Question 1

What is the possibility of getting at least 6 heads, if eight coins are tossed simultaneously?

Accepted Answer

Required probability = Probability of getting 2 tails + Probability of getting 1 tail + Probability of getting no tail. = 8C2 x 1/256 + 8C1 x 1/256 + 8C0 x 1/256 = 37/256

Question 2

A and B aim a target. The Probability that A hits the target is 5/7 and B hits the target in 7/10. What is the probability that exactly one of them hits the target?

Accepted Answer

Required probability = P(A). P(B) + P (A ) . P(B) = 5/7 . 3/10 + 2/7 . 7/10 = 29/70

Question 3

A bag contains 5 white, 7 red and 8 black balls. If 4 balls are drawn one by one with replacement, what is the probability that all are white?

Accepted Answer

Probability that first ball drawn is white = 5C1 = 1/4 Since, balls are drawn with replacement, hence all the four events will have equal probability. Therefore, required probability = 1/4 x 1/4 x 1/4 x 1/4 = 1/256

Question 4

There are 8 blue and 4 white balls in a bag . A ball is drawn at random. Without replacing it another ball is drawn. Find the probability that both the balls drawn are blue?

Accepted Answer

Total number of balls in the bag = 12 Probability of drawing one blue ball in the first draw P(E1) = 8C1 / 12C1 = 8/12 = 2/3 After the first drawn of a blue ball, now there are 7 blue and 4 white ball in the bag. Total number of the balls in the bag is 11. Probability of drawing one blue ball in the second drawn = P(E2) = 7C1 /11C1 = 7/11 ∴ Probability that both are blue P(E1 ∩ E2) = P(E1) x P(E2) = 2/3 x 7/11 = 14 / 33

Question 5

Two dice are thrown simultaneously. What is the probability of getting a number other than 4 on any dice?

Accepted Answer

Here, n(S) = 6 x 6 = 36 and E = Event of getting a number other than 4 on any dice = {(1, 1), (1, 2), (1, 3), (1, 5), (1, 6), (2, 1), (2, 2) , (2, 3) , (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 5), (3, 6), (5, 1) , (5, 2) , (5, 3), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 5), (6, 6) } ∴ P(E) = n(E) / n(S) = 25/36

Question 6

Number 1 to 25 are marked on tokens of equals size one on each. A token is drawn at random. Find the probability of getting a number divisible by both 2 and 3?

Accepted Answer

Total number of ways of selecting one number from 25 number is 25C1 ways. Let E be the event of drawing a number divisible by both 2 and 3 i.e., (if a number is divisible by both 2 and 3, it is divisible by 6 also) E = {6, 12, 18, 24} ∴ Required probability = P(E) = n(E) / n(S) = 4/25

Question 7

A complete cycle of a traffic light takes 60 s. During each cycle the light is green for 25 s, yellow for 5 s and red for 30 s. At a randomly chosen time, the probability that the light will not be green is?

Accepted Answer

Time taken for complete cycle = 60 s Probability that light will be green = (Time for which light is green)/(time taken for complete cycle) = 25/60 = 5/12 ∴ Probability that light will not be green = 1- 5/12 = 7/12

Question 8

The probability that a man will live 10 more year is 1/4 and the probability that his wife will live 10 more year is 1/3. Then, the probability that none of them will be alive after 10 yr, is?

Accepted Answer

Let the name of the man be A and that of his wife be B. Then ,P(A will not be alive after 10 yr and B will not be alive after 10yr) = 3/4 x 2/3 = 1/2

Question 9

The letters B, G, I, N, R are rearranged to form the word 'BRING'. Find its probability?

Accepted Answer

The five letters could be arrange in 5! ways. One of them is 'BRING'. ∴ Required probability = 1/5! = 1/(5 x 4 x 3 x 2 x 1) = 1/120

Question 10

If one rolls a fair sides die twice, what is the probability that the die will land on the same number on both the occasions?

Accepted Answer

Let E = Event of getting same number on both the occasions = (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) ∴ n(E) = 6 ∴ Required probability = 6/62 = 1/6

Question 11

One hundred identical coins each with probability p of showing up head are tossed. If 0 < p < 1 and the probability of heads showing on 50 coins is equals to that of heads on 51 coins. Then the value of p, is?

Accepted Answer

Let N be the number of coins showing heads. ∴ Probability when N equal to 50 = Probability when N equal to 51 ⇒ 100C50 x P 50 x (1 - p)50 = 100C51 x (1 - p)49 ⇒ p = 51/101

Question 12

Four whole number taken at random are multiplied together. The chance that the last digit in the product is 1, 3, 7, or 9 is?

Accepted Answer

If the product of the four numbers ends in one of the digits 1, 3, 7 or 9, each number should have the last digit as one of these 4 digits. ∴ The number of favourable cases = 44 Total number of all possible cases = 104 Hence, the required probability =44/104 = 24/54 = 16/625

Question 13

What is the probability that a card drawn at random from a pack of 52 cards is either a king or a spade?

Accepted Answer

Required probability = 3/52 + 13/52 = 16/52 =4/13 [Hint Why 13/52 because there are 13 spades and why 3/52 instead of 4/52 (there are four kings) because one king is already counted in spades.]

Question 14

A card is drawn from a well-shuffled pack of cards. The probability of getting a queen of club or a king of heart is?

Accepted Answer

Total ways = 52 There is one queen of club and one king of heart. ∴ Favorable ways = 1 + 1 = 2 ∴ Required probability = 2/52 = 1/26

Question 15

If three unbiased coins are tossed simultaneously, then the probability of exactly two heads, is

Accepted Answer

n(S) = 23 = 8 Let E = Event of getting exactly two heads = {(H, H, T), (H, T ,H), (T, H, H)} ⇒ n(E) = 3 ∴ required probability = 3/8

Question 16

Let E be the set of all integers with 1 at their unit places. The probability that a number chosen from {2, 3, 4, ............, 50} is an element of E, is?

Accepted Answer

n(S) = 49 Favourable numbers are 11, 21, 31, 41. ∴ Required probability = 4/49

Question 17

A number is selected at random from the set {1, 2, 3, ........, 50}. The probability that it is a prime, is

Accepted Answer

n(s) = 50 Prime numbers are = 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47 ∴ n(E) = 15 ∴ p(E) = 15/50 = 3/10 = 0.3

Question 18

A committee of 3 members is to be selected to be selected out of 3 man and 2 women . What is the probability that the committee has at least one women?

Accepted Answer

Required probability = (2C1 x 3C2 + 2C2 x 3C1) / (5C3) = 9/10

Question 19

An urn contains 3 red and 4 green marbles. If three marbles are picked at random, what is the probability that two are green and one is red?

Accepted Answer

Number of ways to select 3 marbles out of 7 marbles = n(s) = 7C3 = 35 Probability that 2 are green and 1 is red = n(E) = 4C2 x 3C1 = 18 ∴ Required probability = 18/35

Question 20

If two marbles are picked at random, what is the probability that both marbles are red?

Accepted Answer

Total number of marbles = 6 + 4 + 2 + 3 = 15 Ways of selection of two red marbles = n(E) = 6C2 Ways of selection of two marbles = n(S) = 15C2 So, required probability = (6C2) / (15C2) = (6 x 5) / (15 x 14) = 1/7

Probability Questions