Question

IF A and B are two events such that $P\left(A\right)=\frac{1}{4},P\left(B\right)=\frac{1}{2}$
and $P(A\cap B)=\frac{1}{8}$ Find P (not A and not B)

Answer 1

We have $P\left(A\right)=\frac{1}{4},P\left(B\right)=\frac{1}{2},P(A\cap B)=\frac{1}{8}\Rightarrow P\left(A\right)=1-P\left(A\right)=1-\frac{1}{4}=\frac{3}{4}andP\left(B\right)=1-P\left(B\right)=1-\frac{1}{2}=\frac{1}{2}$
As, $P(A\cap B)=\frac{1}{8}=\frac{1}{4}\times \frac{1}{2}=P\left(A\right)\times P\left(B\right)$
THerefore, A and B are independent events.

$\Rightarrow$ A' and B' are also independent events
$\Rightarrow P(A^{\prime }\cap B^{\prime })=P\left(A^{\prime }\right)P\left(B^{\prime }\right)\therefore P\left(notAandnotB\right)=P(A^{\prime }\cap B^{\prime })=P\left(A^{\prime }\right)P\left(B^{\prime }\right)=\frac{3}{4}\times \frac{1}{2}=\frac{3}{8}$
If A and B are independent events, then A' B, A, B' and A' B' are also independent events.