# 7000+ Probability Questions for IBPS Clerk Pdf - 1

Question: 1

Two dice are thrown together. What is the probability that the sum of the numbers on the two faces is divisible by 4 or 6?

(A) $${2}/ {18}$$

(B) $${7}/ {18}$$

(C) $${9}/ {12}$$

(D) $${12}/ {18}$$

Ans: B

Clearly, n(s) = 6 × 6 = 36.

Let E be the event that the sum of the numbers on the two faces is divisible by 4 or 6.

Then E = {(1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3, 5),(4, 2), (4, 4), (5, 1), (5, 3), (6, 2), ((6, 6)}

∴ n(E) = 14.

Hence, P(E) = $${n(E)} / {n(S)}$$ = $${14}/ {36}$$ = $${7}/ {18}$$.

Question: 2

Out of 5 girls and 3 boys, 4 children are to be randomly selected for a quiz contest. What is the probability that all the selected children are girls?

(A) $${1}/{7}$$

(B) $${1}/{14}$$

(C) $${2}/{17}$$

(D) $${5}/{17}$$

Ans: B

Total number of ways of selecting 4 children out of 8

= ^{8}C_{4} = $${8 × 7 × 6 × 5} / {1 × 2 × 3 × 4}$$ = 70

Number of ways of selecting 4 girls out of 5 = ^{5}C_{4} = 5

Required probability = $${5}/{70}$$ = $${1}/{14}$$.

Question: 3

4 coins are tossed once. Find the probability of exactly 3 tails.

(A) $${1}/{4}$$

(B) $${1}/{16}$$

(C) $${5}/{16}$$

(D) $${7}/{2}$$

Ans: A

There are 4 favourable cases TTTH, TTHT, THTT, HTTT.

∴ P(exactly 3 tails) = $${4}/{16}$$ =$${1}/{4}$$.

Question: 4

A group of 4 students is to be formed from among 4 girls and 6 boys. What is the probability that the group has less number of boys than the number of girls.

(A) $${5}/{42}$$

(B) $${9}/{42}$$

(C) $${11}/{42}$$

(D) $${13}/{42}$$

Ans: A

The selection of 1 boy and 3 girls can be done in ^{6}C_{1} × ^{4}C_{3} = 6 × 4 = 24 ways

Selection of 4 girls and no boys can be done in ^{6}C_{0} × ^{4}C_{4} = 1 × 1 = 1 way.

∴ n(E) = 24 + 1 = 25 ways.

Without any restriction the group can be formed in ^{10}C_{4} = $${10 × 9 × 8 × 7} / {1 × 2 × 3 × 4}$$ = 210 ways.

∴ P(E) = $${n(E)} / {n(S)}$$ =$${25} / {210}$$ = $${5}/{42}$$.

Question: 5

A bag contains 5 red and 7 green balls. Two balls are drawn from it provided that the first ball is replaced before the second ball is drawn. What is the probability, if

(A) both balls drawn were red

(B) both balls drawn were green

(C) the first ball was red and second green

(D) the first ball was green and second red

(i) Probability that both balls drawn were red = $${5/12} × {5/12}$$ = $${25}/{144}$$

(ii) Probability that both balls drawn were green = $${7/12} × {7/12}$$ = $${49}/ {144}$$

(iii) Probability that the first ball drawn was red and second green $${5/12} ×{7/12}$$ = $${35}/{144}$$

(iv) Probability that the first ball drawn was green and second red $${7/12} × {5/12}$$ = $${35}/{144}$$

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