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Class 12 (HSC Board)Mathematics & Statistics2026-27 Syllabus

Chapter 7 Probability Distributions Ex 7.1 Solutions

Complete Maharashtra State Board Balbharati & Yuvakbharati textbook solutions for Chapter 7 Probability Distributions Ex 7.1. Step-by-step solved exercises, numerical problems, and digest answers.

16 Solved Questions• 27 Diagrams• 2552 words

Balbharti12th Maharashtra State Board Maths Solutions BookPdf Chapter 7 Probability Distributions Ex 7.1 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.1

Question 1 Maharashtra Board Solution
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 & Step-by-Step Answer:
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 Maharashtra Board Solution
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 & Step-by-Step Answer:
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 Maharashtra Board Solution
State which of the following are not the probability mass function of a random variable. Give reasons for your
Solution & Step-by-Step Answer:
Solution: P.m.f. of random variable should satisfy the following conditions: (a) 0 ≤ pi ≤ 1 (b) Σpi = 1.

(i)

(a) Here 0 ≤ pi≤ 1
(b) Σpi= 0.4 + 0.4 + 0.2 = 1
Hence, P(X) can be regarded as p.m.f. of the random variable X.

(ii)

P(X = 3) = -0.1, i.e. Pi< 0 which does not satisfy 0 ≤ Pi≤ 1
Hence, P(X) cannot be regarded as p.m.f. of the random variable X.

(iii)

(a) Here 0 ≤ pi≤ 1
(b) ∑pi= 0.1 + 0.6 + 0.3 = 1
Hence, P(X) can be regarded as p.m.f. of the random variable X.

(iv)

Here ∑pi= 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)

Here ∑pi= 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)

(a) Here 0 ≤ pi≤ 1
(b) ∑pi= 0.3 + 0.4 + 0.3 = 1
Hence, P(X) can be regarded as p.m.f. of the random variable X.

Question 4 Maharashtra Board Solution
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 & Step-by-Step Answer:
(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. ∴ P[X = 0] = P(0) = P[X = 1] = P(1) = = P[X = 2] = P(2) = ∴ the required probability distribution is

(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.
∴ P[X = 0] = P(0) =
P[X = 1] = P(1) =
P[X = 2] = P(2) =
P[X = 3] = P(3) =
∴ the required probability distribution is

(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) =
When X = 1, then
X = {HTTT, THTT, TTHT, TTTH}
∴ n(X) = 4
∴ P(X = 1) =
When X = 2, then
X = {HHTT, HTHT, HTTH, THHT, THTH, TTHH}
∴ n(X) = 6
∴ P(X = 2) =
When X = 3, then
X = {HHHT, HHTH, HTHH, THHH}
∴ n(X) = 4
∴ P(X = 3) =
When X = 4, then X = {HHHH}
∴ n(X) = 1
∴ P(X = 4) =
∴ the probability distribution of X is as follows:

Question 5 Maharashtra Board Solution
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 & Step-by-Step Answer:
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. ∴ P(X = 0) = 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 ∴ P(X = 1) = 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 ∴ P(X = 2) = ∴ the required probability distribution is

Question 6 Maharashtra Board Solution
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 & Step-by-Step Answer:
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) = = P(X = 1) = P (1) = P (1 defective and 3 non-defective bulbs) P(X = 2) = P(2) = P(2 defective and 2 non-defective) P(X = 3) = P(3) = P(3 defectives and 1 non-defective) P(X = 4) = P(4) = P(all 4 defectives) = = ∴ the required probability distribution is

Question 7 Maharashtra Board Solution
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 & Step-by-Step Answer:
Given a biased coin such that heads is 3 times as likely as tails. ∴ P(H) = and P(T) = The coin is tossed twice. Let X can be the random variable for the number of tails. Then X can take the value 0, 1, 2. ∴ P(X = 0) = P(HH) = P(X = 1) = P(HT, TH) = P(X = 2) = P(TT) = ∴ the required probability distribution is

