NCERT Solutions Class 11th Mathematics. FREE Flip-BOOK by Study Innovations. 614 Pages

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145-155 10

Answer : Number of boys fi Mid-point xi fi xi 25.3 227.7
The following table is formed. 15.3 198.9
9 100 900 5.3 137.8
Height in cms 13 110 1430 4.7 141
26 120 3120 14.7 176.4
95-105 30 130 3900 24.7 247
105-115 12 140 1680
115-125 10 150 1500
125-135
135-145
145-155

Here,

Q11 :

Calculate the mean deviation about median age for the age distribution of 100 persons given below:

Age Number

16-20 5

21-25 6

26-30 12

31-35 14

36-40 26

41-45 12

46-50 16

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51-55 9

Answer :

The given data is not continuous. Therefore, it has to be converted into continuous frequency distribution by
subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval.

The table is formed as follows.

Age Number fi Cumulative frequency (c.f.) Mid-point xi |xi – Med.| fi|xi – Med.|

15.5-20.5 5 5 18 20 100

20.5-25.5 6 11 23 15 90

25.5-30.5 12 23 28 10 120

30.5-35.5 14 37 33 5 70

35.5-40.5 26 63 38 0 0

40.5-45.5 12 75 43 5 60

45.5-50.5 16 91 48 10 160

50.5-55.5 9 100 53 15 135

100 735

The class interval containing the or 50th item is 35.5 – 40.5.

Therefore, 35.5 – 40.5 is the median class.

It is known that,

Here, l = 35.5, C = 37, f = 26, h = 5, and N = 100
Thus, mean deviation about the median is given by,

Exercise 15.2 : Solutions of Questions on Page Number : 371

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Q1 :
Find the mean and variance for the data 6, 7, 10, 12, 13, 4, 8, 12

Answer :
6, 7, 10, 12, 13, 4, 8, 12

Mean, –3 9
The following table is obtained. –2 4
–1 1
xi 9
3 16
6 4 25
7 –5 1
10 –1 9
12 3 74
13
4
8
12

Q2 :
Find the mean and variance for the first n natural numbers
Answer :
The mean of first n natural numbers is calculated as follows.

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Q3 :
Find the mean and variance for the first 10 multiples of 3

Answer :
The first 10 multiples of 3 are
3, 6, 9, 12, 15, 18, 21, 24, 27, 30
Here, number of observations, n = 10

The following table is obtained.

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www.ncrtsolutions.in 182.25
110.25
xi 56.25
20.25
3 –13.5
6 –10.5 2.25
9 –7.5 2.25
12 –4.5 20.25
15 –1.5 56.25
18 1.5 110.25
21 4.5 182.25
24 7.5 742.5
27 10.5
30 13.5

Q4 : 30
Find the mean and variance for the data 3

xi 6 10 14 18 24 28 338
f i 2 4 7 12 8 4 324
175
Answer :
The data is obtained in tabular form as follows.

xi f i fixi

62 12 –13 169
10 4 40 –9 81
14 7 98 –5 25

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18 12 216 –1 1 12
24 8 192 5 25 200
28 4 112 9 81 324
30 3 90 11 121 363
760 1736
40

Here, N = 40,

Q5 : 104 109
Find the mean and variance for the data 3 3

xi 92 93 97 98 102 192
fi 3 2 3 2 6 98
27
Answer : 8
The data is obtained in tabular form as follows. 24
48
xi f i fixi

92 3 276 –8 64
93 2 186 –7 49
97 3 291 –3 9
98 2 196 –2 4
102 6 612 4
104 3 312 2 16
4

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109 3 327 9 81 243
22 2200 640

Here, N = 22,

Q6 :
Find the mean and standard deviation using short-cut method.

xi 60 61 62 63 64 65 66 67 68
fi 2 1 12 29 25 12 10 4 5

Answer : –4 y f y2 fy2
The data is obtained in tabular form as follows. –3 i ii ii
–2
xi fi –1 16 –8 32
9 –3 9
60 2 0 4 –24 48
61 1 1 1 –29 29
62 12 2 00 0
63 29 3 1 12 12
64 25 4 4 20 40
65 12 220 9 12 36
66 10 16 20 80
67 4 286
68 5 0

100

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Mean,

Q7 :

Find the mean and variance for the following frequency distribution.

Classes 0-30 30-60 60-90 90-120 120-150 150-180 180-210
5 2
Frequencies 2 3 5 10 3

Answer : Frequency fi Mid-point xi y f y2 fy2
i ii ii
Class 2 15
3 45 –3 9 –6 18
0-30 5 75 –2 4 –6 12
30-60 10 105 –1 1 –5 5
60-90 3 135 00 0
90-120 5 165 0 13 3
120-150 2 195 1 4 10 20
150-180 30 2 9 6 18
180-210 3
2 76

Mean,

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Q8 :

Find the mean and variance for the following frequency distribution.

