Question

For the three events $A,B\&C,$ probability of exactly one of the events $A$ or $B$ occurs = probability of exactly one of the events $C$ or $A$ occurs $=p$ $\&$ $P$ (all the three events occur simultaneously) $=p$${}^2$, where

• 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

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

$P$(exactly one of $A$ or $B$ occurs)
$=P\left(A\right)+P\left(B\right)-2P(A\cap B)$
Therefore, $P\left(A\right)+P\left(B\right)-2P(A\cap B)=p$ ...(1)
Similarly, $P\left(B\right)+P\left(C\right)-2P(B\cap C)=p$ ...(2)
and $P\left(C\right)+P\left(A\right)-2P(C\cap A)=p$ ...(3)
Adding $\left(1\right),\left(2\right)$ and $\left(3\right)$ and dividing by $2$
$\Rightarrow P\left(A\right)+P\left(B\right)+P\left(C\right)-P(A\cap B)-P(B\cap C)-P(C\cap A)=\frac{3p}{2}$ ...(4)
Now,$P$(at least one of $A,B$ and $C$)
$=\frac{3p}{2}+p^2=\frac{3p+{2p}^2}{2}$
Put this equal to $\frac{11}{18}$ and solve for $p$.