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# NCERT Solutions class 12 Maths Exercise 13.4

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## NCERT Solutions class 12 Maths Probability

1. State which of the following are not the probability distributions of a random variable. Give reasons for your answer.

(i)

 X 0 1 2 P (X) 0.4 0.4 0.2

(ii)

 X 0 1 2 3 4 P (X) 0.1 0.5 0.2 – 0.1 0.3

(iii)

 Y – 1 0 1 P (Y) 0.6 0.1 0.2

(iv)

 Z 3 2 1 0 – 1 P (Z) 0.3 0.2 0.4 0.1 0.05

Ans. (i) P (0) + P (1) + P (2) = 0.4 + 0.4 + 0.2 = 1

Therefore, it is a probability distribution.

(ii) P (3) = –0.1 which is not possible.

Therefore, it is not a probability distribution.

(iii) P (–1) + P (0) + P (1) = 0.6 + 0.1 + 0.2 = 0.9  1

Therefore, it is not a probability distribution.

(iv) P (3) + P (2) + P (1) + P (0) + P (–1) = 0.3 + 0.2 + 0.4 + 0.1 + 0.05 = 1.05  1

Therefore, it is not a probability distribution.

2. An urn contains 5 red and 2 black balls. Two balls are randomly selected. Let X represents the number of black balls. What are the possible values of X? Is X a random variable?

Ans. There two balls may be selected as BR, RB, BR, BB, where R represents red ball and B represents black ball.

Variable X has the value 0, 1, 2, i.e., there may be no black ball, may be one black ball or both the balls are black.

3. Let X represents the difference between the number of heads and the number of tails obtained when a coin is tossed 6 times. What are possible values of X?

Ans. For one coin, S = {H, T} = 2

Let A represents head  = 1

=  and

P (X = 0) =  =

P (X = 1) = 6 P (X)  =

P (X = 2) = 15 [P (X)]2  =

P (X = 3) = 20 [P (X)]3  =

P (X = 4) = 15 [P (X)]4  =

P (X = 5) = 6 [P (X)]5  =

P (X = 6) =  =

4. Find the probability distribution of:

(i) Number of heads in two tosses of a coin.

(ii) Number of tails in the simultaneous tosses of three coins.

(iii) Number of heads in four tosses of a coin.

Ans. (i) S = {T, H} = 2

Let A represents head  = 1

=  and

P (X = 0) =  =

P (X = 1) = 2 P (A)  =

P (X = 2) = P (A). P (A) =

Probability distribution

 0 1 2

(ii) Three coins tossed once = one coin tossed three times

S = {T, H} = 2

=  and

P (X = 0) =  =

P (X = 1) = 3 P (A)  =

P (X = 2) = 3 P (A). P (A).  =

P (X = 3) = P (A). P (A). P (A) =

Probability distribution

 0 1 2 3

(iii) S = {H, T} = 2

Let A represents head  = 1

=  and

P (X = 0) =  =

P (X = 1) = 4 P (A)  =

P (X = 2) = 6 P (A). P (A).  =

P (X = 3) = 4 P (A). P (A). P (A).  =

P (X = 4) = P (A). P (A). P (A). P (A) =

Probability distribution

 0 1 2 3 4

5. Find the probability distribution of the number of success in two tosses of a die where a success is defined as:

(i) number greater than 4.

(ii) six appears on at least one die.

Ans. S = {1, 2, 3, 4, 5, 6} = 6

(i) Let A be the set of favourable events.  = 1

=  and

P (X = 0) =  =

P (X = 1) = 2 P (A)  =

P (X = 2) = P (A). P (A) =

Probability distribution

 0 1 2

(ii) Let A represents that 6 appears on one dieA = {6}   = 1

=  and

P (X = 0) =  =

P (X = 1) = 2 P (A)  =

P (X = 2) = P (A). P (A) =

P (at least on six) =

Probability distribution

 0 1

6. From a lot of 30 bulbs which include 6 defectives, a sample of 4 bulbs is drawn at random with replacement. Find the probability distribution of the number of defective bulbs.

Ans.  = 30, A = {6 defective bulbs} = 6

and

=4 (4 bulbs are drawn with replacement),  = 0, 1, 2, 3, 4

P (X = 0) =

P (X = 1) =

P (X = 2) =

P (X = 3) =

P (X = 4) =

Probability distribution:

 0 1 2 3 4

7. A coin is biased so that the head is 3 times as likely to occur as tail. If the coin is tossed twice, find the probability distribution of number of tails.

Ans. Let  represents the appearance of tail and  represents the appearance of head.

Now

Since  and

P (X = 0) =

P (X = 1) =

P (X = 2) =

Probability distribution:

 0 1 2

NCERT Solutions class 12 Maths Exercise 13.4

8. A random variable X has the following probability distribution:

 X 0 1 2 3 4 5 6 7 P (X) 0

Determine:

(i)

(ii) P (X < 3)

(iii) P (X > 6)

(iv) P (0 < X < 3)

Ans. (i) Since, the sum of all the probabilities of a distribution is 1.

P (X = 0) + P (X = 1) + ……. + P (X = 7) = 1

or

or

Since, , therefore  is not possible.

(ii) P (X < 3) = P (X = 0) + P (X = 1) + P (X = 2)

=

=

(iii) P (X > 6) = P (X = 7)

=

(iv) P (0 < X < 3)

= P (X = 1) + P (X = 2)

=

NCERT Solutions class 12 Maths Exercise 13.4

9. The random variable X has a probability distribution  of the following form, where  is some number:

(a) Determine the value of

(b) Find

Ans. Probability distribution:

 0 1 2

(a) P (X = 0) + P (X = 1) + P (X = 2) = 1

(b) P (X < 2) = P (X = 0) + P (X = 1)

=

P (X  2) = P (X = 2) =  =

NCERT Solutions class 12 Maths Exercise 13.4

10. Find the mean number of heads in three tosses of fair coin.

Ans. S {H, T} =2

Let A denotes the appearance of head on a toss.