Classes 0-10 10-20 20-30 30-40 40-50
16 6
Frequencies 58 15

Answer : Frequency Mid-point xi –2 y f y2 fy2
fi –1 i ii ii
Class 5 5
8 15 0 4 –10 20
0-10 15 25 1 1 –8 8
10-20 16 35 2 00 0
20-30 6 45 1 16 16
30-40 50 4 12 24
40-50 68
10

Mean,

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Q9 :

Find the mean, variance and standard deviation using short-cut method

Height No. of children

in cms

70-75 3

75-80 4

80-85 7

85-90 7

90-95 15

95-100 9

100-105 6

105-110 6

110-115 3

Answer : Frequency fi Mid-point xi y f y2 fy2
i ii ii
Class Interval 3 72.5
4 77.5 –4 16 –12 48
70-75 7 82.5 –3 9 –12 36
75-80 –2 4 –14 28
80-85

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85-90 www.ncrtsolutions.in –1 1 –7 7
90-95
95-100 7 87.5 0 00 0
100-105 15 92.5
105-110 9 97.5 1 19 9
110-115 6 102.5
6 107.5 2 4 12 24
Mean, 3 112.5
60 3 9 18 54

4 16 12 48

6 254

Q10 : No. of children
The diameters of circles (in mm) drawn in a design are given below: 15
17
Diameters 21
33-36 22
37-40 25
41-44
45-48
49-52

Answer :

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Class Interval Frequency fi Mid-point xi f f y2 fy2
i ii ii
34.5
32.5-36.5 15 38.5 –2 4 –30 60
36.5-40.5 17 42.5 –1 1 –17 17
40.5-44.5 21 46.5
44.5-48.5 22 50.5 0 00 0
48.5-52.5 25 1
100 2 1 22 22

4 50 100

25 199

Here, N = 100, h = 4
Let the assumed mean, A, be 42.5.

Mean,

Exercise 15.3 : Solutions of Questions on Page Number : 375
Q1 :

From the data given below state which group is more variable, A or B?

Marks 10-20 20-30 30-40 40-50 50-60 60-70 70-80
40 10 9
Group A 9 17 32 33 43 15 7

Group B 10 20 30 25

Answer :

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Firstly, the standard deviation of group A is calculated as follows.

Marks Group A fi Mid-point xi y f y2 fy2
i ii ii
10-20 9 15 –3
25 –2 9 –27 81
20-30 17 35 –1 4 –34 68
45 1 –32 32
30-40 32 55 0 00 0
65 1 1 40 40
40-50 33 75 2 4 20 40
3 9 27 81
50-60 40 342
–6
60-70 10

70-80 9

150

Here, h = 10, N = 150, A = 45

The standard deviation of group B is calculated as follows.

Marks Group B Mid-point y f y2 fy2
i ii ii

fi xi

10-20 10 15 –3 9 –30 90
–2 4 –40 80
20-30 20 25 –1 1 –30 30
00 0
30-40 30 35 0

40-50 25 45

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50-60 43 www.ncrtsolutions.in 1 1 43 43
60-70 15 2 4 30 60
70-80 7 55 3 9 21 63
150 65 366
75 –6

Since the mean of both the groups is same, the group with greater standard deviation will be more variable.
Thus, group B has more variability in the marks.

Q2 :

From the data given below state which group is more variable, A or B?

Marks 10-20 20-30 30-40 40-50 50-60 60-70 70-80
40 10 9
Group A 9 17 32 33 43 15 7

Group B 10 20 30 25

Answer :

Firstly, the standard deviation of group A is calculated as follows.

Marks Group A Mid-point y f y2 fy2
i ii ii

fi xi

10-20 9 15 –3 9 –27 81
–2 4 –34 68
20-30 17 25 –1 1 –32 32

30-40 32 35

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40-50 33 www.ncrtsolutions.in 0 00 0
50-60 40 1 1 40 40
60-70 10 45 2 4 20 40
70-80 9 55 3 9 27 81
150 65 342
Σ 75 –6

Here, h = 10, N = 150, A = 45

The standard deviation of group B is calculated as follows.

Marks Group B Mid-point y f y2 fy2
i ii ii

fi xi

10-20 10 15 –3 9 –30 90
–2 4 –40 80
20-30 20 25 –1 1 –30 30
00 0
30-40 30 35 0 1 43 43
1 4 30 60
40-50 25 45 2 9 21 63
3 366
50-60 43 55 –6

60-70 15 65

70-80 7 75

Σ 150

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Since the mean of both the groups is same, the group with greater standard deviation will be more variable.
Thus, group B has more variability in the marks.

