Engineering
Mathematics
Conditional Probability
Question

Let A, B, C be pairwise independent even such that P(C) > and P(A⋂B⋂C) = 0, then P(A'⋂B'⋂C) is equal to 

P(A) – P(B')

P(A') + P(B')

P(A') – P(B')

P(A') – P(B)

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Solution
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PAcBcC=PAcBCCP(C)

=P((1A)(1B)C)P(C)

=P[(1AB+AB)C]P(C)

=P(C)P(AC)P(CB)+P(ABC)P(C)

=P(C)P(C)P(A)P(C)P(B)+P(A)P(B)P(C)P(C)

=P(C)P(C)P(A)P(C)P(B)+P(ABC)P(C)

1 – P(A) – P(B)

P(Ac) – P(B)

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