Aptitude Questions on Probability for Placements
Welcome to our practice page dedicated to helping you ace Probability questions for placement aptitude tests! This page is packed with carefully crafted questions designed to build your skills in solving a range of questions. Practice, review solutions, and keep challenging yourself. With consistent practice, you’ll develop a solid foundation to handle any Probability problem that comes your way!
Q. 1 A bag contains 5 red balls, 4 blue balls, and 3 green balls. If one ball is picked at random, what is the probability that it is either red or green?
A) $( \frac{5}{12} ) $
B) $( \frac{3}{12} ) $
C) $( \frac{8}{12} ) $
D) $( \frac{7}{12} )$
Check Solution
Ans: C) $( \frac{8}{12} )$
Total balls = ( 5 + 4 + 3 = 12 ).
Favorable outcomes (red or green) = ( 5 + 3 = 8 ).
Probability = $( \frac{8}{12} = \frac{2}{3} )$.
Q. 2 Two dice are rolled. What is the probability of getting a sum of 8?
A) $( \frac{5}{36} ) $
B) $( \frac{6}{36} ) $
C) $( \frac{7}{36} )$
D) $( \frac{4}{36} )$
Check Solution
Ans: A) $( \frac{5}{36} )$
Possible outcomes for a sum of 8 are ( (2,6), (3,5), (4,4), (5,3), (6,2) ).
Number of favorable outcomes = 5.
Total outcomes = $( 6 \times 6 = 36 )$.
Probability = $( \frac{5}{36} )$.
Q. 3 A card is drawn from a standard deck of 52 cards. What is the probability of drawing a queen or a diamond?
A) $( \frac{17}{52} ) $
B) $( \frac{4}{13} )$
C) $( \frac{16}{52} )$
D) $( \frac{15}{52} )$
Check Solution
Ans: B) $( \frac{4}{13} )$
There are 4 queens and 13 diamonds in the deck.
One queen is also a diamond.
Favorable outcomes = ( 4 + 13 – 1 = 16 ).
Probability = $( \frac{16}{52} = \frac{4}{13} )$.
Q. 4 What is the probability of getting at least one head in two tosses of a fair coin?
A) $( \frac{3}{4} )$
B) $( \frac{1}{4} )$
C) $( \frac{1}{2} )$
D) $( \frac{7}{8} )$
Check Solution
Ans: A) $( \frac{3}{4} )$
Probability of no head (both tails) = $( \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} )$.
Probability of at least one head = $( 1 – \frac{1}{4} = \frac{3}{4} )$.
Q. 5 A box contains 10 bulbs, 3 of which are defective. If 2 bulbs are drawn randomly, what is the probability that both are defective?
A) $( \frac{1}{15} )$
B) $( \frac{1}{30} )$
C) $( \frac{1}{10} )$
D) $( \frac{1}{20} )$
Check Solution
Ans: A) $( \frac{1}{15} )$
Total ways to choose 2 bulbs = $( \binom{10}{2} = 45 )$.
Ways to choose 2 defective bulbs = $( \binom{3}{2} = 3 )$.
Probability = $( \frac{3}{45} = \frac{1}{15} )$.
Q. 6 A committee of 3 is to be formed from 5 men and 3 women. What is the probability that the committee has at least one woman?
A) $( \frac{35}{56} )$
B) $( \frac{21}{56} $
C) $( \frac{41}{56} $
D) $( \frac{47}{56} )$
Check Solution
Ans: C) $( \frac{41}{56} )$
Total ways = $( \binom{8}{3} = 56 )$.
Ways to select only men = $( \binom{5}{3} = 10 )$.
Ways with at least one woman = $( 56 – 10 = 46 )$.
Probability = $( \frac{46}{56} = \frac{41}{56} )$.
Q. 7 Three fair coins are tossed. What is the probability of getting exactly two heads?
A) $( \frac{1}{8} )$
B) $( \frac{3}{8} )$
C) $( \frac{1}{4} )$
D) $( \frac{1}{2} )$
Check Solution
Ans: B) $( \frac{3}{8} )$
Total outcomes = $( 2^3 = 8 )$.
Favorable outcomes = $( { HHT, HTH, THH } = 3 )$.
Probability = $( \frac{3}{8} )$.
Q. 8 A committee of 3 is to be formed from 5 men and 3 women. What is the probability that the committee has at least one woman?
A) $\frac{35}{56}$
B) $\frac{21}{56}$
C) $\frac{41}{56}$
D) $\frac{47}{56}$
Check Solution
Ans: C) $\frac{41}{56}$
Total ways = $\binom{8}{3} = 56$
Ways to select only men = $\binom{5}{3} = 10$
Ways with at least one woman = $56−10=46$
Probability = $\frac{46}{56} = \frac{41}{56}$
Q. 9 If the probability of an event happening is $\frac{2}{5}$, what is the probability of the event not happening?
A) $\frac{1}{5}$
B) $\frac{3}{5}$
C) $\frac{4}{5}$
D) $\frac{2}{5}$
Check Solution
Ans: B) $\frac{3}{5}$
Probability of not happening = $\frac{2}{5} = \frac{3}{5}$
Q. 10 A bag contains 4 white and 5 black balls. If two balls are drawn randomly, what is the probability that both are black?
A) $\frac{2}{9}$
B) $\frac{1}{3}$
C) $\frac{5}{18}$
D) $\frac{10}{36}$
Check Solution
Ans: C) $\frac{5}{18}$
Total ways to choose 2 balls = $\binom{9}{2} = 36$
Ways to choose 2 black balls = $\binom{5}{2} = 10$
Probability = $\frac{10}{36} = \frac{5}{18}$
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