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

given, $P(A)=\frac{1}{4}$

$P(B)=\frac{1}{2}$

and $P(A\cap B)=\frac{1}{8}$

we have to find P(not A and not B) e.g., $P(A'\cap B')$

we know, $P(A')=1-P(A)$

so, P(A')=$1-\frac{1}{4}=\frac{3}{4}$

similarly, $P(B')=1-P(B)$

so, P(B) = $1-\frac{1}{2}=\frac{1}{2}$

also use formula,

$P(A'\cup B')=P(A\cap B)'=1-P(A\cap B)$

= $1-\frac{1}{8}=\frac{7}{8}$

now, $P(A'\cap B')=P(A')+P(B')-P(A'\cup B')$

= 3/4 + 1/2 - 7/8

= (6 + 4 - 7)/8

= 3/8