Question

Given two independent events A and B such that $P\left(A\right)=0.3$, $P\left(B\right)=0.6$. Find
(i) P(A and not B) (ii) P(neither A nor B)

Answer 1

We have,

$P\left(A\right)=0.3,P\left(B\right)=0.6$

So,$P\left(\overline{A}\right)=1-0.3=0.7$

And$P\left(\overline{B}\right)=1-0.6=0.4$

Part (a)

P(A and not B)$=P\left(A\right)\times P\left(\overline{B}\right)$

P(A and not B)$=0.3\times 0.4=0.12$

Part (b)

P(neither A and nor B)$=P\left(\overline{A}\right)\times P\left(\overline{B}\right)$

P(A neither A and nor B)$=0.7\times 0.4=0.28$