the following table.

Marks 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70

No. of students 2 5 7 6 3 2

b) Find the median class. 15 – 25 25 – 35 35 – 45 45 – 55
8 17 20 22
x 5 – 15
f3

c) Find the first quartile class from the table given below.

Wages (in Rs) 50 – 65 65 – 80 80 – 95 95 – 110 110 – 125

No. of workers 7 3 6 9 5

d) Find the upper quartile class from the following table.

Marks obtained 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 80 – 90

No. of students 5 8 10 15 7 5

Creative section

3. Compute the median from the data given in the tables.

a) Marks obtained 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60

No. of students 4 6 10 7 3

b) Wages (in Rs) 50 – 60 60 – 70 70 – 80 80 – 90 90 – 100

No. of workers 8 11 10 6 3

c) Class 0 – 30 30 – 60 60 – 90 90 – 120 120 – 150 150 – 180

Frequency 5 10 22 25 14 4

d) Marks obtained 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60

No. of students 5 12 30 10 8 5

4. Find the median from the data given in the tables.

a) Marks obtained 0 – 10 0 – 20 0 – 30 0 – 40 0 – 50
No. of students 4 12 24 44 62

b) X 5 – 15 5 – 25 5 – 35 5 – 45 5 –55 5 – 65
f 5 13 23 32 43 44

5. a) If the median of the given table is 24, calculate the value of x.

Marks obtained 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50
No. of students 4 12 x 9 5

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b) If the median of the series given below in the table is 40, find the value of p.

Marks obtained 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60

No. of students 4 6 15 p 5

6. Compute the first or lower quartile (Q1) from the data given in the following tables.

a) Marks obtained 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70

No. of students 2 5 7 6 3 2

b) Marks obtained 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70
No. of students 4 5 10 8 7 6

c) Wages (in Rs) 50 – 65 65 – 80 80 – 95 95 – 110 110 – 125

No. of workers 7 3 6 9 5

d) Wages (in Rs) 100–200 200–300 300–400 400–500 500–600 600–700

No. of workers 4 9 15 11 8 3

e) Marks obtained 20 – 30 20 – 40 20 – 50 20 – 60 20 – 70 20 – 80
21 24
No. of students 5 9 14 18

7. Compute the third or upper quartile (Q3) from the data given in the following tables.

a) Marks obtained 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70

No. of students 2 5 7 6 3 2

b) Weight (in kg) 30 – 35 35 – 40 40 – 45 45 – 50 50 – 55
2
No. of students 3 5 6 4
60 – 75
c) Marks obtained 0 – 15 15 – 30 30 – 45 45 – 60 7

No. of students 8 6 12 15 56 – 60
3
d) Weight (in kg) 40 – 44 44 – 48 48 – 52 52 – 56 60 – 64
0 – 100 1
No. of students 8 10 14 16 40

e) Marks obtained 0 – 20 0 – 40 0 – 60 0 – 80
No. of students 12 15 26 34

8. a) The daily expenses ( in Rs) of 30 students of a school are given below.

7, 15, 24, 10, 5, 12, 20, 8, 27, 18,

25, 16, 9, 13, 14, 22, 28, 11, 4, 17,

12, 19, 14, 15, 21, 12, 12, 14, 10, 11

Display the data in c.f. table with class interval of length 5 and compute the median
expense.

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b) The following are the marks obtained by students in maths in an examination:

51, 22, 63, 35, 46, 57, 79, 21, 39, 51,

32, 43, 52, 59, 38, 45, 40, 32, 60, 63
Make a frequency table of class interval 10 and compute the first quartile.

c) The marks obtained by 20 students in an examination are given below. Make a
frequency table of class interval of 10 and find the third quartile.
15, 12, 23, 35, 46, 57, 18, 12, 39, 51,
32, 43, 25, 59, 18, 38, 45, 40, 32, 33

17.8 Ogive (Cumulative frequency curve)

The sum of the frequencies of all the values up to a given value is known as
cumulative frequency. It is denoted by c.f.

In the case of a continuous series, if upper limit (or lower limit) of each class interval
is taken as x-coordinate and its corresponding c.f. as y-coordinate and the points are
plotted in the graph, we obtain a curve by joining the points with freehand. Such
curve is known as ogive or cumulative frequency curve.

(i) Less than ogive (or less than cumulative frequency curve)

When upper limit of each class interval is taken as x-coordinate and its
corresponding c.f. as y-coordinate, the ogive so obtained is known as less than
ogive (or less than cumulative frequency curve). Obviously, less than ogive is
an increasing curve sloping upwards from left to right and has the shape of an
elongated S.

(ii) More than ogive (or more than cumulative frequency curve)

When lower limit of each class is taken as x-coordinate and its corresponding
c.f. as y-coordinate, the ogive so obtained is known as more than ogive (or
more than cumulative frequency curve). More than ogive is a decreasing curve
sloping downwards from left to right and has the shape of an elongates upside
down ( ).

17.9 Construction of less than ogive and more than ogive

Study the following steps to construct a less than ogive.

(i) Make a less than cumulative frequency table.

(ii) Choose the suitable scale and mark the upper class limits of each class
interval along x-axis and cumulative frequencies along y-axis.

(iii) Plot the coordinates of (upper limit, less than c.f.) on the graph.

(iv) Join the point by freehand and obtain a less than ogive.

In the case of more than ogive, we should prepare the more than cumulative
frequency table. The lower class limits of each class interval are marked on
x-axis. Then, the process is similar to the construction of less than ogive.

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Worked-out examples

Example 1: The table given below shows the marks obtained by 60 students in
mathematics. Construct: (i) less than ogive and (ii) more than ogive.

Marks 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60

No. of students 4 10 20 15 6 5

Solution:

Less than cumulative frequency table.

Marks No. of students (f) Upper limit c.f.
0 – 10 4 10 4 (less than 10)

10 – 20 10 20 14 (less than 20)
20 – 30 20 30 34 (less than 30)

30 – 40 15 40 49 (less than 40)
40 – 50 6 50 55 (less than 50)
50 – 60 5 60 60 (less than 60)

Now, the coordinates (10, 4), (20, 14), (30, 34), (40, 49), (50, 55), and (60, 60) are plotted in

the graph. The less than ogive
Y

Less than cumulative frequencies 60 (60, 60)

(50, 55)

50

(40, 49)

40

30 (30, 34)

20

(20, 14)

10

(10, 4) X

0 10 20 30 40 50 60

Upper limits

More than cumulative frequency table

Marks No. of students (f) Lower limit c.f.
0 – 10 4 0 60 (more than 0)
10 – 20 10 10 60 – 4 = 56 (more than 10)
20 – 30 20 20 56 – 10 = 46 (more than 20)
30 – 40 15 30 46 – 20 = 26 (more than 30)
40 – 50 6 40 26 – 15 = 11 (more than 40)
50 – 60 5 50 11 – 6 = 5 (more than 50)

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Now, the coordinates (0, 60), (10, 56), (20, 46), (30, 26), (40, 11), and (50, 5) are plotted in
the graph.

The more than ogive

(0, 60)
(10, 56)

(20, 46)

More than cumulative frequencies (30, 26)

(40, 11)
(50, 5)

Lower limits

17.10 Use of less than and more than ogives to find median and
quartiles

If we draw both the less than ogive and more than ogive of a distribution on the
same graph paper, they intersect at a point. The foot of the perpendicular drawn
from the point of intersection of two ogives to x-axis gives the value of median of
the distribution.

