Question

If $P\left(A\right)=\frac{2}{5}$, $P\left(B\right)=\frac{3}{10}$ and $P(A\cap B)=\frac{1}{5}$, then find the value of $P\left(A^{\prime }|B^{\prime }\right).P\left(B^{\prime }|A^{\prime }\right)$.

Answer 1

Given, $P\left(A\right)=\frac{2}{5}$, $P\left(B\right)=\frac{3}{10}$ and $P(A\cap B)=\frac{1}{5}$

$P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$
$=\frac{2}{5}+\frac{3}{10}-\frac{1}{5}=\frac{1}{2}$

Now,
$P\left(A^{\prime }|B^{\prime }\right)=\frac{P(A^{\prime }\cap B^{\prime })}{P\left(B^{\prime }\right)}$
$=\frac{P(A\cup B{)}^{\prime }}{1-P\left(B\right)}$
$=\frac{1-P(A\cup B)}{1-P\left(B\right)}$
$=\frac{1-\frac{1}{2}}{1-\frac{3}{10}}$
$=\frac{5}{7}$

Similarly,
$P\left(B^{\prime }|A^{\prime }\right)=\frac{P(B^{\prime }\cap A^{\prime })}{P\left(A^{\prime }\right)}$
$=\frac{P(B\cup A{)}^{\prime }}{1-P\left(A\right)}$
$=\frac{1-P(A\cup B)}{1-P\left(A\right)}$
$=\frac{1-\frac{1}{2}}{1-\frac{2}{5}}$
$=\frac{5}{6}$

$\therefore P\left(A^{\prime }|B^{\prime }\right).P\left(B^{\prime }|A^{\prime }\right)=\frac{5}{7}\times \frac{5}{6}=\frac{25}{42}$