Question

If A, B. C are three pain-wise independent events such that P(A BC) = 0 then P((B° ~ C°VA) is equal to -

Answer 1

A, B , C are three pain wise independent events such that P(A ∩ B ∩ C) = 0 .

we have to find the value of P(B^c ∩ C^c/A)

solution : according to Bayes’ theorem,

P(X/Y) = P(X ∩ Y)/P(Y)

so, P(B^c ∩ C^c/A) = P(A ∩ (B^c ∩ C^c))/P(A)

= [P(A) - {P(A ∩ B) + P(A ∩ C) - P(A ∩ B ∩ C)}]/P(A)

= [P(A) - P(A ∩ B) - P(A ∩ C) + 0}/P(A) [ given P(A ∩ B ∩ C) = 0 .]

= 1 - P(A ∩ B)/P(A) - P(A ∩ C)/P(A)

= 1 - P(A).P(B)/P(A) - P(A).P(C)/P(A)

={ 1 - P(B)} - P(C) or, {1 - P(C)} - P(B)

= P(B^c) - P(C) or, P(C^c) - P(B)

Therefore the value of P(B^c ∩ C^c/A) = P(C^c) - P(B) [ as it is mentioned in option ]