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Q. If $A$ and $B$ are events such that $P\left(A\right) =\frac{1}{3}$, $P\left(B\right) = \frac{1}{4}$ and $P\left(A \cap B\right) =\frac{1}{12}$,then find $P$(not $A$ and not $B$).

Probability - Part 2

Solution:

Here $P\left(A\right). P\left(B\right) =\frac{1}{3}\times\frac{1}{4} = \frac{1}{12} = P\left(A \cap B\right)$
$\Rightarrow $ events $A$ and $B$ are independent.
$\Rightarrow $ events $\bar{A}$ and $\bar{B}$ are also independent.
Now $p \left(\bar{A} \cap \bar{B} \right) = P\left(\bar{A}\right)P\left(\bar{B}\right)$
($\because\bar{A}$ and $\bar{B}$ are independent events)
$= \left(1 - P\left(A\right)\right)\left(1 - P\left(B\right)\right)$
$=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)= \frac{2}{3}\times\frac{3}{4}$
$= \frac{1}{2}$