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.
A 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 :
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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
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.
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