Q59 of 108 Page 271

If P(A) = 2|5, P(B) = 3|10 and P(A ∩ B) = 1|5, the P(A′|B′).P(B′|A′) is equal to

We have,


P(A’|B’) × P(B’) = P(A’ ∩ B’)


…(i)


P(B’|A’) × P(A’) = P(B’ ∩ A’)


…(ii)


On multiplying eq. (i) and (ii), we get




[∵P(A’ ∩ B’) = P[(A ∪ B)’]= 1 – P(A ∪ B)]




[Additive law of Probability]









Hence, the correct option is C

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