Question

A, B, C are three mutually exclusive and exhaustive events associated with a random experiment. Find $P\left( A \right)$ , it is given that $P\left( B \right) = \dfrac{3}{2}P\left( A \right)$ and $P\left( C \right) = \dfrac{1}{2}P\left( B \right)$

Answer 1

Hint : For the given events A, B, C to be mutually exclusive and exhaustive associated with a random experiment, their union should be equal to S. This can be written as $A \cup B \cup C = S$ . Take probability on both sides of this equation and find $P\left( A \right)$

Complete step-by-step answer :
Given to us that A, B, C are three mutually exclusive and exhaustive events.
Probability of these events would be
$P\left( A \right),P\left( B \right)$ and $P\left( C \right)$
The relation between these probabilities is already given as
$P\left( B \right) = \dfrac{3}{2}P\left( A \right)$ and $P\left( C \right) = \dfrac{1}{2}P\left( B \right)$
From this we can form a relation between
$P\left( A \right)$ and $P\left( C \right)$ as follows.
$P\left( C \right) = \dfrac{1}{2}P\left( B \right) \Rightarrow P\left( C \right) = \dfrac{1}{2} \times \dfrac{3}{2}P\left( A \right)$
And hence this relation now becomes
$P\left( C \right) = \dfrac{3}{4}P\left( A \right)$
Now, the events A, B, C are said to be mutually exclusive and exhaustive if
$A \cup B \cup C = S$
By substituting probability on both sides, we get
$P\left( {A \cup B \cup C} \right) = P\left( S \right)$
Now, by formula this can be written as
$P\left( A \right) + P\left( B \right) + P\left( C \right) = P\left( S \right)$ since $P\left( {X \cup Y} \right) = P\left( X \right) + P\left( Y \right)$
We also know that the value of $P\left( S \right)$ is one so this equation now becomes
$P\left( A \right) + P\left( B \right) + P\left( C \right) = 1$
We can now substitute the given relation between these probabilities. So the equation becomes
$P\left( A \right) + \dfrac{3}{2}P\left( A \right) + \dfrac{3}{4}P\left( A \right) = 1$
By taking $P\left( A \right)$ common from this equation, we get
$\left( {1 + \dfrac{3}{2} + \dfrac{3}{4}} \right)P\left( A \right) = 1$
This can be solved as
$\left( {\dfrac{{4 + 6 + 3}}{4}} \right)P\left( A \right) = 1 \Rightarrow \left( {\dfrac{{13}}{4}} \right)P\left( A \right) = 1$
By solving, we get the value of $P\left( A \right)$ is $\dfrac{4}{{13}}$
So, the correct answer is “ $P\left( A \right)$ is $\dfrac{4}{{13}}$ ”.

Note : It is to be noted that the value of any probability is always less than one i.e. probability of any event only lies between zero and one. In the above solution, the value of $P\left( A \right)$ is less than one. We can also calculate the values of $P\left( B \right)$ and $P\left( C \right)$ from the given relation and their value would also be less than one.