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) \[\dfrac{5}{11}\]
(B) \[\dfrac{9}{11}\]
(C) \[\dfrac{8}{11}\]
(D) \[\dfrac{6}{11}\]

Answer 1

Hint: It is given that A, B, and C are mutually exclusive events and \[P\left( A \right)=2P\left( B \right)=3P\left( C \right)\] . Now, split the equation \[P\left( A \right)=2P\left( B \right)=3P\left( C \right)\] into two equations, \[P\left( A \right)=2P\left( B \right)\] and \[2P\left( B \right)=3P\left( C \right)\] . Now, get the relation between \[P\left( A \right)\] and \[P\left( C \right)\]. Similarly also get the relation between \[P\left( B \right)\] and \[P\left( C \right)\] . We know the property that the summation of the probabilities of all the mutually exclusive events in an experiment is 1. So, \[P\left( A \right)+P\left( B \right)+P\left( C \right)=1\] . Now, using the relation between \[P\left( A \right)\] and \[P\left( C \right)\], and the relation between \[P\left( B \right)\] and \[P\left( C \right)\] , simplify the equation \[P\left( A \right)+P\left( B \right)+P\left( C \right)=1\] . Now, calculate the value of \[P\left( C \right)\] . Then, using the relation between \[P\left( A \right)\] and \[P\left( C \right)\], get the value of \[P\left( A \right)\] .

Complete step-by-step answer:
According to the question, it is given that an experiment can result in only 3 mutually exclusive events A, B, and C. Also,
\[P\left( A \right)=2P\left( B \right)=3P\left( C \right)\] ………………………………..(1)
Since the events A, B, and C are mutually exclusive so, these three events cannot occur at the same time.
We know the property that the summation of the probabilities of all the mutually exclusive events in an experiment is 1.
Here, the events A, B, and C are mutually exclusive. So, \[P\left( A \right)+P\left( B \right)+P\left( C \right)=1\] ……………………………(2)
From equation (1), we have \[P\left( A \right)=2P\left( B \right)=3P\left( C \right)\] .
Now, splitting equation (1), we get
\[P\left( A \right)=2P\left( B \right)\] ………………………………..(3)
\[2P\left( B \right)=3P\left( C \right)\] …………………………..(4)
Now, from equation (3) and equation (4), we get
\[P\left( A \right)=3P\left( C \right)\] ……………………..(5)
Transforming equation (4), we get
\[\Rightarrow 2P\left( B \right)=3P\left( C \right)\]
\[\Rightarrow P\left( B \right)=\dfrac{3P\left( C \right)}{2}\] ……………………………………(6)
From equation (2), we have \[P\left( A \right)+P\left( B \right)+P\left( C \right)=1\] .
Now, putting the values of \[P\left( A \right)\] from equation (5) and \[P\left( B \right)\] from equation (6), in equation (2), we get
 \[\begin{align}
  & \Rightarrow 3P\left( C \right)+\dfrac{3P\left( C \right)}{2}+P\left( C \right)=1 \\
 & \Rightarrow \dfrac{6P\left( C \right)+3P\left( C \right)+2P\left( C \right)}{2}=1 \\
 & \Rightarrow \dfrac{11P\left( C \right)}{2}=1 \\
\end{align}\]
\[\Rightarrow P\left( C \right)=\dfrac{2}{11}\] ………………………………(7)
We have to find the value of \[P\left( A \right)\] . So, we need a relation between \[P\left( A \right)\] and \[P\left( C \right)\] .
From equation (5), we have the relation between \[P\left( A \right)\] and \[P\left( C \right)\] .
Now, putting the value of \[P\left( C \right)\] from equation (7) in equation (5), we get
\[\begin{align}
  & \Rightarrow P\left( A \right)=3P\left( C \right) \\
 & \Rightarrow P\left( A \right)=3\times \dfrac{2}{11} \\
\end{align}\]
\[\Rightarrow P\left( A \right)=\dfrac{6}{11}\] ………………………………(8)
From equation (8), we have the value of \[P\left( A \right)\] .
Therefore, \[P\left( A \right)=\dfrac{6}{11}\] .
Hence, option (D) is the correct one.



Note: Whenever we have the question where the events are mutually exclusive and the relation between the probabilities of the events is given. Then, always use the property that the summation of the probabilities of all mutually exclusive events is 1.