ProbabilityHard
Question
If M and N are any two evets, then the probability that exactly one of them occurs is
Options
A.P(M) + P(N) - 2P(M ∩ N)
B.P(M) + P(N) - P
C.P(
) + P(
) - 2P( ∩ )
) + P() - 2P( ∩ )D.P(M ∩ ) - P( ∩ N)
) - P( ∩ N)Solution
P(exactly one of M, Noccurs)
= P{(M ∩) ∪ ( ∩ N)}
= P(M ∩) ∪ P( ∩ N)
= P(M) - P(M ∩ N) + P(N) - P(M ∩ N)
= P(M) + P(N) - 2P(M ∩ N)
Also, P (exactly one of them occurs)
= {1- P( ∩ } {1 - P( ∪ )}
= P( ∪ ) - P( ∩ )
= P() + P() - 2P( ∩ )
Hence, (a) and (c) are correct answers.
= P{(M ∩
) ∪ ( ∩ N)}= P(M ∩
) ∪ P( ∩ N)= P(M) - P(M ∩ N) + P(N) - P(M ∩ N)
= P(M) + P(N) - 2P(M ∩ N)
Also, P (exactly one of them occurs)
= {1- P(
∩ } {1 - P( ∪ )}= P(
∪ ) - P( ∩ )= P(
) + P() - 2P( ∩ ) Hence, (a) and (c) are correct answers.
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