Probability Solved Questions For IBPS, SBI and Other Bank Exams - Page 4

You will find 14 problems in 4 pages..

Probability Problems Solved Questions (Page 4 of 4)

Question 1

In a school, 45% of the students play football, 30% play volleyball and 15% both. If a student is selected at random, then the probability that he plays football or volleyball is:

a) 2/7 b) 3/5 c) 1/4 d) 4/5

Answer : b) 3/5

Solution :

Given that, 45% play football; that is, P(F) = 45/100 = 9/20
30% play volley ball; that is, P(V) = 30/100 = 6/20

And, 15% play both volleyball and football; that is, P(F And V) = 15/100 = 3/20

Now, we have to find the probability that 1 student plays football or volley ball; that we have to find, P(F or V)

We know that, P(F Or V) = P(F) + P(V) - P(F And V)
= 9/20 + 6/20 - 3/20 = 12/20 = 3/5.

Hence, the required probability 3/5.

Question 2
Two groups, A and B wrote an exam. The probability of A's pass is 2/7 and the probability of B's pass is 2/5. What is the probability that only one of them is passed out?

a) 5/6 b) 1/3 c) 12/43 d) 18/35

Answer : d) 18/35

Solution :

Let A be the event of the group A pass
Let B be the event of the group B pass

Then, A'= Event of the group A's fail and B'= event of the group B's fail.

Therefore, p(A) = 2/7 and p(B) = 2/5,
P(A') = 1 - P(A) = 1- 2/7 = 5/7 and P(B') = 1- P(B) = 1- 1/5 = 4/5

Required probability = P[( A And B') Or (B And A')]
= P[( A And B') Or (B And A')]
= P[( A And B') + (B And A')]
= P[( A And B')] + p[(B And A')]
= p(A) x p(B') + P(A') x P(B)
= (2/7 x 4/5) + (2/5 x 5/7)
= (8/35 + 10/35) = 18/35

Question 3
A boy gets a chance of 55% to win 1st round of a game and a girl gets a chance of 60% to win 2nd round of the game. In what % of cases are they likely to contradict each other, narrating the same incident?

a) 49% b) 54% c) 71% d) 38%

Answer : a) 49%

Solution :

Let A be the event that a boy wins 1st round
Let B be the event that a girl wins 2nd round.

Then, A' = Event that the boy losses 1st round and B' = event that the girl losses 2nd round.
Therefore, P(A) = 55/100 = 11/20, P(B) = 60/100 = 12/20.
P(A') = 1- (11/20) = 9/20 and P(B') = 1- (12/20) = 8/20

First, we have to find the probability that they contradict each other;
That is, P( A And B contradicts each other)
= P[(boy win in 1st round And girl losses in 2nd round) (Or) (boy losses in 1st round And girl wins in 2nd round)
= P[(A And B') Or (A' And B)] = P[(A And B')] + p[(A' And B)]
= P(A) x P(B') + P(A') x P(B)
= 11/20 x 8/20 + 9/20 x 12/20
= 88/400 + 108/400
= 196/400

We have to find the %.
Required % = (196/400)x100 = 49%.

Probability Problems Solved Questions (Page 4 of 4)

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