Question

A and B are two events such that P(A) ≠ 0. Find P (B|A), if
(i) A is a subset of B (ii) A ∩ B = ϕ

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Solution

(i) Since A is a subset of B

therefore space space space space space straight A space intersection thin space straight B space equals space straight A

...(1)
Now

straight P left parenthesis straight B space left enclose straight A right parenthesis end enclose space equals space fraction numerator straight P left parenthesis straight B intersection straight A right parenthesis over denominator straight P left parenthesis straight A right parenthesis end fraction space equals space fraction numerator straight P left parenthesis straight A intersection straight B right parenthesis over denominator straight P left parenthesis straight A right parenthesis end fraction space equals space fraction numerator straight P left parenthesis straight A right parenthesis over denominator straight P left parenthesis straight A right parenthesis end fraction

open square brackets because space of space left parenthesis 1 right parenthesis close square brackets

therefore space space space space straight P left parenthesis straight B space left enclose straight A right parenthesis end enclose space equals space 1

(ii)

straight P left parenthesis straight B space left enclose straight A right parenthesis end enclose space equals space fraction numerator straight P thin space left parenthesis straight B intersection straight A right parenthesis over denominator straight P left parenthesis straight A right parenthesis end fraction space equals space fraction numerator straight P left parenthesis straight A intersection straight B right parenthesis over denominator straight P left parenthesis straight A right parenthesis end fraction

equals space fraction numerator straight P left parenthesis straight ϕ right parenthesis over denominator straight P left parenthesis straight A right parenthesis end fraction space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because space space space straight A space intersection straight B space equals straight ϕ space left parenthesis given right parenthesis close square brackets space equals space fraction numerator 0 over denominator straight P left parenthesis straight A right parenthesis end fraction space equals space 0

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