Q3 :
From the prices of shares X and Y below, find out which is more stable in value:

X 35 54 52 53 56 58 52 50 51 49
Y 108 107 105 105 106 107 104 103 104 101

Answer :
The prices of the shares X are
35, 54, 52, 53, 56, 58, 52, 50, 51, 49
Here, the number of observations, N = 10

The following table is obtained corresponding to shares X. 256
9
xi 1
4
35 –16 25
54 3
52 1
53 2
56 5

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58 7 49
52 1 1
50 –1 1
51 0 0
49 –2 4
350

The prices of share Y are
108, 107, 105, 105, 106, 107, 104, 103, 104, 101

The following table is obtained corresponding to shares Y.

yi

108 3 9
107 2 4
105 0 0
105 0 0
106 1 1
107 2 4
104 –1 1
103 –2 4
104 –1 1
101 –4 16
40

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C.V. of prices of shares X is greater than the C.V. of prices of shares Y.
Thus, the prices of shares Y are more stable than the prices of shares X.

Q4 :

An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives
the following results:

Firm A Firm B

No. of wage earners 586 648

Mean of monthly wages Rs 5253 Rs 5253

Variance of the distribution of wages 100 121

(i) Which firm A or B pays larger amount as monthly wages?
(ii) Which firm, A or B, shows greater variability in individual wages?

Answer :
(i) Monthly wages of firm A = Rs 5253
Number of wage earners in firm A = 586
∴Total amount paid = Rs 5253 x 586
Monthly wages of firm B = Rs 5253
Number of wage earners in firm B = 648
∴Total amount paid = Rs 5253 x 648
Thus, firm B pays the larger amount as monthly wages as the number of wage earners in firm B are more than the
number of wage earners in firm A.

(ii) Variance of the distribution of wages in firm A = 100

∴ Standard deviation of the distribution of wages in firm

A ((σ1) =

Variance of the distribution of wages in firm = 121

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∴ Standard deviation of the distribution of wages in firm

The mean of monthly wages of both the firms is same i.e., 5253. Therefore, the firm with greater standard deviation
will have more variability.

Thus, firm B has greater variability in the individual wages.

Q5 :

The following is the record of goals scored by team A in a football session:

No. of goals scored 01234

No. of matches 19753

For the team B, mean number of goals scored per match was 2 with a standard
deviation 1.25 goals. Find which team may be considered more consistent?

Answer :

The mean and the standard deviation of goals scored by team A are calculated as follows.

No. of goals scored No. of matches fx x2 fx2
ii i ii
01 00 0
91 9
19
14 4 28
27 15 9 45
12 16 48
35 50 130

43

25

Thus, the mean of both the teams is same.

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The standard deviation of team B is 1.25 goals.
The average number of goals scored by both the teams is same i.e., 2. Therefore, the team with lower standard
deviation will be more consistent.
Thus, team A is more consistent than team B.
Q6 :
The sum and sum of squares corresponding to length x (in cm) and weight y
(in gm) of 50 plant products are given below:
Which is more varying, the length or weight?
Answer :
Here, N = 50
∴ Mean,

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Mean,

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Thus, C.V. of weights is greater than the C.V. of lengths. Therefore, weights vary more than the lengths.
Exercise Miscellaneous : Solutions of Questions on Page Number : 380
Q1 :
The mean and variance of eight observations are 9 and 9.25, respectively. If six of the observations are 6, 7,
10, 12, 12 and 13, find the remaining two observations.
Answer :
Let the remaining two observations be x and y.
Therefore, the observations are 6, 7, 10, 12, 12, 13, x, y.

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From (1), we obtain
x2 + y2 + 2xy = 144 …(3)
From (2) and (3), we obtain
2xy = 64 … (4)
Subtracting (4) from (2), we obtain
x2 + y2 – 2xy = 80 – 64 = 16
⇒ x – y = ± 4 … (5)
Therefore, from (1) and (5), we obtain
x = 8 and y = 4, when x – y = 4
x = 4 and y = 8, when x – y = –4
Thus, the remaining observations are 4 and 8.

Q2 :
The mean and variance of 7 observations are 8 and 16, respectively. If five of the observations are 2, 4, 10, 12
and 14. Find the remaining two observations.
Answer :
Let the remaining two observations be x and y.

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The observations are 2, 4, 10, 12, 14, x, y.

From (1), we obtain
x2 + y2 + 2xy = 196 … (3)
From (2) and (3), we obtain
2xy = 196 – 100
⇒ 2xy = 96 … (4)
Subtracting (4) from (2), we obtain
x2 + y2 – 2xy = 100 – 96
⇒ (x – y)2 = 4
⇒ x – y = ± 2 … (5)
Therefore, from (1) and (5), we obtain
x = 8 and y = 6 when x – y = 2
x = 6 and y = 8 when x – y = – 2
Thus, the remaining observations are 6 and 8.

Q3 :
The mean and standard deviation of six observations are 8 and 4, respectively. If each observation is
multiplied by 3, find the new mean and new standard deviation of the resulting observations.

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Answer :
Let the observations be x1, x2, x3, x4, x5, and x6.
It is given that mean is 8 and standard deviation is 4.
If each observation is multiplied by 3 and the resulting observations are yi, then

From (1) and (2), it can be observed that,
Substituting the values of xi and in (2), we obtain

Therefore, variance of new observations =
Hence, the standard deviation of new observations is

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Q4 :

Given that is the mean and σ2 is the variance of n observations x1, x2 … xn. Prove that the mean and

variance of the observations ax1, ax2, ax3 …axn are and a2 σ2, respectively (a ≠ 0).

