Probability

29 April 2020


We have provided the selective and important  Probablity questions and answers for various Aptitude tests and competitive exams. Try to solve maximum questions to increase your performance in exams.

probability is a number that reflects the chance or likelihood that a particular event will occur. It tells about the chance or likelihood of the occurrence of a particular event.


Practice Questions For Probability

Q1) A bag contains 2 yellow, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?

A) 10/21

B) 9/11

C)  1/2

D) 7/11

Solution :


Q2) A die is rolled twice. What is the probability of getting a sum equal to 9?

A) 2/3

B) 2/9

C) 1/9

D) 1/3

Solution :


Q3) Three coins are tossed. What is the probability of getting at most two tails?

A) 1/2

B) 7/8

C) 1/8

D) 1/7

Solution :


Clear all your Aptitude Concepts from Experts with Aptitude Cracker Course!-https://bit.ly/2YFn39h

Q4) When tossing two coins once, what is the probability of heads on both the coins

A) None of these

B) 1/4

C) 1/2

D) 3/4

Solution :


Q5) Sam and Joan are playing a tennis match. If the probability of Sam's win is 0.59, then find the probability of Joan's win. 

A) 0.47 

B) 0.36 

C) 0.41 

D) 0.25 

Solution: 

Let event A = Sam wins and event B = Joan wins. Then, 
P(A) = 0.59 
Since if Sam wins, Joan cannot win and if Joan wins, Sam cannot win, so we can say that the 
events A and B are mutually exclusive. Other than these two events, there are 
no any other possible outcomes. So, 
P(A)+P(B) = 1 
0.59+P(B) = 1 
P(B) = 1-0.59 = 0.41 

Q6) Let A and B be events on the same sample space, with P (A) = 0.6 and P (B) = 0.7. Can these two events be disjoint? 

A) Yes 

B) No 

C) None of the above 

D) All the above 

Solution:

These two events cannot be disjoint because P(A)+P(B) >1. 
P(A?B) = P(A)+P(B)-P(A?B). 
An event is disjoint if P(A?B) = 0. If A and B are disjoint P(A?B) = 0.6+0.7 = 1.3 
And Since probability cannot be greater than 1, these two mentioned events cannot be disjoint. 

Q7) A family has two children. find the probability that both the children are girls given that at least one of them is a girl? 

A) 1/4 

B) 2/3 

C) 1/3 

D) 1/4 

Solution: 

Let b stand for boy and g for girl. The sample space of the experiment is 
S = {(g, g), (g, b), (b, g), (b, b)} 
Let E and F denote the following events : 
E : ‘both the children are girls’ 
F : ‘at least one of the child is a girl’ 
Then E = {(g,g)} and F = {(g,g), (g,b), (b,g)} 
Now E n F = {(g,g)} 
Thus P(F) = 3/4 
and P (E n F )= 1/4 
Therefore P(E|F) = P(E n F)/P(F) = (1/4)/(3/4) = 1/3 

Looking for ALL IN ONE placement preparation package ? Join our Complete Placement Package! -https://bit.ly/2WC7tIL

Q8) 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) 1/10 

B) 2/5 

C) 2/7 

D) 5/7 

Solution

P(getting a prize) = 10 / (10+25) 
‹=› 10 / 35 
‹=› 2 / 7. 

Q9) Determine the probability that a digit chosen at random from the digits 1, 2, 3, …12 will be odd. 

A) 1/2

B) 1/9 

C) 5/9

D) 4/9 

Solution: 

Total no. of Digits = 12. Equally likely cases = 12. 
There are six odd digits. Probability = 6 / 12 = 1 / 2 

Q10) Tickets numbered 1 to 20 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 3 or 5? 

A) 1/2 

B) 2/5 

C) 8/15 

D) 9/20 

Solution: 

Here, S = {1, 2, 3, 4, ...., 19, 20}. 
Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}. 
P(E) = n(E)/n(S) = 9/20. 

To test yourself,  click on the FREE TEST SERIES provided by Talent Battle - https://bit.ly/3dmvTNa 

Ask Us Anything !