These NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-13-ex-13-4/
NCERT Solutions for Class 12 Maths Chapter 13 Probability Exercise 13.4
Question 1.
State which of the following are not the probability distributions of a random variable. Give reasons for your answer.
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|
∴ The possible values of X are 0, 2, 4, 6.
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
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
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}
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
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
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
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)
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:
Question 8
A random variable X has the following probability distribution:
Determine
(i) k
(ii) P(X < 3) (iii) P(X > 6)
(iv) P(0 < X < 3)
Solution:
(i) k
Question 9.
The random variable X has a probability distribution P (X) of the following form, where k is some number
(a) Determine the value of k
(b) Find P(X < 2), P (X ≤ 2), P(X ≥ 2)
Solution:
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
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)
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
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
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
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
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
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)