Question 1

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

Accepted Answer

∵ A coin is tossed 6 times S = {6H and 0T, 5H and 1T, 4H and 2T, 3H and 3T, 2H and 4T, 1H and 5T, 0H and 6T} X: Difference between no. of heads and no. of tails. X = 6 – 0 = 6 X = 5 – 1 = 4 X = 4 – 2 = 2 X = 3 – 3 = 0 X = 2 – 4 = -2 X = 1 – 5 = -4 X = 0 – 6 = -6 X = {-6, -4, -2, 0, 2, 4, 6}

Question 2

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?

Accepted Answer

S : Two bolts are drawn from the Urn S = {RR, RB, BR, BB} X : No. of black balls ∴ X = {0, 1, 2}

Question 3

Question 3: Determine whether each of the following is a probability distribution. Give reasons for your

Accepted Answer

(i) Solution: Here, P(X = x) ≥ 0, ∀ x and = p(0) + p(1) + p(2) = 0.4 + 0.4 + 0.2 = 1 ∴ The function is a p.m.f.

Question 4

Question 4: Find the probability distribution of (i) number of heads in two tosses of a coin, (ii) number of trails in three tosses of a coin, (iii) number of heads in four tosses of a coin.

Accepted Answer

(i) S: Coin is tossed two times S = {HH, HT, TH, TT} n(S) = 4 X: No. of heads Range of X = {0, 1, 2} p.m.f. Table

Question 5

Question 5: Find the probability distribution of the number of successes in two tosses of a die if successes are defined as getting a number greater than 4.

Accepted Answer

S = A die is tossed 2 times 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,

Question 6

Question 6: A sample of 4 bulbs is drawn at random with replacement from a lot of 30 bulbs which includes 6 defective bulbs. Find the probability distribution of the number of defective bulbs.

Accepted Answer

Total no. of bulbs = 30 No. of defective bulbs = 6 A sample of 4 bulbs are drawn from 30 bulbs. ∴ n(S) = ∴ No. of non-defective bulbs = 24 Let X = No. of defective bulbs drawn in sample of 4 bulbs.

Question 7

Question 7: A coin is biased so that the head is 3 times as likely to occur as the tail. Find the probability distribution of a number of tails in two tosses.

Accepted Answer

Here, the head is 3 times as likely to occur as the tail. i.e., If 4 times coin is tossed, 3 times there will be a head and 1 time there will be the tail. ∴ p(H) = and p(T) = Let X : No. of tails in two tosses. And coin is tossed twice. ∴ X = {0, 1, 2} For X = 0, p(0) = p(both heads) = p(H) × p(H) =

Question 8

Question 8: A random variable X has the following probability distribution: Determine (i) k, (ii) P(X < 3), (iii) P(0 < X < 3), (iv) P(X > 4).

Accepted Answer

(i) It is a p.m.f. of r.v. X ∴ Σp(x) = 1 ∴ p(1) + p(2) + p(3) + p(4) + p(5) + p(6) + p(7) = 1 ∴ k + 2k + 2k + 3k + k2 + 2k2 + (7k2 + k) = 1 ∴ 10k2 + 9k = 1 ∴ 10k2 + 9k – 1 = 0 ∴ 10k2 + 10k – k – 1 = 0 ∴ 10k(k + 1) – (k + 1) = 0 ∴ (10k – 1)(k + 1) = 0 ∴ 10k – 1 = 0 or k + 1 = 0 ∴ k = or k = -1 but k

Question 9

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

Accepted Answer

E(X) = Σxp = -0.05 V(X) = Σx2p – (Σxp)2 = 2.25 – (-0.05)2 = 2.25 – 0.0025 = 2.2475

Question 10

Question 10: Find expected value and variance of X, the number on the uppermost face of a fair die.

Accepted Answer

S : A fair die is thrown S = {1, 2, 3, 4, 5, 6} n(S) = 6 X: No obtained on uppermost face of die Range of X = {1, 2, 3, 4, 5, 6} E(X) = Σxp = = 3.5 V(X) = Σx2p – (Σxp)2 = – (3.5)2 = 15.17 – 12.25 = 2.92

Question 11

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

Accepted Answer

S : A coin is tossed 3 times S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} n(S) = 8 Range of X = {0, 1, 2, 3} ∴ Mean = E(X) = Σxp = = 1.5

Question 12

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

Accepted Answer

S : Two dice are thrown S = {(1, 1), (1, 2), (1, 3), ……, (6, 6)} n(S) = 36 Range of X = {0, 1, 2} First 6 positive integers are 1, 2, 3, 4, 5, 6 X = Larger two numbers selected 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)

Question 13

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).

Accepted Answer

First 6 positive integers are 1, 2, 3, 4, 5, 6 X : The larger of the selected two numbers S = {(1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6), (6, 1

Question 14

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

Accepted Answer

S : Two fair dice are rolled S = {(1, 1), (1, 2), (1, 4), ……, (6, 6)} n(S) = 36 X : Sum of the two numbers. Range of X = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} V(X) = Σx2p – (Σxp)2 = = 54.22 – (7)2 = 5.22 SD(X) = √V(X) = √5.22 = 2.28

Question 15

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. If X denotes the age of a randomly selected student, find the probabilit

Accepted Answer

Step-by-step solution provided.

Question 16

Question 16: 70% of the member’s favour and 30% oppose a proposal in a meeting. The random variable X takes the value 0 if a member opposes the proposal and the value 1 if a member is in favour

Accepted Answer

E(X) = Σxp = 0.7 V(X) = Σx2p – (Σxp)2 = 0.7 – (0.7)2 = 0.7 – 0.49 = 0.21

Chapter 8 Probability Distributions Ex 8.1 Solutions