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For the three events $A, B$ and $C, P$ (exactly one of the events $A$ or $B$ occurs) = $P$ (exactly one of the events $B$ or $C$ occurs)= $P$ (exactly one of the events $C$ or $A$ occurs)= $p$ and $P$ (all the three events occur simultaneously) $ = {p^2},$ where $0 < p < 1/2$. Then the probability of at least one of the three events $A, B$ and $C$ occurring is
$\frac{{3p + 2{p^2}}}{2}$
$\frac{{p + 3{p^2}}}{4}$
$\frac{{p + 3{p^2}}}{2}$
$\frac{{3p + 2{p^2}}}{4}$
Solution
(a) We know that $P$ (exactly one of $A$ or $B$ occurs)
= $P(A) + P(B) – 2P(A \cap B)$
Therefore, $P(A) + P(B) – 2P(A \cap B) = p$…..$(i)$
Similarly, $P(B) + P(C) – 2P(B \cap C) = p$…..$(ii)$
and $P(C) + P(A) – 2P(C \cap A) = p$…..$(iii)$
Adding $(i),$ $(ii)$ and $(iii),$ we get
$P(A) + P(B) + P(C) – P(A \cap B) – P(B \cap C) – P(C \cap A) = \frac{{3p}}{2}$
…..$(iv)$
We are also given that $P(A \cap B \cap C) = {p^2}$ …..$(v)$
Now, $P$ (at least one of $A,\,\,B$ and $C)$
$ = P(A) + P(B) + P(C) – P(A \cap B) – P(B \cap C)$
$ – P(C \cap A) + P(A \cap B \cap C)$
$ = \frac{{3p}}{2} + {p^2}$, [By $(iv)$ and $(v)$].