Answer :
The given n observations are x1, x2 … xn.

Mean =
Variance = σ2

If each observation is multiplied by a and the new observations are yi, then

Therefore, mean of the observations, ax1, ax2 … axn, is .
Substituting the values of xiand in (1), we obtain

Thus, the variance of the observations, ax1, ax2 … axn, is a2 σ2.

Q5 :

The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it
was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of
the following cases:
(i) If wrong item is omitted.
(ii) If it is replaced by 12.

Answer :
(i) Number of observations (n) = 20
Incorrect mean = 10

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Incorrect standard deviation = 2

That is, incorrect sum of observations = 200
Correct sum of observations = 200 – 8 = 192

∴ Correct mean

(ii) When 8 is replaced by 12,
Incorrect sum of observations = 200
∴ Correct sum of observations = 200 – 8 + 12 = 204

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Q6 :

The mean and standard deviation of marks obtained by 50 students of a class in three subjects,
Mathematics, Physics and Chemistry are given below:

Subject Mathematics Physics Chemistry

Mean 42 32 40.9

Standard deviation 12 15 20

Which of the three subjects shows the highest variability in marks and which shows the lowest?

Answer :
Standard deviation of Mathematics = 12
Standard deviation of Physics = 15

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Standard deviation of Chemistry = 20

The coefficient of variation (C.V.) is given by .

The subject with greater C.V. is more variable than others.
Therefore, the highest variability in marks is in Chemistry and the lowest variability in marks is in Mathematics.

Q7 :
The mean and standard deviation of a group of 100 observations were found to be 20 and 3, respectively.
Later on it was found that three observations were incorrect, which were recorded as 21, 21 and 18. Find the
mean and standard deviation if the incorrect observations are omitted.
Answer :
Number of observations (n) = 100
Incorrect mean
Incorrect standard deviation

∴ Incorrect sum of observations = 2000
⇒ Correct sum of observations = 2000 – 21 – 21 – 18 = 2000 – 60 = 1940

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NCERT Solutions for Class 11 Maths Chapter 16

Probability Class 11

Chapter 16 Probability Exercise 16.1, 16.2, 16.3, miscellaneous Solutions

Exercise 16.1 : Solutions of Questions on Page Number : 386
Q1 :
Describe the sample space for the indicated experiment: A coin is tossed three times.

Answer :
A coin has two faces: head (H) and tail (T).
When a coin is tossed three times, the total number of possible outcomes is 23 = 8
Thus, when a coin is tossed three times, the sample space is given by:
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

Q2 :
Describe the sample space for the indicated experiment: A die is thrown two times.

Answer :
When a die is thrown, the possible outcomes are 1, 2, 3, 4, 5, or 6.
When a die is thrown two times, the sample space is given by S = {(x, y): x, y = 1, 2, 3, 4, 5, 6}
The number of elements in this sample space is 6 × 6 = 36, while the sample space is given by:
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5),
(3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5),
(6, 6)}

Q3 :
Describe the sample space for the indicated experiment: A coin is tossed four times.

Answer :
When a coin is tossed once, there are two possible outcomes: head (H) and tail (T).
When a coin is tossed four times, the total number of possible outcomes is 24 = 16
Thus, when a coin is tossed four times, the sample space is given by:

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S = {HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH,
TTTT}

Q4 :
Describe the sample space for the indicated experiment: A coin is tossed and a die is thrown.

Answer :
A coin has two faces: head (H) and tail (T).
A die has six faces that are numbered from 1 to 6, with one number on each face.
Thus, when a coin is tossed and a die is thrown, the sample space is given by:
S = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}

Q5 :
Describe the sample space for the indicated experiment: A coin is tossed and then a die is rolled only in case
a head is shown on the coin.

Answer :
A coin has two faces: head (H) and tail (T).
A die has six faces that are numbered from 1 to 6, with one number on each face.
Thus, when a coin is tossed and then a die is rolled only in case a head is shown on the coin, the sample space is
given by:
S = {H1, H2, H3, H4, H5, H6, T}

Q6 :
2 boys and 2 girls are in Room X, and 1 boy and 3 girls in Room Y. Specify the sample space for the
experiment in which a room is selected and then a person.

Answer :
Let us denote 2 boys and 2 girls in room X as B1, B2 and G1, G2 respectively. Let us denote 1 boy and 3 girls in room
Y as B3, and G3, G4, G5 respectively.
Accordingly, the required sample space is given by S = {XB1, XB2, XG1, XG2, YB3, YG3, YG4, YG5}

Q7 :

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One die of red colour, one of white colour and one of blue colour are placed in a bag. One die is selected at
random and rolled, its colour and the number on its uppermost face is noted. Describe the sample space.

