Question

In throwing of a die, let $A$ be the event 'an odd humber 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 find the probability that exactly two of $A,B$ and $C$ occur.

• A. $\frac{2}{3}$
• B. $\frac{1}{6}$
• C. $\frac{1}{3}$
• D. $\frac{5}{6}$

Answer 1

Answer: (B)

$\frac{1}{6}$

Event $A=\{1,3,5\},$ event $B=\{3,6\}$ and event $C=\{1,2,3,4\}$
$\therefore$ $P$(exactly two of $A,B$ and $C$ occur)$=P(A\cap B)+P(B\cap C)+P(C\cap A)-3P(A\cap B\cap C)$
$=\frac{1}{6}+\frac{1}{6}+\frac{2}{6}-3\times \frac{1}{6}=\frac{1}{6}$