Question

If $A$ and $B$ are two events such that $P(A)=\frac{1}{4},P(B)=\frac{1}{2}$ and $P(A∩B)=\frac{1}{8}$,find $P$ (not $A$ and not $B$).

Answer 1

The correct answer is: $\frac{3}{8}$
It is given that; $P(A)=\frac{1}{4}, P(B)=\frac{1}{2},$
P(not on A and not on B)=$P(A'\cap B')$
P(not A and not on B)=$P((A\cup B))'$ $[A'\cap B'=(A\cup B)']$
$=1-P(A\cup B)$
$=1-[P(A)+P(B)-P(A\cap B)]$
$=1-\bigg[\frac{1}{4}+\frac{1}{2}-\frac{1}{8}\bigg]$
$=1-\frac{5}{8}$
$=\frac{3}{8}$