Answer :
A die has six faces that are numbered from 1 to 6, with one number on each face.
Let us denote the red, white, and blue dices as R, W, and B respectively.
Accordingly, when a die is selected and then rolled, the sample space is given by
S = {R1, R2, R3, R4, R5, R6, W1, W2, W3, W4, W5, W6, B1, B2, B3, B4, B5, B6}

Q8 :

An experiment consists of recording boy-girl composition of families with 2 children.
(i) What is the sample space if we are interested in knowing whether it is a boy or girl in the order of their
births?
(ii) What is the sample space if we are interested in the number of girls in the family?

Answer :
(i) When the order of the birth of a girl or a boy is considered, the sample space is given by S = {GG, GB, BG, BB}
(ii) Since the maximum number of children in each family is 2, a family can either have 2 girls or 1 girl or no girl.
Hence, the required sample space is S = {0, 1, 2}

Q9 :

A box contains 1 red and 3 identical white balls. Two balls are drawn at random in succession without
replacement. Write the sample space for this experiment.

Answer :
It is given that the box contains 1 red ball and 3 identical white balls. Let us denote the red ball with R and a white ball
with W.
When two balls are drawn at random in succession without replacement, the sample space is given by
S = {RW, WR, WW}

Q10 :

An experiment consists of tossing a coin and then throwing it second time if a head occurs. If a tail occurs
on the first toss, then a die is rolled once. Find the sample space.

Answer :

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A coin has two faces: head (H) and tail (T).
A die has six faces that are numbered from 1 to 6, with one number on each face.
Thus, in the given experiment, the sample space is given by
S = {HH, HT, T1, T2, T3, T4, T5, T6}

Q11 :
Suppose 3 bulbs are selected at random from a lot. Each bulb is tested and classified as defective (D) or non-
defective (N). Write the sample space of this experiment?

Answer :
3 bulbs are to be selected at random from the lot. Each bulb in the lot is tested and classified as defective (D) or non-
defective (N).
The sample space of this experiment is given by
S = {DDD, DDN, DND, DNN, NDD, NDN, NND, NNN}

Q12 :
A coin is tossed. If the out come is a head, a die is thrown. If the die shows up an even number, the die is
thrown again. What is the sample space for the experiment?

Answer :
When a coin is tossed, the possible outcomes are head (H) and tail (T).
When a die is thrown, the possible outcomes are 1, 2, 3, 4, 5, or 6.
Thus, the sample space of this experiment is given by:
S = {T, H1, H3, H5, H21, H22, H23, H24, H25, H26, H41, H42, H43, H44, H45, H46, H61, H62, H63, H64, H65, H66}

Q13 :
The numbers 1, 2, 3 and 4 are written separately on four slips of paper. The slips are put in a box and mixed
thoroughly. A person draws two slips from the box, one after the other, without replacement. Describe the
sample space for the experiment.

Answer :
If 1 appears on the first drawn slip, then the possibilities that the number appears on the second drawn slip are 2, 3,
or 4. Similarly, if 2 appears on the first drawn slip, then the possibilities that the number appears on the second drawn
slip are 1, 3, or 4. The same holds true for the remaining numbers too.

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Thus, the sample space of this experiment is given by S = {(1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (2, 4), (3, 1), (3, 2), (3, 4),
(4, 1), (4, 2), (4, 3)}

Q14 :
An experiment consists of rolling a die and then tossing a coin once if the number on the die is even. If the
number on the die is odd, the coin is tossed twice. Write the sample space for this experiment.

Answer :
A die has six faces that are numbered from 1 to 6, with one number on each face. Among these numbers, 2, 4, and 6
are even numbers, while 1, 3, and 5 are odd numbers.
A coin has two faces: head (H) and tail (T).
Hence, the sample space of this experiment is given by:
S = {2H, 2T, 4H, 4T, 6H, 6T, 1HH, 1HT, 1TH, 1TT, 3HH, 3HT, 3TH, 3TT, 5HH, 5HT, 5TH, 5TT}

Q15 :
A coin is tossed. If it shows a tail, we draw a ball from a box which contains 2 red and 3 black balls. If it
shows head, we throw a die. Find the sample space for this experiment.

Answer :
The box contains 2 red balls and 3 black balls. Let us denote the 2 red balls as R1, R2 and the 3 black balls as B1, B2,
and B3.
The sample space of this experiment is given by
S = {TR1, TR2, TB1, TB2, TB3, H1, H2, H3, H4, H5, H6}

Q16 :
A die is thrown repeatedly until a six comes up. What is the sample space for this experiment?

Answer :
In this experiment, six may come up on the first throw, the second throw, the third throw and so on till six is obtained.
Hence, the sample space of this experiment is given by
S = {6, (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (1, 1, 6), (1, 2, 6), … , (1, 5, 6), (2, 1, 6), (2, 2, 6), … , (2, 5, 6), … ,(5, 1, 6),
(5, 2, 6), …}

Exercise 16.2 : Solutions of Questions on Page Number : 393

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Q1 :
A die is rolled. Let E be the event “die shows 4†and F be the event “die shows even numberâ€. Are E
and F mutually exclusive?

