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, $P(B)+P(C)-2P(B\cap C)=p$..(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)]
$=\frac{3p+2p^2}{2}$