Question

21. A student has applied for admission to two
management Institutes A and B. He has assessed
the chance of getting admission in A as 0.4 and B
as 0.6. He also believes that the chance of getting
admission in both as 0.2.
Find the probabilities of the following events :
(a)
(b)
(c)
He will be successful in getting admission.
He will get admission in only one of the two
institutes.
He will get admission in B, given that he has
been admitted in A.
He will not get admission in A, given that he
has not been admitted in B.
(d)​

Answer 1

Answer:

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Answer 2

Correct Question

A student has applied for admission to two management Institutes A and B. He has assessed the chance of getting admission in A as 0.4 and B as 0.6. He also believes that the chance of getting admission in both is 0.2.

Find the probabilities of the following events :

(a) He will be successful in getting admission.

(b) He will get admission to only one of the two institutes.

(c) He will get admission to B, given that he has been admitted to A.

(d)​ He will not get admission to A, given that he has not been admitted to B.

Answer

The answers are a) 0.8 b) 0.8 c) 0.5 d) 0.5

Given

• A student has applied for admission to two management Institutes A and B.
• The chance of getting admission to A is 0.4.
• The chance of getting admission to B is 0.6.
• The chance of getting admission to both is 0.2.

To Find

1. He will be successful in getting admission.
2. He will get admission to only one of the two institutes.
3. He will get admission to B, given that he has been admitted to A.
4. ​ He will not get admission to A, given that he has not been admitted to B.

Let X be the probability of getting into institute A

Let Y be the probability of getting into institute B

Solution

Accordig to the probalem,

P(X) = 0.4

P(Y) = 0.6

P(X∩Y) = 0.2

a)

The probability that he is successful in getting an admision

= Probability that he got admitted in A or B

= P(X∪Y)

We know that P(T∪U) = P(T) + P(U) - P(T∩U)

Therefore, P(X∪Y) = P(X∪Y) = P(X) + P(Y) - P(X∩Y)

= 0.4 + 0.6 - 0.2

= 0.8

b)

The probability that he gets admitted to only one of the two

= 1 - Probability of him getting into both.

= 1 - P(X∩Y)

= 1 - 0.2

= 0.8

c)

Here, we need to find the probability of him getting admitted into B given he has been admitted into A.

This is a case of conditional probability.

We know that Conditional probability

P(T/U) = P(T∩U)/P(U)

Therefore P(Y/X) = P(X∩Y)/P(X)

= 0.2/0.4

= 1/2

= 0.5

d)

The probability that he will not get admission in A, given that he

has not been admitted in B = P(notX/notY)

= P(notX∩notY)/P(notY)

According to De Morgan's Law

P(notT∩notU) = 1 - P(T∪U)

Therefore,

P(notX∩notY)/P(notY) = (1 - P(X∪Y))/(1 - P(Y))

= (1 - 0.8)/0.4

= 0.2/0.4

= 1/2

= 0.5

Hence the answers are a) 0.8 b) 0.8 c) 0.5 d) 0.5

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