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  4. If P(A)=(3/10) ; P(B)=(2/5) and P(A∪ B)=(3/5), then P(B| A)+P(A |B ) equals

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Q. If $P\left(A\right)=\frac{3}{10} ; P\left(B\right)=\frac{2}{5}$ and $P\left(A\cup B\right)=\frac{3}{5}$, then $P\left(B| A\right)+P\left(A |B \right)$ equals

Probability - Part 2

Solution:

$P\left(A\right)=\frac{3}{10}$,
$P\left(B\right)=\frac{2}{5}$,
$P\left(A\cup B\right)=\frac{3}{5}$
Now, $P\left(A\cap B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cup B\right)$
$=\frac{3}{10}+\frac{2}{5}-\frac{3}{5}$
$=\frac{3+4-6}{10}$
$=\frac{1}{10}$
Now, $P\left(B| A\right)+P\left(A |B\right)=\frac{P\left(A\cap B\right)}{P\left(A\right)}+\frac{P\left(A\cap B\right)}{P\left(B\right)}$
$=\frac{1 /10}{3/ 10}+\frac{1/ 10}{2 /5}$
$=\frac{1}{3}+\frac{1}{4}$
$=\frac{7}{12}$