Question

If a b c are three experiment with a random experiment

Prove that
P ( A ∪ B ∪ C) = P(A) + P(B)+P(C) - P( A ∩ B) - P( B ∩ C) - P ( A ∩ C) + P ( A ∩ B ∩C) .​

Answer 1

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If a b c are three experiment with a random experiment

Prove that

P ( A ∪ B ∪ C) = P(A) + P(B)+P(C) - P( A ∩ B) - P( B ∩ C) - P ( A ∩ C) + P ( A ∩ B ∩C) .

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Given :-

A B C are three experiment with a random experiment .

To Proof :-

P ( A ∪ B ∪ C) = P(A) + P(B)+P(C) - P( A ∩ B) - P( B ∩ C) - P ( A ∩ C) + P ( A ∩ B ∩C) .

Proof :-

Consider E = B U C so that

P (AUBUC) = P (A U E)

=> P(A)+ P(E)- P(A⋂E)______{1}

Now,

P(E) = P(BUC)

=> P(B) + P(C) - P(B⋂C)_____{2}

Also, A⋂E = A⋂(BUC) = (A⋂B)U(A⋂C ) [Using

distribution property of intersection of set over the union.] Thus

P(A⋂E) = P(A⋂B) + P(A⋂C) - P [(A⋂B)⋂(A⋂C)]

=> P(A⋂B) + P(A⋂C) - P[A⋂B⋂]C ______{3}

Using 2 and 3 in 1, we get

=> P ( A ∪ B ∪ C) = P(A) + P(B)+P(C) - P( A ∩ B) - P( B ∩ C) - P ( A ∩ C) + P ( A ∩ B ∩C).

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