Mathematics - Probability Question with Solution | TestHub

$P(B)=\frac{3}{5},P(A\mid B)=\frac{1}{2}$ and $P(A\cup B)=\frac{4}{5}$
$P(A\cap B)=P(A\mid B)P(B)=\frac{1}{2}\cdot \frac{3}{5}=\frac{3}{10}$
$\mathrm{P}(\mathrm{A}\cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A}\cap \mathrm{B})$
$\mathrm{P}(\mathrm{A})=\frac{4}{5}-\frac{3}{10}=\frac{1}{2}.\mathrm{P}\left({\mathrm{A}}^{\prime }\right)=1-\mathrm{P}(\mathrm{A})=\frac{1}{2}$
We know, $\mathrm{P}(\mathrm{A}\cap \mathrm{B})+\mathrm{P}\left({\mathrm{A}}^{\prime }\cap \mathrm{B}\right)=\mathrm{P}(\mathrm{B})$
[as $A\cap B$ and $A^{\prime }\cap B$ are mutually exclusive events]
$\Rightarrow \frac{3}{10}+\mathrm{P}\left({\mathrm{A}}^{\prime }\cap \mathrm{B}\right)=\frac{3}{5}$
$\Rightarrow \mathrm{P}\left({\mathrm{A}}^{\prime }\cap \mathrm{B}\right)=\frac{3}{5}-\frac{3}{10}=\frac{3}{10}$
Now, $\mathrm{P}\left({\mathrm{A}}^{\prime }\cup \mathrm{B}\right)=\mathrm{P}\left({\mathrm{A}}^{\prime }\right)+\mathrm{P}(\mathrm{B})-\mathrm{P}\left({\mathrm{A}}^{\prime }\cap \mathrm{B}\right)$
$=\frac{1}{2}+\frac{3}{5}-\frac{3}{10}=\frac{5+6-3}{10}=\frac{4}{5}$
$\mathrm{P}\left((\mathrm{A}\cup \mathrm{B}{)}^{\prime }\right)=1-\mathrm{P}(\mathrm{A}\cup \mathrm{B})=1-\frac{4}{5}=\frac{1}{5}$
$\therefore \mathrm{P}\left((\mathrm{A}\cup \mathrm{B}{)}^{\prime }\right)+\mathrm{P}\left({\mathrm{A}}^{\prime }\cup \mathrm{B}\right)=\frac{1}{5}+\frac{4}{5}=1$