# Quantitative Aptitude Questions and Answers - Probability

11. A carton contains 12 green and 8 blue bulbs .2 bulbs are drawn at random. Find the probability that they
are of same colour?

A. (91/47)

B. (47/105)

C. (47/95)

D. (47/145)

View Answer

Ans : C

Explanation: Let S be the sample space

Then n(S) = no of ways of drawing 2 bulbs out of (12+8) = 20c2=20*19/2*1=190

Let E = event of getting both bulbs of same colour

Then, n(E) = no of ways (2 bulbs out of 12) or (2 bulbs out of 8)

=12C2+ 8C2=(132/2)+(56/2) = 66+28 = 94

Therefore, P(E) = n(E)/n(S) = 94/190 = 47/95

12. In a Coupon, there are 30prizes and 75blanks. A Coupon is drawn at random. What is the probability of
getting a prize?

A. (2/7)

B. (5/7)

C. (1/2)

D. (5/12)

View Answer

Ans : A

Explanation: Total number of outcomes possible, n(S) = 30+75 = 105

Total number of prizes, n(E) = 30

P(E)=n(E)/n(S)=30/105=2/7

13. Two dice are rolling simultaneously .What is the probability that the sum of the number on the two faces is
divided by 5 Or 7

A. (13/36)

B. (14/36)

C. (15/36)

D. (11/36)

View Answer

Ans : D

Explanation: Clearly, n(S) = 6 x 6 = 36

Let E be the event that the sum of the numbers on the two faces is divided by 5or 7.

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

n(E) = 11.

Hence, P(E) = n(E)/n(S) = 11/36

14. Consider a pack contains 2black, 9 white and 3 pink pencils. If a pencil is drawn at random from the pack,
replaced and the process repeated 2 more times, What is the probability of drawing 2 black pencils and 1 pink
pencil?

A. (3/49)

B. (3/386)

C. (3/14)

D. (3/545)

View Answer

Ans : B

Explanation: Here, total number of pencils = 14

Probability of drawing 1 black pencil = 2/14

Probability of drawing another black pencil = 2/14

Probability of drawing 1 pink pencil = 3/14

Probability of drawing 2 black pencils and 1 pink pencil = 2/14 * 2/14 * 3/14 = 3/686

15. A box contains 3red, 8 blue and 5 green marker pens. If 2 marker pens are drawn at random from the
pack, not replaced and then another pen is drawn. What is the probability of drawing 2 blue marker pens and
1 red marker pen?

A. (1/20)

B. (3/20)

C. (7/20)

D. (9/20)

View Answer

Ans : A

Explanation: Probability of drawing 1 blue marker pen =8/16

Probability of drawing another blue marker pen = 7/15

Probability of drawing 1 red marker pen = 3/14

Probability of drawing 2 blue marker pens and 1 red marker pen = 8/16*7/15*3/14=1/20

16. Consider the example of finding the probability of selecting a red card or a 9 from a deck of 52 cards

A. (5/13)

B. (6/13)

C. (7/13)

D. (8/13)

View Answer

Ans : C

Explanation: Probability of selecting a Red card = 26/52

Probability of selecting a 9 = 4/52

Probability of selecting both a red card and a 9 = 2/52

P(R or 9) = P(R) + P(9) – P(R and 9)

= 26/52 + 4/52 – 2/52

= 28/52

= 7/13

17. A single coin is tossed 7 times. What is the probability of getting at least one tail?

A. (55/128)

B. (56/128)

C. (127/128)

D. (126/128)

View Answer

Ans : C

Explanation: Consider solving this using complement.

Probability of getting no tail = P(all heads) = 1/128

P(at least one tail) = 1 – P(all heads) = 1 – 1/128 = 127/128

18. Murali and his wife appear in an interview for two vacancies in the same post. The probability of murali's
selection is (1/6) and the probability of wife's selection is (1/4). What is the probability that only one of them is
selected ?

A. (1/2)

B. (1/3)

C. (1/4)

D. (1/6)

View Answer

Ans : B

Explanation: A= Event that the husband is selected

B =Event that the wife is selected

P(A)=1/6,P(B)=1/4

P(Ac)=1-1/6=5/6

P(Bc)=1-1/4=3/4

Required Probability=P[ (A and notB) or(B and not A)]

= P(A). P(Bc) + P(B) P(Ac)

=1/6*3/4 + 1/4 * 5/6 = 1/3

19. Two dice are tossed. The probability that the total score is a prime number is:

A. (5/12)

B. (7/12)

C. (11/12)

D. (1/3)

View Answer

Ans : A

Explanation: Clearly, n(S) = (6 x 6) = 36.

Let E = Event that the sum is a prime number.

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

n(E) = 15.

P(E) = n(E)/n(S)=15/36=>5/12

20. A card is drawn from a pack of 52 cards. The probability of getting a queen of club or a king of heart is:

A. (1/13)

B. (1/26)

C. (3/52)

D. (1/52)

View Answer

Ans : B

Explanation: Here, n(S) = 52.

Let E = event of getting a queen of club or a king of heart.

Then, n(E) = 2.

P(E) = n(E)/n(S)=2/52=>1/26

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