100+ Probability Questions for Bank PO, SO, Clerk Exams - 1

Question: 1

In a simultaneous throw of two dice, what is the probability of getting a total of 7?

(A) $${1}/{2}$$

(B) $${1}/{6}$$

(C) $${2}/{3}$$

(D) $${3}/{4}$$

Ans: B

W know that in a simultaneous throw of two dice, n(S) = 6 × 6 = 36.

Let E = event of getting a total of 7.

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

P(E) = $${n(E)}/{n(S)}$$ = $${6}/{36}$$ = $${1}/{6}$$.

Question: 2

Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?

(A) $${1}/{2}$$

(B) $${2}/{5}$$

(C) $${9}/{20}$$

(D) $${10}/{20}$$

Ans: C

Here, S = {1, 2 , 3, 4, ….., 19, 20}

Let E = event of getting a multiple of 3 or 5 = {3, 6, 9, 12, 15, 18, 5, 10, 20}.

∴ P(E) = $${n(E)}/{n(S)}$$ = $${9}/{20}$$.

Question: 3

From a pack of 52 cards, one card is drawn at random. What is the probability that the card drawn is a ten or a spade?

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

(B) $${1}/{13}$$

(C) $${1}/{26}$$

(D) $${4}/{13}$$

Ans: D

Here, n(S) = 52.

There are 13 spades (including one ten) and there are 3 more tens.

Let E = event of getting a ten or a spade.

Then, n(E) = (13 + 3) = 16.

P(E) = $${n(E)} / {n(S)}$$ = $${16}/{52}$$ = $${4}/{13}$$.

Question: 4

An urn contains 6 red, 4 blue, 2 green and 3 yellow marbles. If three marbles are picked up at random, what is the probability that 2 are blue and 1 is yellow?

(A) $${1}/{5}$$

(B) $${7}/{15}$$

(C) $${3}/{91}$$

(D) $${18}/{455}$$

Ans: D

Total number of marbles = (6 + 4 + 2 + 3) = 15.

Let E be the event of drawing 2 blue and 1 yellow marble.

Then, n(E) = (4C2 × 3C1)

= $${4 × 3} / {2 × 1} × 3$$ = 18.

Also, n(S) = 15C3 = $${15 × 14 × 13} / {3 × 2 × 1}$$ = 455.

∴ P(E) = $${n(E)} / {n(S)}$$ = $${18}/{455}$$.

Question: 5

A box contains 20 electric bulbs, out of which 4 are defective. Two balls are chosen at random from this box. The probability that at least one of them is defective is

(A) $${4}/{19}$$

(B) $${7}/{19}$$

(C) $${12}/{19}$$

(D) $${21}/{95}$$

Ans: B

P(none is defective) = 16C2 / 20C2

= $$({16 × 15} / {2 × 1} × {2 × 1 }/ {20 × 19})$$ = $${12} / {19}$$

P(at least 1 is defective) = $$(1 - {12} /{19})$$ = $${7} / {19}$$.

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