For example,

Marks No. of students (f) Upper limit Less than c.f. Lower More than c.f.
limit
0 – 10 4 10 4 60
10 – 20 10 20 14 0 56
20 – 30 20 30 34 46
30 – 40 15 40 49 10 26
40 – 50 6 50 55 11
50 – 60 5 60 60 20 5

30

40

50

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More than and less than cumulative (0, 60) (60, 60)
frequencies (10, 56) (50, 55)
(40, 49)
(20, 46)

(30, 34)
(30, 26)

(20, 14) (40, 11)
(10, 4) (50, 5)

Lower limits

From the graph, the perpendicular drawn from the point of intersection of two ogives meets
x-axis at 28 units from the origin. So, the required median of the given distribution is 28.

Example 2: The table given below shows the marks obtained by 60 students in
mathematics. Construct a less than ogive and compute median, the first
quartile (Q1) and the third quartile (Q3)

Marks 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60
No. of students
4 10 20 15 6 5

Solution:

Less than cumulative frequency table.

Marks No. of students (f) Upper limit c.f.
0 – 10 4 10 4 (less than 10)

10 – 20 10 20 14 (less than 20)

20 – 30 20 30 34 (less than 30)

30 – 40 15 40 49 (less than 40)

40 – 50 6 50 55 (less than 50)

50 – 60 5 60 60 (less than 60)

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Less than cumulative frequencies The less than ogive

(60, 60)
(50, 55)
(40, 49)
E

(30, 34)
A

C(20, 14) F
(10, 4) D B

Upper limits

Median lies in the 50% of the total data.
? 50% of 60 students = 30 students
A straight line drawn from 30 (frequency) on y-axis parallel to x-axis intersects the curve at
A. A perpendicular drawn from A meets x-axis at B and its x-coordinate is 28.
So, median = 28.
Similarly, Q1 lies in 25% of the total data.
? 25% of 60 students = 15 students
Also, Q3 lies in 75% of the total data.
? 75% of 60 students = 45 students
Drawing straight lines as like in the case of median,

Q1 = 20.5 and Q3 = 37.33

EXERCISE 17.3

General section

1. From each of the following cumulative frequency curves, determine the median class.

a) Y b) Y c)

10
8
Number of Students
Number6
No. of students4

X 2

Marks obtained 0 10 20 30 40 50 60 X

Marks obtained Marks obtained

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2. From each of the following cumulative frequency curves, find the first quartile (Q1) class.
a)

Cumulative frequency No. of students Frequency

20 40 60 80 100 0 5 10 15 20 25 30 35 40 X Class interval
Class interval Marks obtained

3. From each of the following cumulative frequency curves, find their quartile (Q3) class.
a) b)
c)

Number of workers Number of workers Number of students

Daily wages Marks obtained Marks obtained
Marks obtained
4. a) In the graph given alongside, two Number of students
cumulative frequency curves ('is less
than' and 'is more than') are intersecting
at a point. Find the median class and
the value of median.

b) Find the median class and the value of Frequency
median from the adjoining cumulative
frequency curved, where 5 – 10 is a class.

Marks obtained

Creative section

5. a) Draw a ‘less than’ ogive from the data given below.

Data 30 – 35 35 – 40 40 – 45 45 – 50 50 – 55 55 – 60

Frequency 5 8 17 10 7 3

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b) Construct a ‘more than’ ogive from the data given below.

Marks 20–30 30–40 40–50 50–60 60–70 70–80 80–90 90–100
No. of 8 12 2
students 20 15 10 7 4

c) Draw ‘less than’ ogive and ‘more than’ ogive on the same coordinate axes and find
the median of the distribution.

Data 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70

Frequency 6 12 18 20 15 9

d) The table given below shows the daily wages 40 workers of a factory.

Wages (in Rs) 50 – 55 55 – 60 60 – 65 65 – 70 70 – 75 75 – 80

No. of workers 4 6 10 12 5 3

Draw ‘less than’ ogive and ‘more than’ ogive on the same coordinate axes and find
the median of the distribution.

6. a) Draw a 'less than' ogive from the data given below and find the lower quartile (Q1).

Marks obtained 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70

No. of students 4 5 10 8 7 6

b) Draw a 'less than' ogive from the data given in the table below and find the upper

quartile (Q3).

Marks obtained 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70

No. of students 8 15 20 10 7

c) The marks obtained by 200 students are given below. Draw a ‘less than’ ogive and
compute the median, the first quartile (Q1) and the third quartile (Q3) from the
graph.

Marks No. of students

0 - 10 7

10 - 20 9

20 - 30 12

30 - 40 21

40 - 50 39

50 - 60 44

60 - 70 29

70 - 80 18

80 – 90 14

90 - 100 7

7. Project work

Collect the marks obtained by you and your friends in a terminal examination in

mathematics. Tabulate the marks in the appropriate class intervals and find mean,

median, quartiles, and mode.

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Objective Questions
Tick the correct alternatives.

1. If the mean of five observations x, x + 2, x + 4, x + 6, and x + 8 is 6, what is the mean
of first two observations?

(A) 9 (B) 7 (C) 5 (D) 3

2. There are 15 boys and 10 girls in a hostel. If the average weight of boys is 42 kg and that
of all 25 students is 40 kg, what is the average weight of girls?

(A) 37 kg (B) 38 kg (C) 39 kg (D) 41 kg

3. In a stationary, the average sale a week was Rs 2000. If sale of Saturday was excluded
then the average sale of remaining days would be Rs 1950, what was the sale of Saturday?

(A) Rs 1800 (B) Rs 2150 (C) Rs 2300 (D) Rs 2475

4. In a continuous data, if 6f = 10 and 6fm = 6f + 50, what is the mean of the data?

(A) 5 (B) 6 (C) 7 (D) 10

5. In a continuous series, if 6f is 5% of 6fm, what is the mean of the series?

(A) 5 (B) 10 (C) 15 (D) 20

6. In a continuous data, the mean of 40 observations was 25. If the class interval 50 – 60
with frequency 10 is included then what will be new mean?

(A) 31 (B) 36 (C) 28 (D) 26

7. In a continuous series, the mean of 28 observations was 20. If the class interval 0 – 10
with frequency 8 is discarded then what will be new mean?

(A) 24 (B) 25 (C) 26 (D) 27

8. In a continuous data, the mean of 50 observations was 40. If the class interval 30 – 40
with frequency 10 was mistakenly copied as class interval 30 – 50 with frequency 10,
what is the correct mean of the data?

(A) 37 (B) 39 (C) 29 (D) 51

9. What is the name of the quartile which divides the data above 25%?

(A) Q1 (B) Q2 (C) Q3 (D) both (A) and (C)

10. In what ratio does Q1divide the data arranged in ascending order from bottom?
(A) 1:2 (B) 1:3 (C) 3:1 (D) 2:1

11. What percent of observations are there below Q3 of a data?
(A) 25% (B) 50% (C) 60% (D) 75%

12. What percent of items of observations are there above Q3 of a data?
(A) 25% (B) 45% (C) 50% (D) 75%

13. In a grouped data, if the position of upper quartile is 18th class, what is the number of
observations in the data?

(A) 36 (B) 18 (C) 12 (D) 72

14. Which of the following is the formula to calculate median of a continuous data?

(A) L + N – c.f. ui (B) L + N – c.f. u i (C) L + 3N – c.f. ui (D) L – N – c.f. ui
2 4 4 2

f ff f

15. If the lower quartile lies in the class of 24 – 32 and the total number of observations (N)
is four times the cumulative frequency (c.f.) of pre-lower quartile class, what is the value
of lower quartile?

