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  4. If P(A)=(2/5), P(B)=(3/10) and P(A∩ B)=(1/5), then P(A' | B').P(B' |A') is equal to

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Q. 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 $P\left(A' | B'\right).P\left(B' |A'\right)$ is equal to

Probability - Part 2

Solution:

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