Question

If $P\left(B\right)=\frac{3}{5}$, $P\left(A|B\right)=\frac{1}{2}$ and $P\left(A\cup B\right)=\frac{4}{5}$, then $P{\left(A\cup B\right)}^{\prime }+P\left(A^{\prime }\cup B\right)=$

• A. $\frac{1}{5}$
• B. $\frac{4}{5}$
• C. $\frac{1}{2}$
• D. $1$

Answer 1

Answer: (D)

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