(A) 32 (B) 28 (C) 30 (D) 24

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Unit Probability

18

18.1 Definitions of basic terms on probability – review

(i) Random experiments

Any experiment whose outcome cannot be predicted or determined in advance
is called a random experiment. For example, tossing a coin, rolling a die, etc. are
random experiments.

(ii) Outcomes

The results of a random experiment are called outcomes. For example, while
tossing a coin, the occurrence of head or tail is the outcome.

(iii) Sample space

The set of all possible outcomes of a random experiment is known as sample
space. Usually, it is denoted by S. For example,
When a coin is tossed, S = {H, T}
When a die is thrown, S = {1, 2, 3, 4, 5, 6}

(iv) Event

A subset of the sample space S related to an experiment is known as event.
For example,
While tossing a coin 3 times, S = {HHH. HHT, HTH, HTT, THH, THT, TTH, TTT}
Let A = {HHH, HHT, HTH, THH},
A is the event in which at least two heads are obtained.

(v) Exhaustive cases

The total number of all possible outcomes of a random experiment is known as
exhaustive cases. For example,
While tossing a coin, S = {H, T}. So, exhaustive cases = 2
While tossing two coins simultaneously, S = {HH, HT, TH, TT}
So, exhaustive cases = 4.

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Probability

(vi) Favourable cases

The outcomes in an random experiment which are desirable (or expected) to us
are called favourable cases. For example,
While tossing a coin, S = {H, T}
Here, the favourable number of case of head is 1 and tail is also 1.
The favourable number of cases of getting a ‘king’ when a card is drawn from a
well shuffled pack of 52 cards is 4.

(vii) Equally likely events

Two or more events are said to be equally likely if the chance of occurring any
one event is equal to the chance of occurring other events. For example, while
throwing a die, the chance of coming up the numbers 1 to 6 is equal. So, they are
equally likely events.

(viii) Independent events

Two or more events are said to be independent if the occurrence or none-occurrence
of one of the events does not affect the occurrence or none-occurrence of the other
events. For example, in the random experiment of tossing a coin twice or more,
the occurrence of any one event in the first trial does not affect the occurrence of
any other event in the second trial. So, they are independent events.

(ix) Dependent events

Two or more events are said to be dependent if the occurrence of one of the
events affects the occurrence of the other events. For example, while drawing a
ball in two successive trials from a bag containing 2 red and 3 blue balls without
a replacement, getting any one coloured ball in the first trial affects to get other
coloured ball in the second trial. So, these are the dependent events.

(x) Mutually exclusive events

Two ore more events of a sample space S are said to be mutually exclusive if the
occurrence of any one event excludes the occurrence of the other events. For
example, while tossing a coin, the occurrence of head excludes the occurrence of
tail or vice versa. So, they are mutually exclusive events.

Furthermore, let’s consider the experiment of throwing a die. Let A be the event,
‘the number obtained is even’.

Then A = {2, 4, 6}. A S
Again, let B be the event, ‘the number obtained is odd’. B

Then, B = {1, 3, 5}.

Here, A ˆ B = I

A ˆ B = I

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Probability

Thus, the joint occurrence of A and B is an impossible event. In this case, the
events A and B are said to be the mutually exclusive events.
In general, if A and B are any two events on a sample space S and A ˆ B = I, the
events are said to be mutually exclusive.

(xi) None-mutually exclusive events S
AB
If A and B are any two events on a sample space S and
A ˆ B z 0, the events are said to be none-mutually A ˆ B z I
exclusive events. For example,
Again, let's consider the experiment of throwing a die. S
Let A be the event 'the number obtained is even' and
B be the event ' the number obtained is less than 5'. A6 2 B
1
Then, A = {2, 4, 6}
B = {1, 2, 3, 4} 43
5
? A ˆ B = {2, 4}

It means, ‘an even number from 1 to 6’ and ‘a number
less than 5’ can occur simultaneously in a single throw
of a die. So, A and B are said to be none-mutually
exclusive events.

18.2 Law of addition of mutually exclusive events

If A and B are any two mutually exclusive events of B S
a sample space S, the probability of the occurrence A
of either A or B is equal to the sum of the individual
probabilities of the occurrence of A and B.

i.e. P (A or B) = P (A) + P (B)
i.e. P (A ‰ B) = P (A) + P (B)

Further more, if A, B and C are three mutually exclusive events, then,
P (A or B or C) = P (A ‰ B ‰ C) = P (A) + P (B) + P (C) and so on.

Worked-out examples

Example 1: One card is drawn at random from the numbered cards, numbered from
Solution: 10 to 21. Find the probability that the card may be prime numbered or
even numbered card.

Let, n(S) be the exhaustive cases, n(A) be the number of events of prime numbered cards,
and n(B) be the number of events of even numbered cards.

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Probability

Here, S = {10, 11, 12, ... 21}. So, n(S) = 12

A = {11, 13, 17, 19}. So, n(A) = 4

B = {10, 12, 14, 16, 18, 20}. So, n(B) = 6

Now, P(A) = n(A) = 4 = 1
n(S) 12 3

P(B) = n(B) = 6 = 1
n(S) 12 2

By using the law of addition of probabilities of mutually exclusive events,

P (A or B) = P (A ‰ B) = P(A) + P(B) = 1 + 1 = 5 .
3 2 6
5
So, the probability that the card may be prime numbered or even numbered is 6 .

Example 2: From a pack of 52 cards, a card is drawn at random. What is the
probability of getting a face card or an ace?

Solution:

Let n (S) be the exhaustive events, n(A) the number of events of face card and n(B) the
number of events of ace.

Here, n(S) = 52, n(A) = 12 and n(B) = 4

Now, by using the law of addition of probabilities of mutually exclusive events,

P(A or B) = P(A ‰ B) = P(A) + P(B)

= n(A) + n(B)
n(S) n(S)

= 12 + 4 = 16 = 4
52 52 52 13

So, the probability that the card may be a face card or an ace is 4 .
13

Example 3: A bag contains 3 red, 5 blue, 7 black, and 9 white identical balls. If a ball
is drawn randomly from the bag, find the probability that it may be either
red, or blue or white.

Solution:

Let, n(S) be the exhaustive cases, n(A) the number of cases of red balls, n(B) the number of
cases of blue balls, and n(C) the number of cases of white balls.

Here, n(S) = 3 + 5 + 7 + 9 = 24
n(A) = 3, n(B) = 5 and n(C) = 9

Now, by using the law of addition of probabilities of mutually exclusive events,
P (A or B or C) = P (A ‰ B ‰ C) = P(A) + P(B) + P(C)

= n(A) + n(B) + n(C)
n(S) n(S) n(S)

= 3 + 5 + 9 = 17
24 24 24 24

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18.3 Law of addition of none-mutually exclusive events S
B
If A and B are any two none-mutually exclusive events of a
sample space S, then the probability of occurrence of either A A ˆ B
A or B in single trial is given by

P(A or B) = P(A ‰ B) = P(A) + P (B) – P(A ˆ B)

Example 4: From the pack of cards, a card is drawn randomly. Find the probability of
Solution: getting it red or face card.

