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show that theorem 2.6, the additive law of probability, holds for conditional probabilities. that is, if a, b, and c are events such that p(c) > 0, prove that p(a ∪ b|c)



Answer :

P(AUB|C)=P(A|C) + P(B|C) - P(AnB|C) is the correct answer.

From distributive law we have,

P((AUB)nC)=P(AnC) + P(BnC)-P(AnBnC)

Dividing both sides by P(C)

P((AUB)nC)/P(C)=P(AnC)/P(C) + P(BnC)/P(C) - P(AnBnC)/P(C)

P(AUB|C)=P(A|C) + P(B|C) - P(AnB|C)

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