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 #39871 – Math – Statistics and Probability

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?

A="selected in firm X", B="selected in firm Y", C="selected at least in one of the firms",

$D = (\bar{A}\cap B)\cup (A\cap \bar{B})\cup (\bar{A}\cap \bar{B}) = "$ at least one of his applications is rejected".

By assignment statement, $\Pr (D) = 0.6$. Then

$$\Pr (A \cap B) = 1 - \Pr (D) = 1 - 0.6 = 0.4$$

Probability that at least one of applications will be selected:

$$\Pr (C) = \Pr (A \cup B) = \Pr (A) + \Pr (B) - \Pr (A \cap B) = 0.7 + 0.5 - 0.4 = 0.8$$

Probability that exactly one of application will be selected

$$\Pr (E) = \Pr ((A \cap \bar{B}) \cup (\bar{A} \cap B)) = \Pr (A \cap \bar{B}) + \Pr (\bar{A} \cap B) = \Pr (A / \bar{B}) \Pr (\bar{B}) + \Pr (\bar{A} / B) \Pr (B)$$

We cannot calculate this probability without additional assumptions. If we assume that events A and B are independent, then $\Pr(A / \bar{B}) = \Pr(A)$, $\Pr(\bar{A} / B) = \Pr(\bar{A})$.

Finally, with additional assumption we have that probability that exactly one of application will be selected

$$\Pr (E) = \Pr (A / \bar{B}) \Pr (\bar{B}) + \Pr (\bar{A} / B) \Pr (B) = \Pr (A) \cdot \Pr (\bar{B}) + \Pr (\bar{A}) \cdot \Pr (B) = 0.7 \cdot 0.5 + 0.3 \cdot 0.5 = 0.5$$