Let, n(S) be the exhaustive cases, n(A) be the number of events of red card, and n(B) be the
number of events of face cards.

Here, n(S) = 52, n(A) = 26, and n(B) = 12.

Also, n(A ˆ B) = 6 3 faced card of heart + 3 faced card of diamond

Now, using the law of addition of probabilities of non-mutually exclusive events,

P(A or B) = P (A ‰ B) = P(A) + P(B) – P(A ˆ B)

= n(A) + n(B) – n(A ˆ B)
n(S) n(S) n(S)

= 26 + 12 – 6 = 26 + 12 – 6 = 32 = 8
52 52 52 52 52 13

So, the probability of getting the card red or face is 8 .
13

Example 5: Two unbiased dice are rolled. Find the probability that the sum of the
Solution: numbers on the two faces is either divisible by 3 or divisible by 4.

Two dice can be thrown in 6n = 62 = 36 ways.
(6 is the exhaustive cases of a die and n is the number of rolling dice.)
Let n(S) be the exhaustive cases, n(A) the number of events that the sum is divisible by 3 and
n(B) the number of events that the sum is divisible by 4.
Here, A = {(1, 2), (2, 1), (1, 5), (2, 4), (3, 3), (4, 2), (5, 1), (3, 6), (4, 5), (5, 4), (6, 3), (6, 6)}
? n(A) = 12
Also, B = {(1, 3), (2, 2), (3, 1), (2, 6), (3, 5), (4, 4), (5, 3), (6, 2), (6, 6)}
? n(B) = 9
The event A ˆ B happens when the sum is divisible by 3 as well as 4, i.e., when the sum is
12 which is (6, 6).
? n(A ˆ B) = 1.
Now, by using the law of addition of probabilities of none-mutually exclusive events,

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Probability

P(A or B) = P(A ‰ B) = P(A) + P(B) – P(A ˆ B)

= n(A) + n(B) – n(A ˆ B)
n(S) n(S) n(S)

= 12 + 9 – 1 = 20 = 5 .
36 36 36 36 9

So, the probability that the sum of the numbers on the two faces is either divisible by 3 or

by 4 is 5 .
9

EXERCISE 18.1

General section

1. Define the following terms with examples.

a) Sample space b) Exhaustive cases

c) Independent and dependent events d) Mutually exclusive events

e) Non-mutually exclusive events

2. a) If A and B are two mutually exclusive events, what is the probability of the
occurrence of either A or B?

b) If X, Y and Z are three mutually exclusive events, what is the probability of the
occurrence of either X or Y or Z ?

c) If A and B are two non-mutually exclusive events, what is the probability of the
occurrence of either A or B?

3. a) A and B are mutually exclusive events. If n(A) = 4, n(B) = 6 and n(S) = 12, find
P (A ‰ B).

b) If P(A) = 2 , P (B) = 94, where A and B are mutually exclusive events, find
9
P (A ‰ B).

c) If P(A ‰ B) = 7 and P(A) = 4 , where A and B are mutually exclusive events,
13 13
find P(B).

d) A and B are mutually exclusive events. If P(A ‰ B) = 0.8 and P(B) = 0.3, find P(A).

e) Two events Q and R are mutually exclusive with P(Q) = 3 and P(R) = 1 . Find the
5 5

probabilities of the following events. (i) P(Q ‰ R) (ii) P( Q ‰ R )

4. a) A and B are non-mutually exclusive events. If n(A) = 5, n(B) = 7, n(A ˆ B) = 2 and

n(S) = 20, find P(A ‰ B).

b) If P(A) = 4 , P(B) = 1 and P(A ˆ B) = 2 , where A and B are none-mutually
25 5 25

exclusive events, find P(A ‰ B).

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c) A and B are non-mutually exclusive events. If P (A ‰ B) = 9 , P(A) = 3 and
16 16
3
P(B) = 4 , find P(A ˆ B).

d) If P(A ‰ B) = 0.7, P(A ˆ B) = 0.1 and P(A) = 0.35, find P(B).

Creative Section

5. From a well-shuffled pack of 52 cards, a card is drawn at random. Find the probabilities
of the following events.

a) It is either a king or a queen. b) It is a black ace or a red king.

c) It is a king or an ace. d) It is either a non-faced card or a jack.

e) It is either a black king or a red king. f) It is either a red or a club.

g) It is either a spade or a heart. h) It is a queen or a black ace.

i) It is neither a king nor an ace. j) It is neither a black jack nor a red queen.

6. a) When a fair die is thrown, find the probabilities of the following events.

(i) Getting either 1, or 6

(ii) Getting either 1, 3, or 5

(iii) Getting an even number less than 6 or an odd number less than 5.

(iv) Getting a multiple of 2 or a multiple of 3

(v) Getting either 1 or a prime number.

b) Two unbiased dice are rolled. Find the probability that the sum of the numbers on
the two faces is either divisible by 5 or divisible by 6.

7. a) From the number cards numbered from 1 to 30, a card is drawn at random. Find the
probabilities of the following events.

(i) The number is either divisible by 5 or by 7.

(ii) The number is either divisible by 5 or by 9.

(iii) The number is either a multiple of 7 or a multiple of 8.

(iv) The number is either a prime number less than 20 or a composite number
more than 20.

b) A number card is drawn randomly from the set of numbered cards, numbered
from 6 to 39. Find the probability that the card may be a prime number or a cubed
number.

c) From the number cards, numbered from 7 to 27 a card is drawn at random. Find the
probability of getting the card of a prime or an even number.

8. a) A box contains 4 yellow, 6 green, and 8 red balls. A ball is drawn at random from
the box. Find the probabilities of the following events.

(i) The ball is either yellow or red.

(ii) The ball is either green or yellow.

(iii) The ball is neither red nor green.

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Probability

b) A basket contains 5 yellow, 3 blue, and 2 green balls. If a ball is drawn randomly
from the basket, find the probability of not getting a blue ball.

9. From a well-shuffled pack of 52 cards, a card is drawn at random. Find the probabilities
of the following events.

a) It is either a black or a face card. b) It is either a spade or a queen card.

c) It is either a red card or an ace. d) It is either a heart or a jack.

10. a) From the number cards numbered from 1 to 30, a card is drawn at random. Find the
probabilities of the following events.

(i) The number is either divisible by 4 or by 5.

(ii) The number is either divisible by 5 or by 6.

(iii) The number is either a multiple of 3 or a multiple of 4.

(iv) The number is either a prime or an even number.

b) Find the probability of getting a square number or an even number card while
drawing a flash card from the set of cards numbered from 5 to 15.

c) Find the probability of occurring a cube number or an odd number card while
drawing a flash card from the set of cards numbered from 7 to 17.

d) A natural number is chosen at random from the first 30 natural numbers. What is
the probability that the number chosen is neither divisible by 4 nor by 6?

e) Two dice are thrown together. What is the probability that the sum of the numbers
on the two faces is either less than 6 or divisible by 4?

