Question

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 occurs simultaneously) =$p^2, \,$ where $\, \, 0 < p < \frac{1}{2}$ Then, the probability of at least one of the three events $A, B$ and $C$ occurring is

• A. $\frac{3p+2p^2}{2}$
• B. $\frac{p+3p^2}{4}$
• C. $\frac{p+3p^2}{2}$
• D. $\frac{3p+2p^2}{4}$

Answer 1

Answer: (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 ) - 2 P ( B \cap C ) = p $..(iI)
and $ P(C) + P(A) - 2P(C \cap A) = p $.(iii)
On adding Eqs. (i), (ii) and (iii), we get
$2 [P(A) + P(B) + P(C) - P(A \cap B) $
$ -P(B \cap C) -P(C \cap A)] =3p$
$\Rightarrow P(A) + P(B) + P(C) - P(A \cap B) $
$-P(B \cap C)-P(C \cap A)=\frac{3p}{2}$ ....(v)
It also given that, $P(A \cap B \cap C) =p^2) $...(iv)
$\therefore P$ (at least one of the events $A, B$, and $C$ occurs)
$ =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 $ [from Eqs. (iv) and (v)]
$ =\frac{3p+2p^2}{2} $