Question

If ${{E}_{1}}$ and ${{E}_{2}}$ are two mutually exclusive events of an experiment with $P\left( \overline{{{E}_{2}}} \right)=0.6=P\left( {{E}_{1}}\cup {{E}_{2}} \right)$ then $P\left( {{E}_{1}} \right)$ is equal to
1. 0.1
2. 0.3
3. 0.4
4. 0.2

Answer 1

In this problem, we are given If ${{E}_{1}}$ and ${{E}_{2}}$ are two mutually exclusive events of an experiment with $P\left( \overline{{{E}_{2}}} \right)=0.6=P\left( {{E}_{1}}\cup {{E}_{2}} \right)$ then we have to find the value of $P\left( {{E}_{1}} \right)$. We know that if two events are mutually exclusive, then $P\left( {{E}_{1}}\cap {{E}_{2}} \right)=0$. We can find the value of $P\left( {{E}_{2}} \right)$ from the given data, we can then substitute the values we get in the formula $P\left( {{E}_{1}}\cup {{E}_{2}} \right)=P\left( {{E}_{1}} \right)+P\left( {{E}_{2}} \right)-P\left( {{E}_{1}}\cap {{E}_{2}} \right)$ and find the required value.

Complete step by step answer:
Here we have to find the value of $P\left( {{E}_{1}} \right)$.
We are given ${{E}_{1}}$ and ${{E}_{2}}$ are two mutually exclusive events of an experiment and $P\left( \overline{{{E}_{2}}} \right)=0.6$, $P\left( {{E}_{1}}\cup {{E}_{2}} \right)=0.6$ …….. (1)
We know that if two events are mutually exclusive, then
$P\left( {{E}_{1}}\cap {{E}_{2}} \right)=0$ …… (2)
We know that,
$P\left( {{E}_{2}} \right)=1-P\left( \overline{{{E}_{2}}} \right)$
We can now substitute the given value, we get
$\Rightarrow P\left( {{E}_{2}} \right)=1-0.6=0.4$…….. (3)
We know that,
$P\left( {{E}_{1}}\cup {{E}_{2}} \right)=P\left( {{E}_{1}} \right)+P\left( {{E}_{2}} \right)-P\left( {{E}_{1}}\cap {{E}_{2}} \right)$
We can now substitute (1), (2), (3) in the above formula, we get
$\Rightarrow 0.6=P\left( {{E}_{1}} \right)+0.4-0$
We can now simplify and solve the above step, we get
$\Rightarrow P\left( {{E}_{1}} \right)=0.6-0.4=0.2$

So, the correct answer is “Option 4”.

Note: We should always remember that if the given two events are mutually exclusive events then $P\left( {{E}_{1}}\cap {{E}_{2}} \right)=0$, where the two events do not occur at same time and only one event can occur. We should also remember that $P\left( {{E}_{1}}\cup {{E}_{2}} \right)=P\left( {{E}_{1}} \right)+P\left( {{E}_{2}} \right)-P\left( {{E}_{1}}\cap {{E}_{2}} \right)$. We should also remember the formula $P\left( {{E}_{2}} \right)=1-P\left( \overline{{{E}_{2}}} \right)$. We should make the simplifications in a correct order to get the required answer.