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