Question

An experiment can result in only 3 mutually exclusive events A, B and C. If $P\left(A\right)=2P\left(B\right)=3P\left(C\right),$ then $P\left(A\right)$ equals

• A. $\frac{5}{11}$
• B. $\frac{9}{11}$
• C. $\frac{8}{11}$
• D. $\frac{6}{11}$

Answer 1

The correct option is D $\frac{6}{11}$

since the events A,B and C are mutually exclusive
Hence $P(A\cap B)=P(B\cap C)=P(C\cap A)=P(A\cap B\cap C)=0AlsoP(A\cup B\cup C)=1$
$\because P(A\cup B\cup C)=P\left(A\right)+P\left(B\right)+P\left(C\right)-P(A\cap B)-P(B\cap C)-P(C\cap A)-P(A\cap B\cap C)\Rightarrow 1=P\left(A\right)+\frac{P\left(A\right)}{2}+\frac{P\left(A\right)}{3}\Rightarrow \frac{11}{6}P\left(A\right)=1\Rightarrow P\left(A\right)=\frac{6}{11}$