Answer :
When a die is rolled, the sample space is given by
S = {1, 2, 3, 4, 5, 6}
Accordingly, E = {4} and F = {2, 4, 6}
It is observed that E ∩ F = {4} ≠Φ
Therefore, E and F are not mutually exclusive events.

Q2 :
A die is thrown. Describe the following events:
(i) A: a number less than 7 (ii) B: a number greater than 7 (iii) C: a multiple of 3
(iv) D: a number less than 4 (v) E: an even number greater than 4 (vi) F: a number not less than 3
Also find

Answer :
When a die is thrown, the sample space is given by S = {1, 2, 3, 4, 5, 6}.
Accordingly:
(i) A = {1, 2, 3, 4, 5, 6}
(ii) B = Φ
(iii) C = {3, 6}
(iv) D = {1, 2, 3}
(v) E = {6}
(vi) F = {3, 4, 5, 6}

A ∪ B = {1, 2, 3, 4, 5, 6}, A ∩ B = Φ
B ∪ C = {3, 6}, E ∩ F = {6}
D ∩ E =Φ, A – C = {1, 2, 4, 5}
D – E = {1, 2, 3},

Q3 :

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An experiment involves rolling a pair of dice and recording the numbers that come up. Describe the following
events:
A: the sum is greater than 8, B: 2 occurs on either die
C: The sum is at least 7 and a multiple of 3.
Which pairs of these events are mutually exclusive?
Answer :
When a pair of dice is rolled, the sample space is given by

It is observed that
A ∩ B =Φ
B ∩ C =Φ

Hence, events A and B and events B and C are mutually exclusive.

Q4 :
Three coins are tossed once. Let A denote the event ‘three heads showâ€, B denote the event “two heads
and one tail showâ€. C denote the event “three tails show†and D denote the event ‘a head shows on the
first coinâ€. Which events are
(i) mutually exclusive? (ii) simple? (iii) compound?
Answer :
When three coins are tossed, the sample space is given by

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S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
Accordingly,
A = {HHH}
B = {HHT, HTH, THH}
C = {TTT}
D = {HHH, HHT, HTH, HTT}
We now observe that
A ∩ B =Φ, A ∩ C =Φ, A ∩ D = {HHH} ≠Φ
B ∩ C =Φ, B ∩ D = {HHT, {HTH} ≠Φ
C ∩ D = Φ
(i) Event A and B; event A and C; event B and C; and event C and D are all mutually exclusive.
(ii) If an event has only one sample point of a sample space, it is called a simple event. Thus, A and C are simple
events.
(iii) If an event has more than one sample point of a sample space, it is called a compound event. Thus, B and D are
compound events.

Q5 :
Three coins are tossed. Describe
(i) Two events which are mutually exclusive.
(ii) Three events which are mutually exclusive and exhaustive.
(iii) Two events, which are not mutually exclusive.
(iv) Two events which are mutually exclusive but not exhaustive.
(v) Three events which are mutually exclusive but not exhaustive.

Answer :
When three coins are tossed, the sample space is given by
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
(i) Two events that are mutually exclusive can be
A: getting no heads and B: getting no tails
This is because sets A = {TTT} and B = {HHH} are disjoint.
(ii) Three events that are mutually exclusive and exhaustive can be
A: getting no heads
B: getting exactly one head
C: getting at least two heads

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i.e.,
A = {TTT}
B = {HTT, THT, TTH}
C = {HHH, HHT, HTH, THH}
This is because A ∩ B = B ∩ C = C ∩ A = Φand A ∪ B ∪ C = S
(iii) Two events that are not mutually exclusive can be
A: getting three heads
B: getting at least 2 heads
i.e.,
A = {HHH}
B = {HHH, HHT, HTH, THH}
This is because A ∩ B = {HHH} ≠Φ
(iv) Two events which are mutually exclusive but not exhaustive can be
A: getting exactly one head
B: getting exactly one tail
That is
A = {HTT, THT, TTH}
B = {HHT, HTH, THH}
It is because, A ∩ B =Φ, but A ∪ B ≠S
(v) Three events that are mutually exclusive but not exhaustive can be
A: getting exactly three heads
B: getting one head and two tails
C: getting one tail and two heads
i.e.,
A = {HHH}
B = {HTT, THT, TTH}
C = {HHT, HTH, THH}
This is because A ∩ B = B ∩ C = C ∩ A = Φ, but A ∪ B ∪ C ≠S

Q6 :
Two dice are thrown. The events A, B and C are as follows:
A: getting an even number on the first die.
B: getting an odd number on the first die.

