Question

Prove that P(A) = `"P"("A" ∩ "B") + "P"("A" ∩ bar"B")`

Answer 1

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