# Top 1,000+ Probability Questions and Answers Pdf - 1

Question: 1

A basket contains 6 blue, 2 red, 4 green and 3 yellow balls. If three balls are picked up at random, what is the probability that none is yellow?

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

(B) $${4}/{5}$$

(C) $${44}/{91}$$

(D) $${3}/{455}$$

Ans: C

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

Let E be the event of drawing 3 non yellow balls.

Then, n(E) = ^{12}C_{3} = $${12 × 11 × 10} / {3 × 2 × 1}$$ = 220.

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

∴ P(E) = $${n(E)} / {n(S)}$$ = $${220}/{455}$$ = $${44}/{91}$$.

Question: 2

A basket contains 4 red, 5 blue and 3 green marbles. If three marbles are picked up at random what is the probability that at least one is blue?

(A) $${5}/{12}$$

(B) $${7}/{12}$$

(C) $${7}/{44}$$

(D) $${37}/{44}$$

Ans: D

Total number of marbles = (4 + 5 + 3) = 12.

Let E be the event of drawing 3 marbles such that none is blue.

Then, n(E) = number of ways of drawing 3 marbles out of 7

= ^{7}C_{3} = $${7 × 6 × 5}/ {3 × 2 × 1}$$ = 35.

And, n(S) = ^{12}C_{3} = $${12 × 11 × 10} / {3 × 2 × 1}$$ = 220.

∴ P(E) = $${n(E)} / {n(S)}$$ = $${35}/{220}$$ = $${7}/{44}$$.

Required probability = 1 – P(E) = $$(1 – {7}/{44})$$ = $${37}/{44}$$

Question: 3

The probability that a card drawn from a pack of 52 cards will be a diamond or a king, is

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

(B) $${1}/{52}$$

(C) $${2}/{13}$$

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

Ans: D

Here, n(S) = 52.

There are 13 cards of diamond (including one king) and there are 3 more kings.

Let E = event of getting a diamond or a king.

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

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

Question: 4

Three unbiased coins are tossed. What is the probability of getting at most two heads?

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

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

(C) $${3}/{8}$$

(D) $${7}/{8}$$

Ans: D

Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}.

Let E = event of getting at most two heads.

Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}

∴ P(E) = $${n(E)} / {n(S)}$$ = $${7}/{8}$$.

Question: 5

Four persons are chosen at random from a group of 3 men, 2 women and 4 children. The chance that exactly 2 of them are children, is

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

(B) $${1}/{9}$$

(C) $${1}/{12}$$

(D) $${10}/{21}$$

Ans: D

n(S) = number of ways choosing 4 persons out of 9

= ^{9}C_{4} = $${9 × 8 × 7 × 6} / {4 × 3 × 2 × 1}$$ = 126.

n(E) = Number of ways of choosing 2 children out of 4 and 2 persons out of (3 + 2) persons.

n(E) = (^{4}C_{2} × ^{5}C_{2}) = $$({4 × 3} / {2 × 1} × {5 × 4} / {2 × 1})$$ = 60.

P(E) = $${n(E)} / {n(S)}$$ = $${60} / {126}$$ = $${10}/{21}$$.

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