Question

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

Answer 1

To prove, P(A ∪ B) = `"P"("A" ∩ "B") + "P"("A" ∩ bar"B") + "P"(bar"A" ∩ bar"B")`

R.H.S. = `"P"("A")."P"("B") + "P"("A")."P"(bar"B") + "P"(bar"A")."P"("B")`

= P(A).P(B) + P(A)[1 – P(B)] + [1 – P(A)].P(B)

= P(A).P(B) + P(A) – P(A).P(B) + P(B) – P(A).P(B)

= P(A) + P(B) – P(A ∩ B)

= P(A ∪ B)

= L.H.S.

Hence proved.