Question

A Chartered accountant applied for a job in two firms X and Y. The probability of his being selected in firm X is 0. 7 and being rejected at Y is 0.5 and the probability of at least one of his applications being rejected is 0.6. What is the probability that he will be selected at least one of the firms?

Options

• 0.8

• 0.2

• 0.4

• 0.7

Answer 1

0.8
Explanation:
Given  `P(X) = 0.7, P(barY)= 0.5`

⇒`P(barX)= 0.3,  P(Y)=0.5`

Also, `P(barX∪barY)=0.6`

⇒ 1 - P(X∩Y) = 0.6

⇒ P(X∩Y) = 0.4

∴ Required probability, P(X∪Y)

= P(X) + P(Y) - P(X∩Y)

= 0.7 + 0.5 - 0.4 = 0.8