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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