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

Hint: 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.