Question

If $E1$ and $E2$ are two mutually exclusive events of an experiment with$P(notE2)=0.6=P(E1\cup E2)$. Then, $P\left(E1\right)$ is equal to

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

Answer 1

The correct option is D

$0.2$

Step 1. Find the value of $\mathit{P}\mathbf{\left(}\mathit{E}\mathbf{1}\mathbf{\right)}$:

Given that $A$ and $B$ are mutually exclusive events

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

Given:

$\begin{array}{rcl}P(notE2)& =& 0.6\end{array}$

$\Rightarrow$ $\begin{array}{rcl}P\left(E2\right)& =& 1-0.6\\ & =& 0.4\end{array}$

Now,

$\begin{array}{rcl}P(E1\cup E2)& =& 0.6\end{array}$

Step 2.Substitute the values in the formula, we get

$\begin{array}{rcl}0.6& =& P\left(E1\right)+0.4–0\end{array}$ $\begin{array}{rcl}\mathbf{P}\mathbf{(}\mathbf{E}\mathbf{1}\mathbf{}\mathbf{\cup }\mathbf{}\mathbf{E}\mathbf{2}\mathbf{)}\mathbf{}& \mathbf{=}& \mathbf{}\mathbf{P}\mathbf{\left(}\mathbf{E}\mathbf{1}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{P}\mathbf{\left(}\mathbf{E}\mathbf{2}\mathbf{\right)}\mathbf{-}\mathbf{P}\mathbf{(}\mathbf{E}\mathbf{1}\mathbf{\cap }\mathbf{E}\mathbf{2}\mathbf{)}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}P\left(E1\right)& =& 0.6–0.4\end{array}$

$\Rightarrow$$\begin{array}{rcl}P\left(E1\right)& =& 0.2\end{array}$

Hence, the correct answer is option (D).