# 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. 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

3. 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

4. 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.

5. 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

6. 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

7. 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

8. 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

9. 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

10. 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

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