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 $P\left(A\right)=$

• A. $\frac{6}{11}$
• B. $\frac{5}{11}$
• C. $\frac{9}{11}$
• D. None

Answer 1

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

Given that $P\left(A\right)=2P\left(B\right)=3P\left(C\right)$

Let $P\left(C\right)=x$ , we get $P\left(B\right)=\frac{3}{2}x$ and $P\left(A\right)=3x$

Given that there are only these three events.

So we have $P\left(A\right)+P\left(B\right)+P\left(C\right)=1$

$\Rightarrow 3x+\frac{3x}{2}+x=1$

$\Rightarrow x=\frac{2}{11}$

Therefore the value of $P\left(A\right)=3x=\frac{6}{11}$

So the correct option is $A$.