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Q.

An electric component manufactured by a company is tested for its defectiveness by a sophisticated device. Let ‘A’ denote the event “the device is defective” and ‘B’ the event “the testing device reveals the component to be defective” Suppose P(A)=α and P(BA)=P(B¯A¯)=1α . Where 0<α<1 . If it is given that the testing device reveals it to be defective, then the probability that the component is not defective is 

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a

14

b

34

c

0.7

d

0.5

answer is D.

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Detailed Solution

P(A¯B)=P(A¯).P(B/A¯)P(A).P(B/A)+P(A¯).P(B/A¯)             =(1α)αα(1α)+(1α)α=12

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