Question

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A charted accountant applies for a job in two finns .X and Y. lle estimates that 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 application being rejected is 0.6. Now the probability tha
Lar 0.6
(c) 0.3
he will be selected in at least one of the firm is
(b) 0.7
(d) 0.8​

Answer 1

Step-by-step explanation:

Note:

P(A)= probability that he gets selected in firm A; P(~A)= probability that he gets rejected at firm A.

All other symbols have the usual meaning.

Solution:

P(X)= 0.7 => P(~X)= 0.3

P(~Y)= 0.5 => P(Y)= 0.5

Now, it is given that P(~X U ~Y)= 0.6

=> P(~X)+P(~Y)-P(~X ⋂ ~Y)= 0.6

=> 0.3+0.5- P(~X ⋂ ~Y) =0.6

=> P(~X ⋂ ~Y)= 0.2

This means the probability that the person is rejected by both the firms is 0.2.

Hence, the probability that the person would be selected by at least one of the firm is 0.8