If $M$ and $N$ are any two events, the probability that the exactly one of them occurs is

P (M) + P (N) - 2 P (M \cap N)

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P (M) + P (N) - P {(M \cap N)}^{c}

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P (M^{c}) + P (N^{c}) - 2 P (M^{c} \cap N^{c})

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P (M \cap N^{c}) + P (M^{c} \cap N) .

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Solution

The correct options are
A $P (M) + P (N) - 2 P (M \cap N)$
C $P (M^{c}) + P (N^{c}) - 2 P (M^{c} \cap N^{c})$
D $P (M \cap N^{c}) + P (M^{c} \cap N) .$
$P$ (exactly one of $M, N$ occurs)
$= P {(M \cap N^{C}) \cup (M^{C} \cap N)} = P (M \cap N^{C}) + (M^{C} \cap N)$
$= P (M) - P (M \cap N) + P (N) - P (M \cap N) = P (M) + P (N) - 2 P (M \cap N)$
also, $P$ (exactly one of them occurs)
$= {1 - P (M^{C} \cap N^{C})} {1 - P (M^{C} \cup N^{C})}$
$= P (M^{C} \cup N^{C}) - P (M^{C} \cap N^{C}) = P (M^{C}) + P (N^{C}) - 2 P (M^{C} \cap N^{C})$

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Similar questions

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If M and N are any two events, the probability that the exactly one of them occurs is

If $M$ and $N$ are any two events, the probability that the exactly one of them occurs is