Question

An experiment can result in only 3 mutually exclusive events A, B and C. If P(A) = 2P(B) =3P(C), then P(A) =
[a] $\dfrac{6}{11}$
[b] $\dfrac{5}{11}$
[c] $\dfrac{9}{11}$
[d] None of the above

Answer 1

Use the fact that if A and B are mutually exclusive events then $P\left( A\bigcup B \right)=P\left( A \right)+P\left( B \right)$. Use the fact that the probability of Sample space is 1, i.e. P(S) = 1. Since A, B and C are exhaustive use the fact that $S=A\bigcup B\bigcup C$ and hence determine the value of P(A), P(B) and P(C). Alternative use law of total probability which states that if A, B and C are mutually exclusive and exhaustive events then $P\left( E \right)=P\left( E|A \right)P\left( A \right)+P\left( E|B \right)P\left( B \right)+P\left( E|C \right)P\left( C \right)$. Hence determine the value of P(A), P(B) and P(C).

Complete step-by-step answer :
We know that if ${{A}_{1}},{{A}_{2}},\cdots ,{{A}_{n}}$ are mutually exclusive events, then $P\left( \bigcup\limits_{i=1}^{n}{{{A}_{i}}} \right)=\sum\limits_{i=1}^{n}{P\left( {{A}_{i}} \right)}$
Since A, B, C are mutually exclusive events, we have
$P\left( A\bigcup B\bigcup C \right)=P\left( A \right)+P\left( B \right)+P\left( C \right)$
Also, since A, B and C are exhaustive events, we have
$A\bigcup B\bigcup C=S$
Hence, we have
$P\left( S \right)=P\left( A \right)+P\left( B \right)+P\left( C \right)$
We know that the Sample space is a sure event.
Hence, we have
$P\left( A \right)+P\left( B \right)+P\left( C \right)=1$
Let P(A) = 6x
Hence, we have
2P(B) = 6x i.e. P(B) = 3x and 3P(C) = 6x i.e. P(C) = 2x
Hence, we have
$\begin{aligned} & 6x+3x+2x=1 \\\ & \Rightarrow 11x=1 \\\ & \Rightarrow x=\dfrac{1}{11} \\\ \end{aligned}$
Hence, we have
$P\left( A \right)=6x=\dfrac{6}{11}$
Hence option [a] is correct.

Note : Alternative solution:
We know that if ${{A}_{1}},{{A}_{2}},\cdots ,{{A}_{n}}$ are mutually exclusive and exhaustive events and E is any event, then $P\left( E \right)=\sum\limits_{i=1}^{n}{P\left( E|{{A}_{i}} \right)P\left( {{A}_{i}} \right)}$
Hence, we have
$P\left( S \right)=P\left( S|A \right)P\left( A \right)+P\left( S|B \right)P\left( B \right)+P\left( S|C \right)P\left( C \right)$
We know that $P\left( S|E \right)=1\forall E\subset S,P\left( E \right)\ne 0$
Hence, we have
$\begin{aligned} & 1=1\times P\left( A \right)+1\times P\left( B \right)+1\times P\left( C \right) \\\ & \Rightarrow P\left( A \right)+P\left( B \right)+P\left( C \right)=1 \\\ \end{aligned}$
Proceeding similarly as above, we have
$P\left( A \right)=\dfrac{6}{11}$
Hence option [a] is correct.