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  4. If P(B)=(3/5), P(A | B)=(1/2) and P(A∪ B)=(4/5), then P(A∪ B)' +P(A'∪ B)=

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Q. 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 $P\left(A\cup B\right)' +P\left(A'\cup B\right)=$

Probability - Part 2

Solution:

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