Chapter 7 Probability Distributions Ex 7.1

Chapter 7 Probability Distributions Ex 7.1

Question 1.
Let X represent the difference between a number of heads and the number of tails when a coin is tossed 6 times. What are the possible values of X?
Solution:

When a coin is tossed 6 times, the number of heads can be 0, 1, 2, 3, 4, 5, 6.
The corresponding number of tails will be 6, 5, 4, 3, 2, 1, 0.
∴ X can take values 0 – 6, 1 – 5, 2 – 4, 3 – 3, 4 – 2, 5 – 1, 6 – 0
i.e. -6, -4, -2, 0, 2, 4, 6.
∴ X = {-6, -4, -2, 0, 2, 4, 6}.

Question 2.
An urn contains 5 red and 2 black balls. Two balls are drawn at random. X denotes the number of black balls drawn. What are the possible values of X?
Solution:

The urn contains 5 red and 2 black balls.
If two balls are drawn from the urn, it contains either 0 or 1 or 2 black balls.
X can take values 0, 1, 2.
∴ X = {0, 1, 2}.

Question 3.
State which of the following are not the probability mass function of a random variable. Give reasons for your answer.

Solution:
P.m.f. of random variable should satisfy the following conditions:
(a) 0 ≤  ≤ 1
(b) Σ = 1.

(i)
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q3.1
(a) Here 0 ≤  ≤ 1
(b) Σ = 0.4 + 0.4 + 0.2 = 1
Hence, P(X) can be regarded as p.m.f. of the random variable X.

(ii)
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q3.2
P(X = 3) = -0.1, i.e.  < 0 which does not satisfy 0 ≤  ≤ 1
Hence, P(X) cannot be regarded as p.m.f. of the random variable X.

(iii)
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q3.3
(a) Here 0 ≤  ≤ 1
(b) ∑ = 0.1 + 0.6 + 0.3 = 1
Hence, P(X) can be regarded as p.m.f. of the random variable X.

(iv)
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q3.4
Here ∑ = 0.3 + 0.2 + 0.4 + 0 + 0.05 = 0.95 ≠ 1
Hence, P(Z) cannot be regarded as p.m.f. of the random variable Z.

(v)
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q3.5
Here ∑ = 0.6 + 0.1 + 0.2 = 0.9 ≠ 1
Hence, P(Y) cannot be regarded as p.m.f. of the random variable Y.

(vi)
Maharashtra Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.1 Q3.6
(a) Here 0 ≤  ≤ 1
(b) ∑ = 0.3 + 0.4 + 0.3 = 1
Hence, P(X) can be regarded as p.m.f. of the random variable X.

Question 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.
Solution:

(i) For two tosses of a coin the sample space is {HH, HT, TH, TT}
Let X denote the number of heads in two tosses of a coin.
Then X can take values 0, 1, 2.
word image 21601 8
∴ the required probability distribution is
word image 21601 9

(ii) When three coins are tossed simultaneously, then the sample space is
{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
Let X denotes the number of tails.
Then X can take the value 0, 1, 2, 3.
word image 21601 10
∴ the required probability distribution is
word image 21601 11

(iii) When a fair coin is tossed 4 times, then the sample space is
S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, HTTH, THHT, THTH, TTHH, HTTT, THTT, TTHT, TTTH, TTTT}
∴ n(S) = 16
Let X denotes the number of heads.
Then X can take the value 0, 1, 2, 3, 4
When X = 0, then X = {TTTT}
∴ n(X) = 1
∴ P(X = 0) = n(X)n(S)=116
When X = 1, then
X = {HTTT, THTT, TTHT, TTTH}
∴ n(X) = 4
word image 21601 12

∴ the probability distribution of X is as follows:
word image 21601 13

Question 5.
Find the probability distribution of a number of successes in two tosses of a die, where success is defined as a number greater than 4 appearing on at least one die.
Solution:

When a die is tossed twice, the sample space s has 6 × 6 = 36 sample points.
∴ n(S) = 36
The trial will be a success if the number on at least one die is 5 or 6.
Let X denote the number of dice on which 5 or 6 appears.
Then X can take values 0, 1, 2.
When X = 0 i.e., 5 or 6 do not appear on any of the dice, then
X = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (4, 4)}
∴ n(X) = 16.
word image 21601 14
When X = 1, i.e. 5 or 6 appear on exactly one of the dice, then
X = {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (6, 1), (6, 2), (6, 3), (6, 4)}
∴ n(X) = 16
word image 21601 15
When X = 2, i.e. 5 or 6 appear on both of the dice, then
X = {(5, 5), (5, 6), (6, 5), (6, 6)}
∴ n(X) = 4
word image 21601 16
∴ the required probability distribution is
word image 21601 17

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:

