Question

If $P\left(A\right)=\frac{2}{5},P\left(B\right)=\frac{3}{10}$ and $P\left(A\cap B\right)=\frac{1}{5}$, then $P\left(A^{\prime }|B^{\prime }\right).P\left(B^{\prime }|A^{\prime }\right)$ is equal to

• A. $\frac{5}{6}$
• B. $\frac{5}{7}$
• C. $\frac{25}{42}$
• D. $1$

Answer 1

Answer: (C)

$P\left(A\right)=\frac{2}{5},P\left(B\right)=\frac{3}{10},P\left(A\cap B\right)=\frac{1}{5}$
$P\left(A^{\prime }\right)=1-P\left(A\right)=1-\frac{2}{5}=\frac{3}{5}$
$P\left(B^{\prime }\right)=1-P\left(B\right)=1-\frac{3}{10}=\frac{7}{10}$
$P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)=\frac{2}{5}+\frac{3}{10}-\frac{1}{5}=\frac{1}{2}$
$P\left(A^{\prime }\cap B^{\prime }\right)=P{\left(A\cup B\right)}^{\prime }=1-P\left(A\cup B\right)=1-\frac{1}{2}=\frac{1}{2}$
$\therefore P\left(A^{\prime }|B^{\prime }\right)\cdot P\left(B^{\prime }|A^{\prime }\right)=\frac{P\left(A^{\prime }\cap B^{\prime }\right)}{P\left(B^{\prime }\right)}\cdot \frac{P\left(A^{\prime }\cap B^{\prime }\right)}{P\left(A^{\prime }\right)}$
$=\frac{1/2}{7/10}\cdot \frac{1/2}{3/5}=\frac{1}{4}\times \frac{50}{21}=\frac{25}{42}$