18.4 Multiplication law of probability for independent events

Two or more events are said to be independent if the occurrence or none-occurrence
of any one event does not affect the occurrence or none-occurrence of the other
events.

Suppose, a bag contains 1 red ball and 2 black balls.

If a ball is drawn and again it is replaced to draw another ball, the outcomes of these
two trials are independent events. Study the following possible outcomes of two
trials shown in the table.

R B1 B2
R RR RB1 RB2
B1 B1R B1 B1 B1 B2
B2 B2R B2 B1 B2 B2

Here, the probability of Red ball, P(R) = 1
3

The probability of Black ball, P(B) = 2
3

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Now, from the table, = 1 = 1 u 1 = P(R) × P(R)
(i) Probability of Red balls in both trials P(R and R) 9 3 3

(ii) Probability of Black balls in both trials P(B and B) = 4 = 2 u 2 = P(B) u P(B)
9 3 3
(iii) Probability of Red at 1st and Black in 2nd trial P(R and B)
= 2 = 1 u 2 = P(R) × P(B)
(iv) Probability of Black at 1st and Red in 2nd trial P(B and R) 9 3 3

= 2 = 2 u 1 = P(B) u P(R)
9 3 3

From the above illustrations, we can verify the following probability relationships between
any two independent events.

(i) P(R and R) = P(R ˆ R) = P(R) u P(R)
(ii) P(B and B) = P(B ˆ B) = P(B) u P(B)
(iii) P(R and B) = P(R ˆ B) = P(R) u P(B)
(iv) P(B and R) = P(B ˆ R) = P(B) u P(R)

Thus, if A and B are any two independent events, P(A and B) = P(A ˆ B) = P(A) u P(B). It is
called the multiplication law of probability for independent events.

Furthermore, if A, B and C are any three independent events,
P(A ˆ B ˆ C) = P(A) u P(B) u P(C)

Worked-out examples

Example 1: Two cards are drawn one after another with replacement from a well
shuffled pack of 52 cards. Find the probability that both cards are king.

Solution:

There are 4 kings in a pack of 52 cards.

Let K denotes the event of king.

Since the card which is drawn at first is replaced to draw another card, it is an independent
event.

Now, by using multiplication law of probability for independent events,

P(K and K) = P(K ˆ K)= P(K) u P(K)

= n(K) u n(K) = 4 u 4 = 1 u 1 = 1
n(S) n(S) 52 52 13 13 169

So, the required probability is 1 .
169

Example 2: A bag contains 5 red balls and 3 blue balls. A ball is drawn at random
Solution: and replaced. After that another ball is drawn. Find the probability that
(i) both balls are blue and (ii) none of them are blue.

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Probability

Here, the number of possible outcomes = total number of balls
n(S) = 5 + 3 = 8 balls

Let B denotes the event of blue ball.

Since the ball which is drawn at first is replaced to draw another ball, it is an independent event.

Now, by using multiplication law of probability for independent events,

(i) P(B and B) = P(B ˆ B) = P(B) u P(B)

= n(B) u n(B) = 3 u 3 = 9
n(S) n(S) 8 8 64

So, the probability that both of them are blue is 9 .
64

(ii) P(B and B) = [1 – P(B)] [1 – P(B)] = 1 – 3 1 – 3 = 25
8 8 64

Alternatively, none of them are blue means both of them are red.

So, P(R and R) = P(R) ˆ P (R) = P(R) × P(R) = 5 × 5 = 25
8 8 64

Example 3: A box contains 4 red, 3 blue and 5 white balls. A ball is drawn at random
and it is replaced, then another ball is drawn. Find the probability that
(i) the first is white and the second is blue,
(ii) the first is blue and the second is red,
(iii) both of them are red,
(iv) both of them are of the same colour,
(v) both of them are not blue.

Solution:

Here, the number of possible outcomes = total number of balls
n(S) = 4 + 3 + 5 = 12 balls

Let, R denotes red, B denotes blue and W denotes white balls. Since, the ball which is drawn
at first is replaced to draw another ball, it is an independent event.
Now, by using multiplication law of probability for independent events,

(i) P(W and B) = P(W ˆ B) = P(W) u P(B) = n(W) u n(B) = 5 u 3 = 5
n(S) n(S) 12 12 48

? The probability that the first is white and the second is blue is 5 .
48

(ii) P(B and R) = P(B ˆ R) = P(B) u P(R) = n(B) u n(R) = 3 u 4 = 1
n(S) n(S) 12 12 12

? The probability that the first is blue and the second is red is 1 .
12

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Probability

(iii) P(R and R) = P(R ˆ R) = P(R) u P(R)

= n(R) u nn((RS)) = 4 u 4 = 1 .
n(S) 12 12 9

? The probability that both of them are red is 1 .
9

(iv) P(R, R or B, B or W, W) = [P(R) u P(R)] + [P(B) u P(B)] + [P(W) u P(W)]

= n(R) u n(R) + n(B) u n(B) + n(W) u n(W)
n(S) n(S) n(S) n(S) n(S) n(S)

= 4 u 4 + 3 u 3 + 5 u 5
12 12 12 12 12 12

= 1 + 1 + 25 = 50 = 25 .
9 16 144 144 72

? The probability that both of them are of the same colour is 25 .
72

(v) Here, both of them are not blue means, they may be either RR or RW or WR or WW.
So, P(R, R or R, W or W, R or WW)
= [P(R) u p(R)] + [P(R) u P(W)] + [P(W) u P(R)] + [P(W) u P(W)]

= n(R) u n(R) + n(R) u n(W) + n(W) u n(R) + n(W) u n(W)
n(S) n(S) n(S) n(S) n(S) n(S) n(S) n(S)

= 4 u 4 + 4 u 5 + 5 u 4 + 5 u 5
12 12 12 12 12 12 12 12

= 1 + 5 + 5 + 25 = 9 .
9 36 36 144 16

? The probability that both of them are not blue is 9 .
16

Alternative process

Here, n(S) = 12

n(B) = 3

? P(B) = n(B) = 3 = 1
n(S) 12 4

Now, P ( B ˆ B ) = [1 – P(B)] [1 – P(B)]

= 1 – 1 1 – 1 = 3 × 3 = 9
4 4 4 4 16

So, the probability that both of them are not blue is 9 .
16

Example 4: If two unbiased coins are tossed simultaneously, find the probability of
getting (i) two heads (ii) at least one head.

Solution:

Here, the sample space, S = {HH, HT, TH, TT} Alternative process
n(S) = 4
Here, n(HH) = 1.
(i) Now, by using multiplication law of probability for
P(HH) = n(HH) =1
independent events, ? n(S) 4

P(H and H) = P(H) u P(H) = 1 u 1 = 1
2 2 4

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(ii) Again, P(HH or HT or TH) Alternative process

= P(H) u P(H) + P(H) u P(T) + P(T) u P(H) Here, n(at least one head) n(A) = 3

= 1 u 1 + 1 u 1 + 1 u 1 ? P(A) = n(A) = 3 .
2 2 2 2 2 2 n(S) 4

= 1 + 1 + 1 = 3 .
4 4 4 4

Example 5: Find the probability of getting 3 on the die and head on the coin when a die
is rolled and a coin is tossed simultaneously.

