# 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) = 12C3 = \$\${12 × 11 × 10} / {3 × 2 × 1}\$\$ = 220.

Also, n(S) = 15C3 = \$\${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

= 7C3 = \$\${7 × 6 × 5}/ {3 × 2 × 1}\$\$ = 35.

And, n(S) = 12C3 = \$\${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

= 9C4 = \$\${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) = (4C2 × 5C2) = \$\$({4 × 3} / {2 × 1} × {5 × 4} / {2 × 1})\$\$ = 60.

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

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