Here, the number of defective bulbs is the random variable.
Let the number of defective bulbs be denoted by X.
∴ X can take the value 0, 1, 2, 3, 4.
Since the draws are done with replacement, therefore the four draws are independent experiments.
Total number of bulbs is 30 which include 6 defectives.
∴ P(X = 0) = P(0) = P(all 4 non-defective bulbs)
word image 21601 18
P(X = 1) = P (1) = P (1 defective and 3 non-defective bulbs)
word image 21601 19
P(X = 2) = P(2) = P(2 defective and 2 non-defective)
word image 21601 20
P(X = 3) = P(3) = P(3 defectives and 1 non-defective)
word image 21601 21
P(X = 4) = P(4) = P(all 4 defectives)
word image 21601 22
∴ the required probability distribution is
word image 21601 23

Question 7.
A coin is biased so that the head is 3 times as likely to occur as the tail. If the coin is tossed twice. Find the probability distribution of a number of tails.
Solution:
Given a biased coin such that heads is 3 times as likely as tails.
word image 21601 24
∴ the required probability distribution is
word image 21601 25

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

Determine:
(i) k
(ii) P(X < 3) (iii) P(X > 4)
Solution:

(i) Since P (x) is a probability distribution of x,
⇒ P(0) + P(1) + P(2) + P(3) + P(4) + P(5) + P(6) + P(7) = 1
⇒ 0 + k + 2k + 2k + 3k + k2 + 2k2 + 7k2 + k = 1
⇒ 10k2 + 9k – 1 = 0
⇒ 10k2 + 10k – k – 1 = 0
⇒ 10k(k + 1) – 1(k + 1) = 0
⇒ (k + 1)(10k – 1) = 0
⇒ 10k – 1 = 0 ……..[∵ k ≠ -1]
⇒ k = 1/10

(ii) P(X< 3) = P(0) + P(1) + P(2)
= 0 + k + 2k
= 3k
word image 21601 27

Question 9.
Find expected value and variance of X for the following p.m.f.:

Solution:

We construct the following table to calculate E(X) and V(X):
word image 21601 29
From the table,
word image 21601 30
= 2.25 – (-0.05)2
= 2.25 – 0.0025
= 2.2475
Hence, E(X) = -0.05 and V(X) = 2.2475.

Question 10.
Find expected value and variance of X, where X is the number obtained on the uppermost face when a fair die is thrown.
Solution:

If a die is tossed, then the sample space for the random variable X is
S = {1, 2, 3, 4, 5, 6}
∴ P(X) = 1/6; X = 1, 2, 3, 4, 5, 6.
word image 21601 31
Hence, E(X) = 3.5 and V(X) = 2.9167.

Question 11.
Find the mean number of heads in three tosses of a fair coin.
Solution:

When three coins are tossed the sample space is {HHH, HHT, THH, HTH, HTT, THT, TTH, TTT}
∴ n(S) = 8
Let X denote the number of heads when three coins are tossed.
Then X can take values 0, 1, 2, 3
word image 21601 32

Question 12.
Two dice are thrown simultaneously. If X denotes the number of sixes, find the expectation of X.
Solution:

When two dice are thrown, the sample space S has 6 × 6 = 36 sample points.
∴ n(S) = 36
Let X denote the number of sixes when two dice are thrown.
Then X can take values 0, 1, 2
When X = 0, then
X = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5)}
∴ n(X) = 25
word image 21601 33

Question 13.
Two numbers are selected at random (without replacement) from the first six positive integers. Let X denote the larger of the two numbers. Find E(X).
Solution:

Two numbers are chosen from the first 6 positive integers.
word image 21601 34

Question 14.
Let X denote the sum of numbers obtained when two fair dice are rolled. Find the standard deviation of X.
Solution:

If two fair dice are rolled then the sample space S of this experiment is
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)}
∴ n(S) = 36
Let X denote the sum of the numbers on uppermost faces.
Then X can take the values 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
word image 21601 35
∴ the probability distribution of X is given by
word image 21601 36
word image 21601 37
word image 21601 38

Question 15.
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 student is recorded. What is the probability distribution of the random variable X? Find mean, variance, and standard deviation of X.
Solution:

Let X denote the age of the chosen student. Then X can take values 14, 15, 16, 17, 18, 19, 20, 21.
We make a frequency table to find the number of students with age X:
word image 21601 39
The chances of any student selected are equally likely.
If there are m students with age X, then P(X) = 
Using this, the following is the probability distribution of X:
word image 21601 40
word image 21601 41
word image 21601 42
= 312.2 – 307.3
= 4.9
Standard deviation = √V(X) = √4.9 = 2.21
Hence, mean = 17.53, variance = 4.9 and standard deviation = 2.21.

Question 16.
In a meeting, 70% of the member’s 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 values 0 and 1.
It is given that
word image 21601 43
= 0.7 – 0.49
= 0.21
Hence, E(X) = 0.7 and Var(X) = 0.21.