Solution:

Here, the sample space S1 for a die = {1, 2, 3, 4, 5, 6}. So, n(S1) = 6
The sample space S2 for a coin = {H, T}. So n(S2) = 2
Also, n(3) = 1 and n(H) = 1.
It is the case of independent events.
Now, by using the law of multiplication of probability for independent events,

P(3 and H) = P(3) u P(H) = n(3) u n(H) = 1 u 1 = 1
n(S1) n(S2) 6 2 12

So, the required probability is 1 .
12

Example 6: The probability of solving a mathematical problem by two students A and B

are 1 and 1 respectively. If the problem is given to the both students, find
3 4

the probability of solving.

Solution:

Here, the problem may be solved by both the students. So, it is the case of none-mutually
exclusive event.
? P (A or B) = P(A ‰ B) = P(A) + P(B) – P(A ˆ B)
Also, the problem solving event is independent.

? P (A and B) = P(A ˆ B)

= P(A) u P(B)

= 1 u 1 = 1
3 4 12

Now, P(A ‰ B)= P(A) + P(B) – P(A ˆ B)

= 1 + 1 – 1
3 4 12

= 4+3–1 = 6 = 1
12 12 2

So, the required probability is 1 .
2

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Probability

EXERCISE 18.2

General section
1. a) If A and B are two independent events, find the probability of (A ˆ B).

b) If X, Y and Z are three independent events, find P(X ˆ Y ˆ Z).

c) A ball is drawn from a box and it is replaced to draw another ball. Are the
probabilities of outcomes independent? Why?

2. a) A coin is tossed two times. Find the probability that

(i) both the outcomes are head, (ii) the first is head and the second is tail,

(iii) at least one tail, (iv) at most one head,

b) What is the probability of getting 3 on the die and head on the coin when a die is
rolled and a coin is tossed simultaneously?

c) If a card is drawn at random from a pack of 52 cards and at the same time a marble
is drawn at random from a bag containing 2 red marbles and 3 blue marbles. Find
the probability of getting a blue marble and a king.

d) A card is drawn randomly from a pack of 52 cards and a die is thrown once.
Determine the probability of not getting a card king as well as 6 on the die.

3. a) Two cards are drawn one after another with replacement from a well-shuffled pack
of 52 cards. Find the probability that both the cards are ace.

b) A card is drawn from a well-shuffled pack of playing cards. If another card is drawn
after replacing the first card, find.

(i) the probability that both of them are red queen,

(ii) the probability that both of them are not king,
(iii) the probability that one is jack and other is ace,
(iv) the probability that one is not a king and other is ace,
(v) the probability that both of them are of the same colour.

4. a) A bag contains 6 red balls and 4 white balls. A ball is drawn at random and it is
replaced to draw another ball.

(i) Find the probability that both of them are red.

(ii) Find the probability that both of them are white.
(iii) Find the probability that the first is red and the second is white.
(iv) Find the probability that none of them are red.
(v) Find the probability that none of them are white.

b) A box contains 5 black, 7 blue and 4 yellow balls. A ball is drawn at random and it
is replaced, then another ball is drawn. Find the probability that
(i) the first is blue and the second is black, (ii) both of them are yellow,
(iii) both of them are of the same colour,
(iv) the first is black and the second is yellow. (v) both of them are not black.

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c) A bag contains 6 black and 3 white balls. Another bag contains 5 black and 4 white
balls. If one ball is drawn from each bag, find the probability that these two balls
are of the same colour.

5. a) The probability that Bishwant gets scholarship is 0.9 and Sunayana will get
is 0.8. What is the probability that at least one of them gets the scholarship?

[Hint: P(B or S) = P(B) + P(S) – P(B ˆ S)]

b) The probability of solving a mathematical problem by two students A and B are 1
2
1
and 5 respectively. If the problem is given to both the students, find the probability

of solving the problems.

18.5 Probability tree diagram

The probabilities of all possible outcomes of a series of trials of random experiment
can be shown by a diagram called probability tree diagram. Every branch of a tree
diagram shows the probability of respective event in a particular trial.

From a tree diagram, the probability of a particular event of a random experiment
can be obtained as the product of the probabilities of the event in different trials
along the path of the event.
Let’s consider an example of tossing of a coin 3 times and calculate the probabilities
of the events.

Toss 1 Toss 2 Toss 3 1 1 1 1
2 2 2 8
) = 12 P(HHH) = u u =

H P(H 3

=12 T P(T3) = 1 P(HHT) = 1 u 1 u 1 = 1
2 2 2 2 8
)
P(H
2

H P(H3) = 1 P(HTH) = 1 u 1 u 1 = 1
2 2 2 2 8
T H
=12 P(T ) = 1
2 2 T

P(H ) P(T ) = 21 P(HTT) = 1 u 1 u 1 = 1
3 2 2 2 8
1

H ) = 12 P(THH) = 1 u 1 u 1 = 1
T 2 2 2 8
P(H 3
H
P(T ) =21 H P(H2) = 1 T P(T3) = 1 P(THT) = 1 u 1 u 1 = 1
1 2 2 2 2 2 8

T P(T 1 1 1 1 1
2 2 2 2 2 8
) =21 P(H3) = P(TTH) = u u =

H P(T
T 3

) = 21 P(TTT) = 1 u 1 u 1 = 1
2 2 2 8

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Probability

From the tree diagram = P(HHH) = 1 u 1 u 1 = 1
(i) The probability of getting 3 heads 2 2 2 8

(ii) The probability of getting 2 heads and 1 tail = P(HHT or HTH or THH)

= P(HHT) + P(HTH) + P(THH)

= 1 1 1 = 3
8 8 8 8

(iii) The probability of getting at least 2 tails = P(HTT or THT or TTH or TTT)

= P(HTT) + P(THT) + P(TTH) + P(TTT)

= 1 + 1 + 1 + 1 = 4 = 1
8 8 8 8 8 2

(iv) The probability of getting at most 1 head = P(HTT or THT or TTH or TTT)

= P(HTT) + P(THT) + P(TTH) + P(TTT)

= 1 + 1 + 1 + 1 = 4 = 1
8 8 8 8 8 2

(v) The probability of getting exactly 1 tail = P(HHT, HTH, THH)

= P(HHT) + P(HTH) + P(THH)

= 1 + 1 + 1 = 3
8 8 8 8

18.6 Multiplication law of probability of dependent events

Two or more events are said to be dependent if the occurrence of any one event

affects the occurrence of the other events. Suppose, a box contains 5 red and 3 white

balls. Let a ball be drawn at random and not replaced. 5 red
3 White
Here, in the first trial, P(R) = 5 and P(W) = 3
8 8

Let, the outcome in the first trial be red ball. 4 red
3 White
Then, in the second trial, P(R) = 4 and P(W) = 3
7 7

Let, the outcome in the first trial be white ball. 5 red
2 White

Then, in the second trial, P(R) = 5 and P(W) = 2
7 7

Now, let’s denote the occurrence of red ball by R1, or the occurrence of white ball by
W1 in the first trial; the occurrence of red ball by R2 or the occurrence of white ball by
W2 in the second trial.

Then, from the above illustrations, 4
7
P(R2/R1) [read as ‘probability of R2 given that R1 has occurred] =

P(W2/W1) [read as ‘probability of W2 given that W1 has occurred] = 2
7

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P(R2/W1) [read as ‘probability of R2 given that W1 has occurred] = 5
P(W2/R1) 7

[read as ‘probability of W2 given that R1 has occurred] = 3
7

Now, the multiplication law of probability in the case of dependent events can be

stated as following.

