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Question

An MBM applies 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. The probability of at least one of his applications being rejected is 0.6. The probability that he will be selected in one of the firms, is

A
0.6
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B
0.4
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C
0.8
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D
None of these
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Solution

The correct option is C 0.8
Let A and B denote the events that the person is selected in firms X and Y respectively.
Then in the usual notations, we are given:
P(A)=0.7P(¯¯¯¯A)=10.7=0.3P(¯¯¯¯B)=0.5P(B)=10.5=0.5
and P(¯¯¯¯A¯¯¯¯B)=0.6
The probability that the persons will be selected in one of the two firms X or Y is given by:
P(AB)=1P(¯¯¯¯A¯¯¯¯B)=1(P(¯¯¯¯A)+P(¯¯¯¯B)P(¯¯¯¯A¯¯¯¯B))=1(0.3+0.50.6)=0.8

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