Probability Aptitude Questions and Answers
This section focuses on "Probability" in Quantitative Aptitude. These Multiple Choice Questions (mcq) should be practiced to improve the Quantitative Aptitude skills required for various interviews (campus interviews, walk-in interviews, company interviews), placements, entrance exams and other competitive examinations.
1. When a single die is rolled, the sample space is {1,2,3,4,5,6}.What is the probability of getting a 3 when a
die is rolled?
A. (1/2)
B. (1/6)
C. (1/4)
D. (1/3)
View Answer
Ans : B
Explanation: No. of ways it can occur = 1
Total no. of possible outcomes = 6
So the probability of rolling a particular number (3) when a die is rolled = 1/6.
2. 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
3. Tickets numbered 1 to 37 are mixed up and then a ticket is drawn at random. What is the probability that
the ticket drawn has a number which is a multiple of 4 or 10?
A. (11/37)
B. (12/37)
C. (13/37)
D. (14/37)
View Answer
Ans : A
Explanation: Here, S = {1, 2, 3, 4, ...., 36, 37}.
Let E = event of getting a multiple of 4 or 10= {4,8,12,16,20,24,28,32,36,10, 30}.
P(E) = n(E)/n(S) = 11/37
4. 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
5. A Receptacle contains 3violet, 4purple and 5 black balls. Three balls are drawn at random from the
receptacle. The probability that all of them are purple, is:
A. (3/55)
B. (4/55)
C. (1/55)
D. (9/55)
View Answer
Ans : C
Explanation: Let S be the sample space.
Then, n(S) = number of ways of drawing 3 balls out of 12 = 12C3 = 220
Let E = event of getting all the 3 purple balls.
n(E) = 4C3= 4
P(E) = n(E)/n(S) = 4/220 = 1/55
6. 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
7. one rupee coin is tossed twice. What is the probability of getting two consecutive heads ?
A. (1/2)
B. (1/3)
C. (1/4)
D. (4/3)
View Answer
Ans : C
Explanation: Probability of getting a head in one toss = 1/2
The coin is tossed twice.
So 1/2 * 1/2 = 1/4 is the answer.
Here’s the verification of the above answer with the help of sample space.
When a coin is tossed twice, the sample space is {(H,H), (H,T), (T,H), (T,T)}.
Our desired event is (H,H) whose occurrence is only once out of four possible outcomes and hence, our
answer is 1/4.
8. 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
9. A box contains 5 cone and 4 chocobar ice-creams. Preethi eats 3 of them, by randomly choosing. What
is the probability of choosing 1 chocobar and 2 cone
ice-creams?
A. (10/63)
B. (20/63)
C. (30/63)
D. (40/63)
View Answer
Ans : A
Explanation: Probability of choosing 1 cone= 5/9
After taking out 1 cone, the total number is 8 .
Probability of choosing 2nd cone = 4/8
Probability of choosing 1chocobaricecream out of a total of 7 = 4/7
So the final probability of choosing 2 cone and 1chocobar ice cream = 5/9*1/2*4/7 =10/63
10. 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
11. What is the probability of drawing a jack and a queen consecutively from a deck of 52 cards, without
replacement?
A. (1/663)
B. (2/663)
C. (3/663)
D. (4/663)
View Answer
Ans : D
Explanation: Probability of drawing a jack = 4/52 = 1/13
After drawing one card, the number of cards are 51.
Probability of drawing a queen = 4/51.
Now, the probability of drawing a jack and queen consecutively is 1/13 * 4/51 = 4/663
12. 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
13. In MSM college, 35% of the students study Tamil and English. 40% of the students study English. What is
the probability of a student studying Tamil given he/she is already studying english?
A. 0.75
B. 0.815
C. 0.825
D. 0.875
View Answer
Ans : D
Explanation: P(T and E) = 0.35
P(E) = 0.40
P(T/E) = P(T and E)/P(E) = 0.35/0.40 = 0.875
14. 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
15. A cartoon contains 15 torch lights out of which 3 are defective. Two torch light are chosen at random from
this cartoon. The probability that at least one of these is defective is.
A. (11/35)
B. (12/35)
C. (13/35)
D. (14/35)
View Answer
Ans : C
Explanation: P (none is defective) = 12c2/15c2
= (12*11/2*1)/(15*14/2*1)
= 66/105=22/35
P (at least one is defective) = (1 – 22/35)
=13/35
16. 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
17. Three unbiased coins are tossed. What is the probability of getting at most two heads?
A. (5/8)
B. (6/8)
C. (3/8)
D. (7/8)
View Answer
Ans : D
Explanation: 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
18. 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
19. In a lottery, there are 10 prizes and 25 blanks. A lottery is drawn at random. What is the probability of getting a prize?
A. (2/7)
B. (3/7)
C. (4/7)
D. (5/7)
View Answer
Ans : A
Explanation: P (getting a prize) =10/35 => 2/7
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
Discussion