Question

In throwing a die, let $A$ be the event 'an odd number turns up', $B$ be the event 'a number divisible by $3$ turns up' and $C$ be the event 'a number $\le 4$ turns up', then the probability that exactly one of events $A,B,C$ occur is :

• A. $1$
• B. $\frac{1}{6}$
• C. $\frac{1}{3}$
• D. $\frac{2}{3}$

Answer 1

The correct option is D $\frac{2}{3}$

Event $A=\{1,3,5\},B=\{3,6\},C=\{1,2,3,4\}$
$\Rightarrow A\cap B=\left\{3\right\},B\cap C=\left\{3\right\},A\cap C=\{1,3\}$ and $A\cap B\cap C=\left\{3\right\}$
So, P (exactly one of $A,B,C$) $=P\left(A\right)+P\left(B\right)+P\left(C\right)-2P(A\cap B)-2P(B\cap C)-2P(A\cap C)+3P(A\cap B\cap C)$
$=\frac{3}{6}+\frac{2}{6}+\frac{4}{6}-2\cdot \frac{1}{6}-2\cdot \frac{1}{6}-2\cdot \frac{2}{6}+3\cdot \frac{1}{6}=\frac{2}{3}$