P(R1 and R2) = P(R1 ˆ R2) = 5 u 4 = P(R1) u P(R2/R1)
8 7

P(W1 and W2) = P(W1 ˆ W2) = 3 u 2 = P(W1) u P(W2/W1)
8 7

P(R1 and W2) = P(R1 ˆ W2) = 5 u 3 = P(R1) u P(W2/R1)
8 7

P(W1 and R2) = P(W1 ˆ R2) = 3 u 5 = P(W1) u P(R2/W1)
8 7

Study the following tree diagram to illustrate these probabilities.

P(R 2/R 1) = 4 RR o P(RR) = 5 u 4 = 5
7 8 7 14

P(R ) = 58 4 red P(W2/R1) = 3 RW o P(RW) = 5 u 3 = 15
3 white 7 8 7 56
1
5 red
5 red P(W 2 white P(R 2/W 1) = 5 WR o P(WR) = 3 u 3 = 15
3 white 1 7 8 7 56

) = 83

P(W /W ) = 2 WW o P(WW) = 3 u 3 = 3
2 1 7 8 7 28

Worked-out examples

Example 1: Two children were given birth by a couple at the interval of four years.
Illustrate the probabilities of son or daughter by drawing a tree diagram
and calculate the probability that both are girls.

Solution:
Let S be the event of son and D be the event of daughter.

Probability tree diagram:

P(S2) = 12 P(SS) = 1 u 1 = 1
2 2 4

1S 1 1 1
P(S 1) = 2 D 2 2 4
P(D ) = 1 P(SD) = u =
2 2

S

D P(S2) = 1 P(DS) = 1 1 = 1
2 2 2 4
P(D ) = 1 S u
1 2
D
P(DD) = 1 1 = 1
P(D ) = 1 2 u 2 4
2 2

From the tree diagram, 1 1 1
2 2 4
The probability of both are girls = P(DD) = P(D) u P(D) = u =

So, the probability of both are girls is 1 .
4

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Example 2: There were three black and five white balls in a box. Two balls were drawn
at random one by one. Find the probability that both balls are of the same
colour. (no replacement)

Solution:

Let B1 or W1 denote the occurrence of black or white ball in the first trial and let B2 or W2
denote the occurrence of black or white ball in the second trial.

Here, the number of possible outcomes = total number of balls = 3 + 5 = 8 balls.

Since the ball which is drawn at first is not replaced to draw another ball, it is a dependent
event.

Now, by using multiplication law of probability for a dependent event,

P(B1B2) = P(B1) u P(B2/B1) = 3 u 2 = 3
8 7 28

Also, P(W1,W2) = P(W1) u P(W2/W1) = 5 u 4 = 5
8 7 14

Again, the probability of both balls are same colour is

P(B1B2 or W1W2) = P(B1B2) + P(W1W2) = 3 + 5 = 3 + 10 = 13
28 14 28 28

So, the probability that both balls are of the same colour is 13 .
28

Example 3: There are 5 black balls and 4 white balls in a bag. Two balls are drawn
randomly one after the other. Show the probability in a tree diagram. Also
find the probability of (i) getting both white balls and (ii) getting none of
white balls.

Solution:

Let W1 and B1 denote the occurrence of white or black ball in the first trial and let W2 and B2
denote the occurrence of white or black ball in the second trial.

Here, the number of possible outcomes = total number of balls = 5 + 4 = 9 balls

Clearly, the balls are drawn one after the other indicates that there is no replacement. So, it
is a dependent event.

Probability tree diagram 4 P(B1 B2) = 5 × 4 = 5
8 9 8 18
P(B 2/B 1) =

5 4 black P(W /B1 ) = 4 P(B1 W2) = 5 × 4 = 5
P(B 1) = 9 4 white 2 8 9 8 18

5 black P(W 5 black P(B 2/W 1) = 5 P(W1 B2) = 4 × 5 = 5
4 white 1 3 white 8 9 8 18

) = 94

P(W /W ) = 3 P(W1W2) = 4 × 3 = 1
2 1 8 9 8 6

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Probability

(i) From the tree diagram,

the probability of getting two white balls = P(W1W2) = 1
6

(ii) Also, probability of getting non of white balls

P( W1 ˆ W2 ) = [1 – P(W1)] [1 – P (W2 / B1)]

= 1 – 4 1 – 4 = 5 × 4 = 5
9 8 9 8 18

Example 4: Two cards are drawn from a well-shuffled deck of 52 cards without
replacing the first card. By drawing a tree-diagram, find the probability
that both card are jacks.

Solution:

Let J1 or J1 denote the occurrence of jack or the card except jack in the first trial and J2 or J2
denotes the occurrence of jack or the card except jack in the second trial.

Here, the number of possible outcomes = 52.

Since the card which is drawn at first is not replaced, it is a dependent event.
Probability tree diagram:

P(J /J ) = 531 P(J1 J2) = 4 u 3 = 1
52 51 221
2 1

P(J1) = 542 3J P(J1, J1) = 4 u 48 = 16
48J1 P(J2/J1) = 5418 52 51 221
4J

48J1 P(J1 ) = 5428 4J P(J2/J1) = 4 P(J1 J2) = 48 4 = 16
51 P(J1 J2) 52 51 221
u
47J1

P(J2 /J1 ) = 5417 = 48 u 47 = 188
52 51 221

From the tree diagram, 1
221
The probability that both cards are jack = P(J1J2) = .

Example 5: There is one red, one blue and one white ball in a bag. A ball is drawn
Solution: randomly and not replaced, then another ball is drawn. Write the sample
space of the experiment using a tree diagram and find the probability of
getting at least one red ball.

Let R, B and W denote the red, blue and white ball.

Here, the number of possible outcomes = 1 + 1 + 1 = 3

When the first ball is drawn, it is not replaced to draw another ball. So, it is a dependent event.

Probability tree diagram

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P(B) = 12 P(RB) = 1 u 1 = 1
3 2 6

1B 1 P(RW) = 1 u 1 = 1
1W 2 3 2 6
P(R) =13 P(W) =
1R
1R P(B) = 1 1W P(R) = 1 P(BR) = 1 u 1 = 1
1B 3 2 3 2 6
1W 1R
1B P(W) = 1 P(BW) = 1 u 1 = 1
P(W) =31 2 3 2 6

P(R) = 1 P(WR) = 1 1 = 1
2 3 2 6
u

P(B) =21 P(WB) = 1 u 1 = 1
3 2 6

From the tree diagram, = P(RB or RW or BR or WR)
The sample space, S = {RB, RW, BR, BW, WR, WB}
Also, the probability of getting at least one red ball

= P(RB) + P(RW) + P(BR) + P(WR)

= 1 + 1 + 1 + 1 = 4 = 2
6 6 6 6 6 3

Alternative process

The sample space, n(S) = 6

The number of events where there is at least one red ball, n(A) = 4

? Probability of getting at least one red ball = n(A) = 4 = 2 .
n(S) 6 3

EXERCISE 18.3

General section

1. a) A coin is tossed two times in succession, write down the sample space by drawing
the tree diagram.

b) A fair coin is tossed three times. Draw a probability tree diagram to show all the
possible outcomes and determine,

(i) the probability of getting 3 heads
(ii) the probability of getting 1 head and 2 tails
(iii) the probability of getting at least 2 heads
(iv) the probability of getting at most 1 tail
(v) the probability of getting exactly 1 head

c) A bag contains 3 red and 5 white balls. A ball is drawn at random and replaced,
then another ball is drawn. Draw a probability tree diagram and find the probability
that both of them are not of the same colour.

d) A bag contains 6 black and 4 white marbles. A marble is drawn at random from
the bag and it is replaced to draw another marble. By using a tree diagram, find the
probability of getting

(i) both of them of the same colour. (ii) Both of them of the different colours.

