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Asked Oct 3, 2019
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Prove that if two events, A and B, are mutually exclusive, then P(A | A U B ) = P(A) / ( P(A) + P(B) )

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Step 1

If two events are mutually exclusive, then the two events are disjoint event.

In other words, if A a...

P(An B) 0
We need to proof
P(A)
P(A)P(B)
A
P
AUB
From conditional probability
Р(MON)
P
P(N)
P(An (AU B))
P(AUB)
A
=> P
A U B
=> P(An (A U B)) P(A)
P(A) + P(B) P(A n B)= P(A) + P(B) 0 = P(A) + P(B)
=> P(AUB)
Using above
P(An (A U B))
P(A)
P(AUB)
P(A)P(B)
\A U B.
Hence proof
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Image Transcriptionclose

P(An B) 0 We need to proof P(A) P(A)P(B) A P AUB From conditional probability Р(MON) P P(N) P(An (AU B)) P(AUB) A => P A U B => P(An (A U B)) P(A) P(A) + P(B) P(A n B)= P(A) + P(B) 0 = P(A) + P(B) => P(AUB) Using above P(An (A U B)) P(A) P(AUB) P(A)P(B) \A U B. Hence proof

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