Question

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

• A. $P(M) + P(N) - 2P(M \cap N)$
• B. $P(M) + P(N) - P(\overline{(M \cup N)}$
• C. $P(\overline{M})+P(\overline{N})-2P(\overline{M} \cap \overline{N})$
• D. $P(M \cap \overline{N})-P(\overline{M}\cap N)$

Answer 1

Answer: (C)

$P(\overline{M})+P(\overline{N})-2P(\overline{M} \cap \overline{N})$

Answer 2

P(exactly one of M, N occurs)
$=P \\{(M \cap \overline{N}) \cup ( \overline{M} \cap N \\} = P(M \cap\overline{N})+P(\overline{M} \cap N)$
$= P(M) - P(M \cap N ) + P (N ) - P(M \cap N )$
$= P(M ) + P(N) - 2P(M \cap N )$
Also, P(exactly one of them occurs)
=$\\{ 1-P(\overline{M} \cap \overline{N})\\} \\{1-P(\overline{M} \cup \overline{N})\\}$
$= P(\overline{M} \cup \overline{N})-P(\overline{M} \cap \overline{N})=P(\overline{M})+P(\overline{N})-2P(\overline{M}\cap \overline{N})$
Hence, (a) and (c) are correct answers.