Question

Solve this:
Q.11. If P(A') = 0.3, P(B) = 0.4 and $P\left(A\cap B\text{'}\right)=0.5$, then $P\left(B\setminus A\cup B\text{'}\right)$ is
(A) 0.75
(B) 0.65
(C) 0.35
(D) 0.25

Answer 1

$$\begin{gathered} P(A\text{'})=0.3 \\ P\left(B\right)=0.4 \\ P(A\cap B\text{'})=0.5 \\ Weknow \\ P(A\cap B\text{'})=P\left(A\right)-P(A\cap B)=0.5 \\ 1-P(A\text{'})-P(A\cap B)=0.5 \\ P(A\cap B)=1-0.3-0.5=0.2 \\ \mathit{P}\mathbf{(}\mathit{A}\mathbf{\cap }\mathit{B}\mathbf{)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{2} \\ Weknow\left(A\text{'}\cap B\right)\text{'}=A\cup B\text{'}[demorganslaw(A\cap B)\text{'}=A\text{'}\cup B\text{'}] \\ P(A\text{'}\cap B)=P\left(B\right)-P(A\cap B)=0.4-0.2=0.2 \\ P(A\text{'}\cap B)=0.2 \\ P(A\text{'}\cap B)\text{'}=1-P(A\text{'}\cap B)=1-0.2 \\ P(A\text{'}\cap B)\text{'}=0.8 \\ \mathit{P}\mathbf{(}\mathit{A}\mathbf{\cup }\mathit{B}\mathbf{\text{'}}\mathbf{)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{8} \\ P\left(B\right|A\cup B\text{'})=\frac{P(B\cap (A\cup B\text{'})}{P(A\cup B\text{'})}=\frac{P\left(\right(B\cap A)\cup (B\cap B\text{'}\left)\right)}{P(A\cup B\text{'})}=\frac{P\left(\right(B\cap A)\cup \phi )}{P(A\cup B\text{'})}=\frac{P(A\cap B)}{P(A\cup B\text{'})} \\ \mathit{P}\mathbf{(}\mathit{B}\mathbf{|}\mathit{A}\mathbf{\cup }\mathit{B}\mathbf{\text{'}}\mathbf{)}\mathbf{=}\frac{\mathbf{P}\mathbf{(}\mathbf{A}\mathbf{\cap }\mathbf{B}\mathbf{)}}{\mathbf{P}\mathbf{(}\mathbf{A}\mathbf{\cup }\mathbf{B}\mathbf{\text{'}}\mathbf{)}}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{2}}{\mathbf{0}\mathbf{.}\mathbf{8}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{25} \end{gathered}$$

Option d