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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