# NCERT Solutions class 12 Maths Exercise 13.4 (Ex 13.4) Chapter 13 Probability

## NCERT Solutions for Class 12 Maths Exercise 13.4 Chapter 13 Probability – FREE PDF Download

Free PDF download of NCERT Solutions for Class 12 Maths Chapter 13 Exercise 13.4 (Ex 13.4) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 12 Maths Chapter 13 Probability Exercise 13.4 Questions with Solutions to help you to revise complete Syllabus and Score More marks.

# NCERT Solutions for Class 12 Maths Chapter 13 Probability (Ex 13.4) Exercise 13.4

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 of a random variable.

(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.

Since, X is a number whose values are defined on the outcomes of a random experiment, therefore, X is a random variable.

### 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. Let hh denotes the number of heads and tt denotes the number of tails, when a coin is tossed 6 times. Then,

X = difference between hh and tt = |hth−t|

Now,                    hh : 0          1          2          3          4          5          6

Therefore,           tt  : 6          5          4          3          2          1          0

And, hence         X : 6          4          2          0          2          4          6

Therefore, the possible values of X are 6, 4, 2, 0.

### 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) The sample space of the random experiment ‘a coin is tossed twice’ is S = {HH, HT, TH, TT} = 4

Let X denotes the random variable ‘number of heads’, then A can take the values 0, 1 or 2.

P (X = 0) =  =

P (X = 1) = 2 P (A)  = 1212

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

Probability distribution

 0 1 2 1212

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

S = {(HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} = 8

Let X denotes the random variable ‘number of heads’, then A can take the values 0, 1, 2 or 3.

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) A coin is tossed four times = S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, HTTH, THHT, THTH, TTHH, HTTT, THTT, TTHT, TTTH, TTTT} = 16

Let X denotes the random variable ‘number of heads’, then A can take the values 0, 1, 2, 3 or 4.

P (X = 0) =  =

P (X = 1) = 4 P (A)  =  = 1414

P (X = 2) = 6 P (A). P (A).  =  = 3838

P (X = 3) = 4 P (A). P (A). P (A).  =  = 1414

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

Probability distribution

 0 1 2 3 4 1414 3838 1414

### 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

Let s denotes success and f denotes the failure. Then

P (s) = P (6 appears on at least one die)

= P (6 appears on one die or 6 appears on both dice)

= P (6 appears on first dice and does not appear on second dice) + P (6 does not appear on first dice and 6 appears on second dice) + P (6 appears on both dice)

=  16×56+56×16+16×16=113616×56+56×16+16×16=1136

Let X denotes the numebr of success in two tosses of a dice, then X can take value 0 or 1.

P (X = 0) = P (no success) = P(f) =  =

P (X = 1) = P (one success) = P(s) = 11361136

Probability distribution

 X 0 1 P(X)

### 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

### 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)

### 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) =  =

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

Ans. =8

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 =  =

### 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 x 6 = 36

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) =

### 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.

 2 3 4 5 6 2 4 6 8 10 30

E (X) =  =

### 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) =

 2 3 4 5 6 7 8 9 10 11 12

Mean =

Now  =

Variance =  =  = 54.83 – 49 = 5.83

Standard deviation =  (nearly)

### 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) =

 14 15 16 17 18  19 20 21 2  1 2 3 1  2 3 1 15

Mean =  = 17.53

Variance =  =

= 312.20 – 307.42 = 4.78

Standard deviation =

### 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.

 0 1 0 0

E (X) = Mean =  = 0.7
Variance (X) =

### 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.

 1 2 5

Therefore, option (B) is correct.

### 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) =

 0 1 2

P (X = 1) =
P (X = 2) =
Now E (X) =
Therefore, option (D) is correct.