Probability Ques 46
If $M$ and $N$ are any two events, then the probability that exactly one of them occurs is
(a) $P(M)+P(N)-2 P(M \cap N)$
(b) $P(M)+P(N)-P(\overline{M \cup N})$
(c) $P(\bar{M})+P(\bar{N})-2 P(\bar{M} \cap \bar{N})$
(d) $P(M \cap \bar{N})-P(\bar{M} \cap N)$
(1984, 3M)
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Answer:
Correct Answer: 46.(a, c)
Solution:
Formula:
Set theoretical notation of probability and some important results:
- $P$ (exactly one of $M, N$ occurs)
$=P\{(M \cap \bar{N}) \cup(\bar{M} \cap N)\}=P(M \cap \bar{N})+P(\bar{M} \cap N)$
$=P(M)-P(M \cap N)+P(N)-P(M \cap N)$
$=P(M)+P(N)-2 P(M \cap N)$
Also, $P$ (exactly one of them occurs)
$=\{1-P(\bar{M} \cap \bar{N})\}\{1-P(\bar{M} \cup \bar{N})\}$
$=P(\bar{M} \cup \bar{N})-P(\bar{M} \cap \bar{N})=P(\bar{M})+P(\bar{N})-2 P(\bar{M} \cap \bar{N})$
Hence, (a) and (c) are correct answers.
BETA