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C: getting the sum of the numbers on the dice ≤ 5
Describe the events
(i) (ii) not B (iii) A or B
(iv) A and B (v) A but not C (vi) B or C
(vii) B and C (viii)
Answer :
When two dice are thrown, the sample space is given by

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(vii) B and C = B ∩ C

Q7 :
Two dice are thrown. The events A, B and C are as follows:
A: getting an even number on the first die.
B: getting an odd number on the first die.
C: getting the sum of the numbers on the dice ≤ 5
State true or false: (give reason for your answer)

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(i) A and B are mutually exclusive
(ii) A and B are mutually exclusive and exhaustive
(iii)
(iv) A and C are mutually exclusive
(v) A and are mutually exclusive
(vi) are mutually exclusive and exhaustive.

Answer :

(i) It is observed that A ∩ B = Φ
∴ A and B are mutually exclusive.
Thus, the given statement is true.
(ii) It is observed that A ∩ B = Φ and A ∪ B = S
∴ A and B are mutually exclusive and exhaustive.
Thus, the given statement is true.
(iii) It is observed that

Thus, the given statement is true.
(iv) It is observed that A ∩ C = {(2, 1), (2, 2), (2, 3), (4, 1)} ≠Φ
∴A and C are not mutually exclusive.
Thus, the given statement is false.
(v)

∴A and are not mutually exclusive.
Thus, the given statement is false.

(vi) It is observed that ;

However,

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Therefore, events are not mutually exclusive and exhaustive.

Thus, the given statement is false.

Exercise 16.3 : Solutions of Questions on Page Number : 403
Q1 :

Which of the following can not be valid assignment of probabilities for outcomes of sample space S

= ω1 ω2 ω3 ω4 ω5 ω6 ω7
0.1 0.01 0.05 0.03 0.01 0.2 0.6
Assignment
(a)

(b)

(c) 0.1 0.2 0.3 0.4 0.5 0.6 0.7

(d) –0.1 0.2 0.3 0.4 –0.2 0.1 0.3

(e)

Answer :
(a)

ω1 ω2 ω3 ω4 ω5 ω6 ω7
0.03 0.01 0.2 0.6
0.1 0.01 0.05

Here, each of the numbers p(ωi) is positive and less than 1.
Sum of probabilities

Thus, the assignment is valid.
(b)

ω1 ω2 ω3 ω4 ω5 ω6 ω7

Here, each of the numbers p(ωi) is positive and less than 1.

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Sum of probabilities

Thus, the assignment is valid.
(c)

ω1 ω2 ω3 ω4 ω5 ω6 ω7
0.1 0.2 0.3 0.4 0.5 0.6 0.7

Here, each of the numbers p(ωi) is positive and less than 1.
Sum of probabilities

Thus, the assignment is not valid. ω2 ω3 ω4 ω5 ω6 ω7
(d) 0.2 0.3 0.4 –0.2 0.1 0.3

ω1 ω3 ω4
–0.1

Here, p(ω1) and p(ω5) are negative.

Hence, the assignment is not valid.

(e) ω2 ω5 ω6 ω7

ω1

Here,
Hence, the assignment is not valid.

Q2 :
A coin is tossed twice, what is the probability that at least one tail occurs?

Answer :
When a coin is tossed twice, the sample space is given by

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S = {HH, HT, TH, TT}
Let A be the event of the occurrence of at least one tail.
Accordingly, A = {HT, TH, TT}

Q3 :
A die is thrown, find the probability of following events:
(i) A prime number will appear,
(ii) A number greater than or equal to 3 will appear,
(iii) A number less than or equal to one will appear,
(iv) A number more than 6 will appear,
(v) A number less than 6 will appear.
Answer :
The sample space of the given experiment is given by
S = {1, 2, 3, 4, 5, 6}
(i) Let A be the event of the occurrence of a prime number.
Accordingly, A = {2, 3, 5}

(ii) Let B be the event of the occurrence of a number greater than or equal to 3. Accordingly, B = {3, 4, 5, 6}

(iii) Let C be the event of the occurrence of a number less than or equal to one. Accordingly, C = {1}

(iv) Let D be the event of the occurrence of a number greater than 6.
Accordingly, D = Φ

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(v) Let E be the event of the occurrence of a number less than 6.
Accordingly, E = {1, 2, 3, 4, 5}

Q4 :
A card is selected from a pack of 52 cards.
(a) How many points are there in the sample space?
(b) Calculate the probability that the card is an ace of spades.
(c) Calculate the probability that the card is (i) an ace (ii) black card.
Answer :
(a) When a card is selected from a pack of 52 cards, the number of possible outcomes is 52 i.e., the sample space
contains 52 elements.
Therefore, there are 52 points in the sample space.
(b) Let A be the event in which the card drawn is an ace of spades.
Accordingly, n(A) = 1

(c) (i)Let E be the event in which the card drawn is an ace.
Since there are 4 aces in a pack of 52 cards, n(E) = 4

(ii)Let F be the event in which the card drawn is black.
Since there are 26 black cards in a pack of 52 cards, n(F) = 26

Q5 :

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A fair coin with 1 marked on one face and 6 on the other and a fair die are both tossed. Find the probability
that the sum of numbers that turn up is (i) 3 (ii) 12
Answer :
Since the fair coin has 1 marked on one face and 6 on the other, and the die has six faces that are numbered 1, 2, 3,
4, 5, and 6, the sample space is given by
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
Accordingly, n(S) = 12
(i) Let A be the event in which the sum of numbers that turn up is 3.
Accordingly, A = {(1, 2)}

