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 probabilty 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 – 39295 – 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?

Solution

Let event X – he will be selected in firm X;

Event Y – he will be selected in firm Y.

Then we have: $P(X) = 0.7$, $P(\bar{X}) = 0.3$, $P(\bar{Y}) = P(Y) = 0.5$, $P(\bar{X} \cup \bar{Y}) = 0.6$.

Using following equalities

$$P (\bar {X} \cup \bar {Y}) = P (\bar {X}) + P (\bar {Y}) - P (\bar {X} \cap \bar {Y}),$$

And

$$P (X \cup Y) = 1 - P (\overline {{X \cup Y}}) = 1 - P (\bar {X} \cap \bar {Y}).$$

We obtain

$$0.6 = 0.3 + 0.5 - P (\bar {X} \cap \bar {Y}),$$ $$P (\bar {X} \cap \bar {Y}) = 0.2.$$

From the second equality

$$P (X \cup Y) = 1 - 0.2 = 0.8.$$

Answer: 0.8.