NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4

These NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 12 Maths Chapter 13 Probability Exercise 13.4

NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4

Question 1.
State which of the following are not the probability distributions of a random variable. Give reasons for your answer.
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4 1
Solution:
P (0) + P (1) + P (2) = 0.4 + 0.4 + 0.2 = 1
It is a probability distribution.

(ii) P (3) = – 0.1 which is not possible.
Thus it is not a probability distribution.

(iii) P(-1)+P(0)+P(1) = 0.6 + 0.1 + 0.2 = 0.9 ≠ 1
Thus 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
Hence it is not a probability distribution.

Question 2.
An urn contains 5 red and 2 black balls. Two balls are randomly drawn. Let X represent the number of black balls. What are the possible values of X ?. Is X a random variable?
Solution:
These two balls may be selected as RR, RB, BR, BB, where R represents red and B represents black ball, variable X has the value 0,1,2, i.e., there may be no black balls, may be one black ball, or both the balls are.black. Yes , X is a random variable.

Question 3.
Let X represent 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?
Solution:
Let ‘w’ denote the number of heads and ‘n’ the number of tails when a coin is tossesd six times
X is the difference between m and n
∴ X = |m – n|
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4 2
∴ The possible values of X are 0, 2, 4, 6.

NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4

Question 4.
Find the probability distribution of
(a) number of heads in two tosses of a coin.
(b) number of tails in the simultaneous tosses of three coins.
(c) number of heads in four tosses of a coin.
Solution:
i. When a coin is tossed twice, the sample
space S = {HH, HT, TH, TT}
Let X denote the number of heads.
Then X takes the values 0, 1, 2.
P(X = 0) = P(two tails) = P{TT} = \(\frac { 1 }{ 4 }\)
P(X = 1) = P(one head & one tail)
= P{HT,TH} = \(\frac { 2 }{ 4 }\) = \(\frac { 1 }{ 2 }\)
P(X = 2) = P(two heads) = P{HH} = \(\frac { 1 }{ 4 }\)
∴ The probability distribution of X is
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4 3

ii. When a coin is tossed 3 times, the sample space S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
Let X denote the number of tails.
Then X take values 0, 1, 2, 3.
P(X = 0) = P(no tail) = P{HHH} = \(\frac { 1 }{ 8 }\)
P(X = 1) = P(one tail & two heads)
= P{HHT, THH, HTH} = \(\frac { 3 }{ 8 }\)
P(X = 2) = P(two tails & one head)
= P{HTT, THT, TTH} = \(\frac { 3 }{ 8 }\)
P(X = 3) = P(three tails) = P(TTT) = \(\frac { 1 }{ 8 }\)
The probability distribution of X is
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4 4

iii. When four coins are tossed, sample space, S = {HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT}
Let X denote the number of heads in the four tosses of a coin.
Then X can take the values 0, 1,2, 3 and 4
P(H) = \(\frac { 1 }{ 2 }\) ,P(T) = \(\frac { 1 }{ 2 }\)
P(X = 0) = P(TTTT) = \(\frac { 1 }{ 2 }\) x \(\frac { 1 }{ 2 }\) x \(\frac { 1 }{ 2 }\) x \(\frac { 1 }{ 2 }\) = \(\frac { 1 }{ 16 }\)
P(X = 1) = P{HTTT, THTT, TTHT, TTTH}
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4 5

Question 5.
Find the probability distribution of the number of successes in two tosses of a die, where a success is defined as
i. number greater than 4
ii. six appears on atleast one die
Solution:
i. When a die is tossed the sample space S = {1, 2, 3, 4, 5, 6}.
Let A denote the success
A: getting a number greater than 4
A = {5, 6}
P(A) = \(\frac { 2 }{ 6 }\) = \(\frac { 1 }{ 3 }\)
P(A’) = 1 – P(A)= 1 – \(\frac { 1 }{ 3 }\) = \(\frac { 2 }{ 3 }\)
Let X denote the number of successes in ‘two tosses of a die.
Then X takes the values 0, 1, 2
P(X = 0) = P(A’A’) = \(\frac { 2 }{ 3 }\) x \(\frac { 2 }{ 3 }\) = \(\frac { 4 }{ 9 }\)
P(X = 1) = P(A A’ or A’ A)
= \(\frac { 1 }{ 3 }\) x \(\frac { 2 }{ 3 }\) + \(\frac { 2 }{ 3 }\) x \(\frac { 1 }{ 3 }\) = \(\frac { 4 }{ 9 }\)
P(X = 2) = P(AA) = \(\frac { 1 }{ 3 }\) x \(\frac { 1 }{ 3 }\) = \(\frac { 1 }{ 9 }\)
The Probability distribution of X is
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4 6

ii. Let B denote the event of getting 6 on atleast one die
∴ B = {(1, 6), (2, 6), (3, 6), (4, 6), 6), (6, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)}
Let X denote the number of success
Then X takes the values 0 and 1
P(X = 0) = P(B’)= 1 – P(B)
= 1 – \(\frac { 11 }{ 36 }\) = \(\frac { 25 }{ 36 }\)
P(X = 1) = P(B) = \(\frac { 11 }{ 36 }\)
∴ The probability distribution of X is
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4 7

NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4

Question 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.
Solution:
Let D denote a defective bulb
∴ P(D) = \(\frac { 6 }{ 30 }\)
∴ P(D’) = 1 – P(D) = 1 – \(\frac { 1 }{ 5 }\) = \(\frac { 4 }{ 5 }\)
Let X denote the number of defective bulbs in the sample of 4 bulbs.
Then X takes the values 0, 1, 2, 3, 4
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4 8

Another Method:
There are 6 defective bulbs and 24 non defective bulbs
P(getting a defective bulb) = \(\frac { 6 }{ 30 }\) = \(\frac { 1 }{ 5 }\)
P(getting a non defective bulb) = \(\frac { 24 }{ 30 }\) = \(\frac { 4 }{ 5 }\)
Let X denote the number of defective bulbs.
Then X takes the values 0, 1, 2, 3, 4
P(X = 0) = P(no defective bulb)
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4 9

Question 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 tads.
Solution:
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4 10

Question 8
A random variable X has the following probability distribution:
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4 11
Determine
(i) k
(ii) P(X < 3) (iii) P(X > 6)
(iv) P(0 < X < 3)
Solution:
(i) k
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4 12

Question 9.
The random variable X has a probability distribution P (X) of the following form, where k is some number
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4 13
(a) Determine the value of k
(b) Find P(X < 2), P (X ≤ 2), P(X ≥ 2)
Solution:
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4 14

Question 10.
Find the mean number of heads in three tosses of a fair coin.
Solution:
When 3 coins are tossed, the sample space S = {HHH, HHT, FITH, HTT, THH, THT, TTH, TTT}
Let X denote the number of heads
Then X takes the values 0, 1, 2, 3
P(X = 0) = P(no head) = \(\frac { 1 }{ 8 }\)
P(X = 1) = P(one head) = \(\frac { 3 }{ 8 }\)
P(X = 2) = P(two heads) = \(\frac { 3 }{ 8 }\)
P(X = 3) = P(three heads) = \(\frac { 1 }{ 8 }\)
The probability distribution of X is
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4 15

Question 11.
Two dice are thrown simultaneously. If X denotes the number of sixes, find the expectation of X.
Solution:
Let X denote the number of 6’s when two dice are thrown.
Then X takes the values 0, 1, 2
P(X = 0) = P(no six on both dice)
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4 16

Question 12.
Two numbers are selected at random (without replacement) from the first six positive integers. Let X denote the larger of the two numbers obtained. Find E(X).
Solution:
There are six numbers 1, 2, 3, 4, 5, 6 one of them is selected in 6 ways
When one of the numbers has been selected, 5 numbers are left, one number out of 5 may be select in 5 ways
∴ No. of ways of selecting two numbers without replacement out of 6 positive integers = 6 x 5 = 30
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4 17

NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4

Question 13.
Let X denote the sum of the numbers obtained when two fair dice are rolled. Find the variance and standard deviation of X.
Solution:
When two dice are rolled, then the sample space S has 36 simple events.
Let X denote the sum of numbers on the two dice Then X takes the values 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
P(X = 2) = P{(1,1)} = \(\frac { 1 }{ 36 }\)
P(X = 3) = P{(1,2), (2,1)} = \(\frac { 2 }{ 36 }\)
P(X = 4) = P{(1,3),(2,2),(3,1)} = \(\frac { 3 }{ 36 }\)
P(X = 5) = P{(1,4),(2,3),(3,2),(4,1)} = \(\frac { 4 }{ 36 }\)
P(X = 6) = P{(1,5), (2,4), (3,3), (4,2), (5,1)} = \(\frac { 5 }{ 36 }\)
P(X = 7) = P{(1,6), (2,5), (3,4), (4,3), (5,2), (6, 1)} = \(\frac { 6 }{ 36 }\)
P(X = 8) = P{(2,6), (3, 5), (4,4), (5, 3), (6,2)} = \(\frac { 5 }{ 36 }\)
P(X = 9) = P{(3,6), (4,5), (5,4), (6,3)} = \(\frac { 4 }{ 36 }\)
P(X = 10) = P{(4, 6), (5, 5), (6,4)} = \(\frac { 3 }{ 36 }\)
P(X=11) = P{(5,6),(6, 5)} = \(\frac { 52}{ 36 }\)
P(X =12) = P{(6, 6)} = \(\frac { 1 }{ 36 }\)
The probability distribution of X is
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4 18

Question 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?
Solution:
X denotes the age of the student selected.
∴ X takes the values 14, 15, 16, 17, 18, 19, 20, 21
The data can be summarised into the following table
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4 19

Question 15.
In a meeting, 70% of the members favour and 30% oppose a certain proposal. A member is selected at random and we take X = 0, if he opposed, and X = 1, if he is in favour Find E (X) and Var (X).
Solution:
X takes the values 0 and 1
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4 20

Question 16.
The mean of the number 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) \(\frac { 8 }{ 3 }\)
Solution:
Let X denote the number written on the face of the die.
Then X takes the values 1, 2, 5
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4 21

NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4

Question 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) \(\frac { 37 }{ 221 }\)
(b) \(\frac { 5 }{ 13 }\)
(c) \(\frac { 1 }{ 13 }\)
(d) \(\frac { 2 }{ 13 }\)
Solution:
Let X denote the number of aces
Then X can take the values 0, 1, 2
P(X = 0) = P(no ace)
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4 22

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