probability mcq pdf Model Questions & Answers, Practice Test for ibps rrb so officer sclae 2 3 single exam 2024

Question :11

In single cast with two dice the odds against drawing 7 is

Answer: (c)

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

∴ P(E) = $6/{6 × 6} = 1/6$

So, odds against drawing 7

= ${P(\ov{E})}/{P(E)} = {1 - 1/6}/{1/6} = 5/1$

Question :12

The probability that A can solve a problem is $2/3$ and B can solve it is $3/4$. If both attempt the problem, what is the probability that the problem gets solved?

Answer: (c)

The probability that A cannot solve the problem

= $1 - 2/3 = 1/3$

The probability that B cannot solve the problem

= $1 - 3/4 = 1/4$

The probability that both A and B cannot solve the problem

= $1/3 × 1/4 = 1/12$

∴ The probability that at least one of A and B can solve the problem

= $1 - 1/12 = 11/12$

∴ The probability that the problem is solved = $11/12$

Question :13

An urn contains 4 green, 5 blue, 2 red and 3 yellow marbles. If three marbles are drawn at random, what is the probability that at least one is yellow ?

Answer: (a)

According to question,

n(S) = $^14C_3 = {14!}/{(14 - 3)! 3!} = {14 × 13 × 12}/{3 × 2 × 1} = 364$

∴ Required probability

= 1 - $^11C_3/^14C_3 = 1 - 165/364 = {364 - 165}/364 = 199/364$

Question :14

Let 0 < P(A) < 1, 0 < P(B) < 1 and P(A ∪ B) = P(A) + P(B) - P(A) P(B), then :

Answer: (d)

Given P(A) + P(B) – P(A) P(B) = P(A ∪ B)

Comparing with

P(A) + P(B) – P(A ∩ B) = P(A ∪ B)

we get P(A ∩ B) = P(A).P(B)

∴ A and B independent events.

Question :15

The probability of getting head and tail alternately in three throws of a coin (or a throw of three coins), is

Answer: (a)

Total probable ways = 8

Favourable number of ways = HTH, THT

Hence required probability = $2/8 = 1/4$

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