(ii) Let B be the event in which the sum of numbers that turn up is 12.
Accordingly, B = {(6, 6)}

Q6 :
There are four men and six women on the city council. If one council member is selected for a committee at
random, how likely is it that it is a woman?
Answer :
There are four men and six women on the city council.
As one council member is to be selected for a committee at random, the sample space contains 10 (4 + 6) elements.
Let A be the event in which the selected council member is a woman.
Accordingly, n(A) = 6

Q7 :
A fair coin is tossed four times, and a person win Re 1 for each head and lose Rs 1.50 for each tail that turns
up. From the sample space calculate how many different amounts of money you can have after four tosses
and the probability of having each of these amounts.

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Answer :
Since the coin is tossed four times, there can be a maximum of 4 heads or tails.

When 4 heads turns up, is the gain.

When 3 heads and 1 tail turn up, Re 1 + Re 1 + Re 1 – Rs 1.50 = Rs 3 – Rs 1.50 = Rs 1.50 is the gain.

When 2 heads and 2 tails turns up, Re 1 + Re 1 – Rs 1.50 – Rs 1.50 = – Re 1, i.e., Re 1 is the loss.

When 1 head and 3 tails turn up, Re 1 – Rs 1.50 – Rs 1.50 – Rs 1.50 = – Rs 3.50, i.e., Rs 3.50 is the loss.
When 4 tails turn up, – Rs 1.50 – Rs 1.50 – Rs 1.50 – Rs 1.50 = – Rs 6.00, i.e., Rs 6.00 is the loss.

There are 24 = 16 elements in the sample space S, which is given by:

S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTTH, TTHH, HTHT, THTH, THHT, HTTT, THTT, TTHT, TTTH,
TTTT}

∴n(S) = 16

The person wins Rs 4.00 when 4 heads turn up, i.e., when the event {HHHH} occurs.

∴Probability (of winning Rs 4.00) =

The person wins Rs 1.50 when 3 heads and one tail turn up, i.e., when the event {HHHT, HHTH, HTHH, THHH}
occurs.

∴Probability (of winning Rs 1.50) =

The person loses Re 1.00 when 2 heads and 2 tails turn up, i.e., when the event {HHTT, HTTH, TTHH, HTHT, THTH,
THHT} occurs.

∴Probability (of losing Re 1.00)
The person loses Rs 3.50 when 1 head and 3 tails turn up, i.e., when the event {HTTT, THTT, TTHT, TTTH} occurs.

Probability (of losing Rs 3.50) =
The person loses Rs 6.00 when 4 tails turn up, i.e., when the event {TTTT} occurs.

Probability (of losing Rs 6.00) =

Q8 :
Three coins are tossed once. Find the probability of getting
(i) 3 heads (ii) 2 heads (iii) at least 2 heads
(iv) at most 2 heads (v) no head (vi) 3 tails

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(vii) exactly two tails (viii) no tail (ix) at most two tails.
Answer :
When three coins are tossed once, the sample space is given by
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
∴Accordingly, n(S) = 8
It is known that the probability of an event A is given by

(i) Let B be the event of the occurrence of 3 heads. Accordingly, B = {HHH}

∴P(B) =
(ii) Let C be the event of the occurrence of 2 heads. Accordingly, C = {HHT, HTH, THH}

∴P(C) =
(iii) Let D be the event of the occurrence of at least 2 heads.
Accordingly, D = {HHH, HHT, HTH, THH}

∴P(D) =
(iv) Let E be the event of the occurrence of at most 2 heads.
Accordingly, E = {HHT, HTH, THH, HTT, THT, TTH, TTT}

∴P(E) =
(v) Let F be the event of the occurrence of no head.
Accordingly, F = {TTT}

∴P(F) =
(vi) Let G be the event of the occurrence of 3 tails.
Accordingly, G = {TTT}

∴P(G) =
(vii) Let H be the event of the occurrence of exactly 2 tails.

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Accordingly, H = {HTT, THT, TTH}

∴P(H) =
(viii) Let I be the event of the occurrence of no tail.
Accordingly, I = {HHH}

∴P(I) =
(ix) Let J be the event of the occurrence of at most 2 tails.
Accordingly, I = {HHH, HHT, HTH, THH, HTT, THT, TTH}

∴P(J) =

Q9 :
If is the probability of an event, what is the probability of the event ‘not A’.
Answer :
It is given that P(A) = .
Accordingly, P(not A) = 1 – P(A)

Q10 :
A letter is chosen at random from the word ‘ASSASSINATION’. Find the probability that letter is (i) a vowel (ii)
an consonant

Answer :
There are 13 letters in the word ASSASSINATION.
∴Hence, n(S) = 13
(i) There are 6 vowels in the given word.

∴Probability (vowel) =

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