Question

A chartered Accountant applies for a job in two firms X and Y. He estimates that the probability of his being selected in firm X is 0.7 and being rejected in Y is 0.5 and the probability that atleast one of his applications rejected is 0.6. What is the probability that he will be selected in one of the firms?

Answer 1

Answer on Question #39547 – Math – Other

A chartered Accountant applies for a job in two firms X and Y. He estimates that the probability of his being selected in firm X is 0.7 and being rejected in Y is 0.5 and the probability that at least one of his applications rejected is 0.6. What is the probability that he will be selected in one of the firms?

Solution:

Let

$p(x) = 0.7$ - the probability that a chartered accountant will be selected in firm X

$p(y) = 0.5$ - the probability that a chartered accountant will be selected in firm Y

and, using $p(x) = 1 - q(x)$,

$q(x) = 0.3$ - the probability that a chartered accountant will be rejected in firm X

$q(y) = 0.5$ - the probability that a chartered accountant will be rejected in firm Y.

Then the probability that a chartered accountant will be selected in one of the firms X or Y

$p(x \cup y) = p(x) + p(y) - p(x \cap y)$.

From the conditions of the problem

$q(x \cup y) = 0.6$ - the probability that at least one of a chartered accountant’s applications rejected.

We get

$p(x \cap y) = 1 - q(x \cup y) = 1 - 0.6 = 0.4.$

From the equation above

$p(x \cup y) = 0.7 + 0.5 - 0.4 = 0.8.$

Answer: 0.8.