Question

Let $E1$, $E2$ be two mutually exclusive events of an experiment with $P\left(\text{not}E2\right)=0.6=P(E1\cup E2)$, then $P\left(E1\right)=$

• A. $0.1$
• B. $0.3$
• C. $0.4$
• D. $0.2$

Answer 1

The correct option is D

$0.2$

Explanation for correct option:

Finding $P\left(E1\right)$:

Considering the given equation,

$$\begin{gathered} P(\text{not}E2)=0.6 \\ \therefore P\left(E2\right)=1-0.6 \\ =0.4 \end{gathered}$$,

$E1$ and $E2$are mutually exclusive events.

Hence, $P\left(E1\cap E2\right)=0$

And $P\left(E1\cup E2\right)=0.6$ is given

We know that,

$P\left(E1\cup E2\right)=P\left(E1\right)+P\left(E2\right)–P\left(E1\cap E2\right)$

Substitute the all the values in the above equation, we get

$$\begin{gathered} \Rightarrow 0.6=P\left(E1\right)+0.4-0 \\ \Rightarrow P\left(E1\right)=0.6-0.4 \\ \Rightarrow P\left(E1\right)=0.2 \end{gathered}$$

Therefore, the correct answer is option (D).