Question

For three events $A,B$ and $C,P($exactly one of the events $A$ or $B$ occurs$)=P($exactly one of the vents $B$ 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 $0<p<\frac{1}{2}$.
Then the probability of atleast one of the three events $A,B$ and $C$ occurring is

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

Answer 1

Answer: (A)

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

It is given that
$P(A\cap \overline{B})+P(\overline{A}\cap B)=p,P(B\cap \overline{C})+P(\overline{B}\cap C)=p,P(A\cap \overline{C})+P(\overline{A}\cap C)=p$
and $P(A\cap B\cap C)=p^2$

We have to find $P(A\cup B\cup C)$

Now, $P(A\cap \overline{B})+P(\overline{A}\cap B)=p\Rightarrow P\left(A\right)+P\left(B\right)-2P(A\cap B)=p$ ...(1)
Similarly $P(B\cap \overline{C})+P(\overline{B}\cap C)=p\Rightarrow P\left(B\right)+P\left(C\right)-2P(B\cap C)=p$ ...(2)

and, $P(A\cap \overline{C})+P(\overline{A}\cap C)=p\Rightarrow P\left(A\right)+P\left(C\right)-2P(A\cap C)=p$ ...(3)

Adding (1),(2) and (3), we get
$2[P\left(A\right)+P\left(B\right)+P\left(C\right)-P(A\cap B)-P(B\cap C)-P(A\cap C)]=3p$

$\Rightarrow P\left(A\right)+P\left(B\right)+P\left(C\right)-P(A\cap B)-P(B\cap C)-P(A\cap C)=\frac{3p}{2}$

So, required probability
$P(A\cup B\cup C)=P\left(A\right)+P\left(B\right)+P\left(C\right)-P(A\cap B)-P(B\cap C)$

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