A =    =1

and

=3,  = 0, 1, 2, 3

P (X = 0) =

P (X = 1) =

P (X = 2) =

P (X = 3) =

Probability distribution:

 0 1 2 3

Mean =  =

NCERT Solutions class 12 Maths Exercise 13.4

11. Two dice are thrown simultaneously. If X denotes the number of sixes, find expectation of X.

Ans. Two dice thrown simultaneously is the same the die thrown 2 times.

Let S = {1, 2, 3, 4, 5, 6} = 6

Let A denotes the number 6 A = {6}

P (A) = and

= 2, = 0, 1, 2

P (X = 0) =

P (X = 1) = 2 P (A).

P (X = 2) = P (A). P (A) =

E (X) =

NCERT Solutions class 12 Maths Exercise 13.4

12. Two numbers are selected at random (without replacement), from the first six positive integers. Let X denotes the larger of two numbers obtained. Find E (X).

Ans. S = {(1, 2), (2, 1), (3, 1), (4, 1), (5, 1), (6, 1),

(1, 3), (2, 3), (3, 2), (4, 2), (5, 2), (6, 2),

(1, 4), (2, 4), (3, 4), (4, 3), (5, 3), (6, 3),

(1, 5), (2, 5), (3, 5), (4, 5), (5, 4), (6, 4),

(1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 6)}

= 30

Let X denotes the larger of the two numbers obtained.

 23456 246810 30

E (X) =  =

NCERT Solutions class 12 Maths Exercise 13.4

13. Let X denotes the sum of the numbers obtained when two fair dice are rolled. Find the variance and standard deviation of X.

Ans. S = {(1, 1), (2, 1), (3, 1), (4, 1), (5, 1), (6, 1),

(1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2),

(1, 3), (2, 3), (3, 3), (4, 3), (5, 3), (6, 3),

(1, 4), (2, 4), (3, 4), (4, 4), (5, 4), (6, 4),

(1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5),

(1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 6)}

= 36

Let A denotes the sum of the numbers = 2, B denotes the sum of the numbers = 3

C denotes the sum of the numbers = 4, D denotes the sum of the numbers = 5

E denotes the sum of the numbers = 6, F denotes the sum of the numbers = 7

G denotes the sum of the numbers = 8, H denotes the sum of the numbers = 9

I denotes the sum of the numbers = 10, J denotes the sum of the numbers = 11

K denotes the sum of the numbers = 12

A = {1, 1},  = 1, P (A) =

B = {(1, 2), (2, 1)},  = 2, P (A) =

C = {(1, 3), (2, 2), (3, 1)},  = 3, P (A) =

D = {(1, 4), (2, 3), (3, 2), (4, 1)},  = 4, P (A) =

E = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)},

= 5, P (A) =

F = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)},

= 6, P (A) =

G = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)},

= 5, P (A) =

H = {(3, 6), (4, 5), (5, 4), (6, 3)},

= 4, P (A) =

I = {(4, 6), (5, 5), (6, 4)},

= 3, P (A) =

J = {(5, 6), (6, 5)},

= 2, P (A) =

K = {6, 6},  = 1, P (A) =

 2345 6789 101112

Mean =

=

=

=

Now  =

=

=

Variance =  =  = 54.83 – 49 = 5.83

Standard deviation =  (nearly)

NCERT Solutions class 12 Maths Exercise 13.4

14. A class has 15 students whose ages are 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 and 20 years. One student is selected in such a manner that each has the same chance of being chosen and the age X of the selected student is recorded. What is the probability distribution of the random variable X? Find mean, variance and standard deviation of X.

Ans.  = 15, P (A) =

 1415161718192021 21231231 15

Mean =  = 17.5

Variance =  =

= 312.20 – 307.4 = 4.78

Standard deviation =

NCERT Solutions class 12 Maths Exercise 13.4

15. In a meeting 70% of the members favour a certain proposal, 30% being opposed. A member is selected at random and we let X = 0 if the opposed and X = 1 if he is in favour. Find E (X) and Var (X).

Ans.

 01 0 0

E (X) = Mean =  = 0.7

Variance (X) =

NCERT Solutions class 12 Maths Exercise 13.4

Choose the correct answer in each of the following:

16. The mean of the numbers obtained on throwing a die having written 1 on three faces, 2 on two faces and 5 on one face is:

(A) 1

(B) 2

(C) 5

(D)

Ans.

 125

Therefore, option (B) is correct.

NCERT Solutions class 12 Maths Exercise 13.4

17. Suppose that two cards are drawn at random from a deck of cards. Let X be the number of aces obtained. What is the value of E(X)?

(A)

(B)

(C)

(D)

Ans.  = 52,  = 4

P (X = 0) =

 012

P (X = 1) =

P (X = 2) =

Now E (X) =

Therefore, option (D) is correct.

## NCERT Solutions class 12 Maths Exercise 13.4

NCERT Solutions Class 12 Maths PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 12 Maths includes text book solutions from both part 1 and part 2. NCERT Solutions for CBSE Class 12 Maths have total 13 chapters. 12 Maths NCERT Solutions in PDF for free Download on our website. Ncert Maths class 12 solutions PDF and Maths ncert class 12 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide

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### 2 thoughts on “NCERT Solutions class 12 Maths Exercise 13.4”

1. Thanks

2. good solution,
It eplain properly