Question

Let $A,B$ & $C$ be $3$ arbitrary events defined on a sample space $S$ and if, $P\left(A\right)+P\left(B\right)+P\left(C\right)=\frac{1}{2},P(A\cap B)+P(B\cap C)+P(C\cap A)=\frac{1}{4}$ & $P(A\cap B\cap C)=\frac{1}{6}$, then the probability that exactly one of the three events occurs is

• A. $\frac{2}{3}$
• B. $\frac{1}{5}$
• C. $\frac{1}{4}$
• D. $\frac{1}{2}$

Answer 1

The correct option is D $\frac{1}{2}$

Probability that exactly one of event occurs
$=P\left(\text{only}A\right)+P\left(\text{only}B\right)+P\left(\text{only}C\right)$

$P\left(\text{only}A\right)=P\left(A\right)-P(A\cap B)-P(A\cap C)+P(A\cap B\cap C)$
$P\left(\text{only}B\right)=P\left(B\right)-P(A\cap B)-P(B\cap C)+P(A\cap B\cap C)$
$P\left(\text{only}C\right)=P\left(C\right)-P(A\cap C)-P(B\cap C)+P(A\cap B\cap C)$
$\therefore P\left(\text{only}A\right)+P\left(\text{only}B\right)+P\left(\text{only}C\right)=\left(P\right(A)-P(A\cap B)-P(A\cap C)+P(A\cap B\cap C\left)\right)+\left(P\right(B)-P(A\cap B)-P(B\cap C)+P(A\cap B\cap C\left)\right)+\left(P\right(C)-P(A\cap C)-P(B\cap C)+P(A\cap B\cap C\left)\right)$
$=\frac{1}{2}-2\cdot \frac{1}{4}+3\cdot \frac{1}{6}=\frac{1}{2}$