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 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

• 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 $\frac{3p+2p^2}{2}$

We know that
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 (1), (2) and (3) we get
$2\left[P\right(A)+P(B)+P(C)-P(A\cap B)-P(B\cap C)-P(C\cap A\left)\right]=3p$
$\Rightarrow P\left(A\right)+P\left(B\right)+P\left(C\right)-P(A\cap B)-P(B\cap C)-P(C\cap A)=3p/2$ (4)
We are also given that $P(A\cap B\cap C)=p^2$ (5)
Now, P (at least one of $A,B$ and $C$)
$=P\left(A\right)+P\left(B\right)+P\left(C\right)-P(A\cap B)-P(B\cap C)-P(C\cap A)+P(A\cap B\cap C)$

$=\frac{3p}{2}+p^2=\frac{3p+2p^2}{2}$ [using (4) and (5)]