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 $C$ or $A$ occurs) $=P$(exactly one of the events $B$ or $C$ occurs) $=p$ and $P$( all three events occur simultaneously) = $p^2$, where $0<p<\frac{1}{2}$. Then find the probability of atleast one of the three events $A,B$ and $C$ occurring.

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

Answer 1

The correct option is A $\frac{3p+2p^2}{2}$

According to the question,
$P\left(A\right)+P\left(B\right)-2P(A\cap B)=P\left(B\right)+P\left(C\right)-2P(B\cap C)=P\left(A\right)+P\left(C\right)-2P(A\cap C)=p$.
Adding, we get, $2\left(P\right(A)+P(B)+P(C)-P(A\cap B)-P(B\cap C)-P(A\cap C\left)\right)=3p$ $\therefore P\left(A\right)+P\left(B\right)+P\left(C\right)-P(A\cap B)-P(B\cap C)-P(A\cap C)=\frac{3p}{2}$.

Also $P(A\cap B\cap C)=p^2$
$\therefore$ $P(A\cup B\cup C)=P\left(A\right)+P\left(B\right)+P\left(C\right)-P(A\cap B)-P(B\cap C)-P(A\cap C)+P(A\cup B\cup C)$

$=\frac{3p}{2}+p^2=\frac{3p+2p^2}{2}$