Question

If$A,B$ are two events and $P(A\text{'})=0.3,P\left(B\right)=0.4$and $P(A\cap B\text{'})=0.5$, then $P(A\cup B\text{'})$is equal to

• A. $0.5$
• B. $0.8$
• C. $1$
• D. $0.1$

Answer 1

The correct option is B

$0.8$

Find the value of $\mathit{P}\mathbf{(}\mathit{A}\mathbf{}\mathbf{\cup }\mathbf{}\mathit{B}\mathbf{\text{'}}\mathbf{)}\mathbf{}$:

Given that and $P(A\text{'})=0.3,P\left(B\right)=0.4$and $P(A\cap B\text{'})=0.5$

$\Rightarrow P\left(A\right)=1-P(A\text{'})=1-0.3=0.7$

$\Rightarrow P(B\text{'})=1-P\left(B\right)=1–0.4=0.6$

Now, substitute the values, we get

$$\begin{gathered} P(A\cup B\text{'})=0.7+0.6–0.5 \\ \Rightarrow P(A\cup B\text{'})=0.7+0.1 \\ \Rightarrow P(A\cup B\text{'})=0.8. \end{gathered}$$ $\mathbf{}\mathit{P}\mathbf{(}\mathit{A}\mathbf{}\mathbf{\cup }\mathbf{}\mathit{B}\mathbf{\text{'}}\mathbf{)}\mathbf{}\mathbf{=}\mathbf{}\mathit{P}\mathbf{\left(}\mathit{A}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}\mathit{P}\mathbf{(}\mathit{B}\mathbf{\text{'}}\mathbf{)}\mathbf{}\mathbf{–}\mathbf{}\mathit{P}\mathbf{(}\mathit{A}\mathbf{}\mathbf{\cap }\mathbf{}\mathit{B}\mathbf{\text{'}}\mathbf{)}$

Hence, the correct option is (B).