Question 8 Maharashtra Board Solution
A random variable X has the following probability distribution: Determine: (i) k (ii) P(X < 3) (iii) P(X > 4)
Solution & Step-by-Step Answer:
(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 =

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

(iii) P(0 < X < 3) = P (1) + P (2)
= k + 2k
= 3k
= 3()
=

Question 9 Maharashtra Board Solution
Find expected value and variance of X for the following p.m.f.:
Solution & Step-by-Step Answer:
We construct the following table to calculate E(X) and V(X): From the table, Σxipi = -0.05 and = 2.25 ∴ E(X) = Σxipi = -0.05 and V(X) = = 2.25 – (-0.05)2 = 2.25 – 0.0025 = 2.2475 Hence, E(X) = -0.05 and V(X) = 2.2475.

Question 10 Maharashtra Board Solution
Find expected value and variance of X, where X is the number obtained on the uppermost face when a fair die is thrown.
Solution & Step-by-Step Answer:
If a die is tossed, then the sample space for the random variable X is S = {1, 2, 3, 4, 5, 6} ∴ P(X) = ; X = 1, 2, 3, 4, 5, 6. Hence, E(X) = 3.5 and V(X) = 2.9167.

Question 11 Maharashtra Board Solution
Find the mean number of heads in three tosses of a fair coin.
Solution & Step-by-Step Answer:
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 P(X = 0) = P(0) = P(X = 1) = P(1) = P(X = 2) = P(2) = P(X = 3) = P(3) = ∴ mean = E(X) = ΣxiP(xi) = = = = 1.5
Question 12 Maharashtra Board Solution
Two dice are thrown simultaneously. If X denotes the number of sixes, find the expectation of X.
Solution & Step-by-Step Answer:
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 ∴ P(X = 0) = When X = 1, then X = {(1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)} ∴ n(X) = 10 ∴ P(X = 1) = When X = 2, then X = {(6, 6)} ∴ n(X) = 1 ∴ P(X = 2) = ∴ E(X) = ΣxiP(xi) = = =
Question 13 Maharashtra Board Solution
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 & Step-by-Step Answer:
Two numbers are chosen from the first 6 positive integers. ∴ n(S) = = 15 Let X denote the larger of the two numbers. Then X can take values 2, 3, 4, 5, 6. When X = 2, the other positive number which is less than 2 is 1. ∴ n(X) = 1 ∴ P(X = 2) = P(2) = When X = 3, the other positive number less than 3 can be 1 or 2 and hence can be chosen in 2 ways. ∴ n(X) = 2 P(X = 3) = P(3) = Similarly, P(X = 4) = P(4) = P(X = 5) = P(5) = P(X = 6) = P(6) = ∴ E(X) = ΣxiP(xi) = = = =
Question 14 Maharashtra Board Solution
Let X denote the sum of numbers obtained when two fair dice are rolled. Find the standard deviation of X.
Solution & Step-by-Step Answer:
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 ∴ the probability distribution of X is given by

Question 15 Maharashtra Board Solution
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 & Step-by-Step Answer:
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: 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: Variance = V(X) = . P(xi) – [E(X)]2 = 312.2 – (17.53)2 = 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 Maharashtra Board Solution
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 & Step-by-Step Answer:
X takes values 0 and 1. It is given that P(X = 0) = P(0) = 30% = = 0.3 P(X = 1) = P(1) = 70% = = 0.7 ∴ E(X) = Σxi. P(xi) = 0 × 0.3 + 1 × 0.7 = 0.7 Also, = 0 × 0.3 + 1 × 0.7 = 0.7 ∴ Variance = V(X) = = 0.7 – (0.7)2 = 0.7 – 0.49 = 0.21 Hence, E(X) = 0.7 and Var(X) = 0.21.
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Chapter 7 Probability Distributions Ex 7.2