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Probability

e) Two cards are drawn from a well-shuffled deck of 52 cards. Find the probability
that both cards are king (the first card is replaced) by showing a tree diagram.

f) Two children were born from a married couple. Find the probability of having at
least one son by drawing a tree diagram.

g) Two children were given birth by a couple at the interval of five years. Illustrate
the probabilities of son or daughter by drawing a tree diagram and calculate the
probability of both are son.

h) Two children were born in a family. By drawing a tree diagram, find the probability
of having no daughter.

i) Three children were born in a family. By drawing a tree diagram, find the probability
of having at least a girl.

j) Three children were born in a family. By drawing a tree diagram, find the following
probabilities.

(i) at least two daughters (ii) all of them are boys (iii) at least a boy.

2. In the following Wtre2 edednioagtersamblaBc1kaonrdwWh1itdeebnaolltedsrbawlanckinorthwehsietceobnadlltdriraalw. n in the first
trial and B2 and

P(B 2/B 1) = 1 A
2

5 B
9
P(B 1) = W P(W2/B1) = 1 B
2
B
5
W P(W B P(B 2/W 1) = 8 C
1
) = 4
9
W
P(W
2 /W ) = 3 D
1 8

(i) How many balls are there altogether? How many of them are black and white?

(ii) What do you mean by P(B2/B1), P (W2/B1), P (B2/W1) and P (W2/W1)?
(iii) Name the outcomes of the events A, B, C, and D and write the sample space of this

experiment.

(iv) Are the probabilities of the events dependent or independent?

(v) Find the probabilities of the events A, B, C, and D.

3. a) 8 black and 5 white pens of the same size are kept in a bag. Two pens are drawn
one after the other without replacement. Show the probabilities of all possible
outcomes in a tree diagram.

b) A bag contains 3 red and 4 blue balls of the same size. Two balls are drawn one after
another without replacement. Show all the probabilities in a tree diagram.

4. a) A bag contains 7 black and 5 white balls. Two balls are drawn randomly one by one
without replacing the first one.

(i) Find the probability that the first is black and the second is white.

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Probability

(ii) Find the probability that both of them are not black.
(iii) Find the probability that both of them are of the different colour.
(iv) Find the probability that both of them are of the same colour.
(v) Show all these probabilities in a tree diagram.
b) There are 3 red, 5 blue and 2 yellow identical pencils in a box. Two pencils are
drawn at random and without replacing the first one.

(i) Find the probability that both of them are of blue colour.
(ii) Find the probability that both of them are not red colour.
(iii) Find the probability that at least one is yellow.
(iv) Find the probability that none of them are blue and yellow.
(v) Find the probability that all of them are of the same colour.
(vi) Show all these probabilities in a tree diagram.

5. a) A card is drawn from a packet of numbered cards numbered from 1 to 20. If the
second card is also drawn without replacing the first one, find the probability of
each of the following events,

(i) Both of them are of the multiples of 5.
(ii) The first is of even number and the second prime number.
(iii) None of them are of multiples of 3.
(iv) Show these probabilities in a tree diagram.

b) From a well-shuffled pack of 52 cards two cards are drawn at random without
replacing the first one.

(i) Find the probability that both of them are king.

(ii) Find the probability that both of them are faced cards.
(iii) Find the probability that both of them are black none-faced cards.
(iv) Find the probability that the first is ace and the second is queen of heart.
c) From a packet of cards, three cards are drawn in succession at random without
replacement. Find the probability that the cards so drawn are no face cards.

Objective Questions

Tick the correct alternatives. (C) 0 ≤ P (E) ≤ 1 (D) 1 ≤ P (E) ≤ 0
1. The probability scale of any event E is

(A) 0 < P (E) < 1 (B) 1 < P (E) < 0

2. For any event E, P ( E ) is equal to (C) P (E) – 1
(A) - P (E) (B) 1 – P (E) (D) P (E)

3. Three unbiased coins are tossed simultaneously. What is the probability of getting at

most two heads? 3 7
1 8 8 1
(A) 8 (B) (C) (D) 2

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4. A box contains the lottery tickets numbered from 3 to 32. If a ticket is drawn at random,
what is the probability of the ticket bearing square or cube number?

(A) 1/5 (B) 1/10 (C) 1/15 (D) 2/15

5. What is the probability of a 53 Sundays in a leap year?

(A) 1 (B) 2 (C) 7 (D) 53
7 7 366 366
6. A dice is rolled once. What is the probability that the digit turned off is a prime number
or composite number?

(A) 1 (B) 1 (C) 1 (D) 5
3 2 6 6
7. An ace of diamond is lost from a deck of 52 playing cards and a card is drawn at random.

What is the probability of getting black faced card or ace?

(A) 10 (B) 3 (C) 4 (D) 16
51 17 13 51

8. If A and B are two mutually exclusive events then which of the following relations is

NOT correct?

(A) Probability of solving the problem = P (A) + P (B) – P (AˆB)

(B) Probability of solving the problem by both= P (A) × P (B)]

(C) Probability of solving the problem by at least one of them = 1 – [P ( A ) × P ( B )]

(d) Probability of solving the problem by at least one of them = 1 – [P (A) × P (B)]

9. Three children were born in the interval of five years. What is the probability of having
the children of alternate sexes?

(A) 0.125 (B) 0.25 (C) 0.5 (D) 0.375

10. A die is rolled twice. What is the probability that the digits turn off on the top face have
the sum 9 or 11?

(A) 1 (B) 1 (C) 1 (D) 1
36 18 9 6
11. Two events A and B are independent events such that P (A) = 0.4 and P (B) = 0.2, what

is the value of P ( AB )?

(A) 0.6 (B) 0.8 (C) 0.12 (D) 0.32

12. Two events A and B are mutually exclusive such that P (A) = 0.5 and P (B) = 0.4, what
is the value of P ( A ∪ B )?

(A) 1 (B) 0.9 (C) 0.8 (D) 0.1

13. The probability of solving a mathematical problem by two students A and B are 0.2 and
0.25 respectively, then the probability of solving the problem is

(A) 0.2 (B) 0.25 (C) 0.4 (D) 0.45

14. Three students A, B and C are to take part in a swimming competition. The probability
of A’s winning is twice the probability of B’s winning and thrice the probability of C’s
winning. If there is no tie then the probability of A’s winning is

(A) 2/11 (B) 3/11 (C) 6/11 (D) 10/11

15. The probability of winning a game by a mathematical problem by two students A and B
are 0.2 and 0.25 respectively, then the probability of solving the problem is

(A) 0.2 (B) 0.25 (C) 0.4 (D) 0.45

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