Question

A student cannot qualify for the interview if he fails in subjects A and B. The probabilities that he fails in A and B are known to be 0.01 and 0.03 respectively. It is also known that he is more likely to fail in subject B with probability 0.06 if he failed in A. What is the probability that he (i) cannotqualify for the interview and (ii) fails in A given that he failed in B.

Answer 1

(i) Probability that he cannot qualify for the interview is 0.0006.

(ii) Probability that he fails in A given that he failed in B is 0.02.

Step-by-step explanation:

We are given that a student cannot qualify for the interview if he fails in subjects A and B.

The probabilities that he fails in A and B are known to be 0.01 and 0.03 respectively. It is also known that he is more likely to fail in subject B with probability 0.06 if he failed in A.

Let the Probability that student failed in subject A = P(A) = 0.01

Probability that student failed in subject B = P(B) = 0.03

Also, Probability that student fail in subject B give that he failed in A = P(B/A) = 0.06

(i) Probability that he cannot qualify for the interview = Probability that he failed in both subjects A and B = $P(A \bigcap B)$

Now, As we know that the conditional probability is given by;

             P(B/A) =  $\frac{P(A \bigcap B)}{P(A)}$

So,  $P(A \bigcap B)$  =  $P(B/A) \times P(A)$

                        =  $0.06 \times 0.01$

                        = 0.0006

Hence, Probability that he cannot qualify for the interview is 0.0006.

(ii) Probability that he fails in A given that he failed in B  =  P(A/B)

Since,   P(A/B) =  $\frac{P(A \bigcap B)}{P(B)}$

                        =  $\frac{0.0006}{0.03}$

                        =  0.02

Hence, probability that he fails in A given that he failed in B is 0.02.