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Question
Prove that P(A) = `"P"("A" ∩ "B") + "P"("A" ∩ bar"B")`
Solution
To prove, P(A) = `"P"("A" ∩ "B") + "P"("A" ∩ bar"B")`
R.H.S. = `"P"("A" ∩ "B") + "P"("A" ∩ bar"B")`
= `"P"("A")."P"("B") + "P"("A")."P"(bar"B")`
= `"P"("A") ["P"("B") + "P"(bar"B")]`
= P(A) . 1
= P(A)
= L.H.S.
Hence proved
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