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.