Question

Given two independent events $A$ and $B$ such that $P\left(A\right)=0.3,P\left(B\right)=0.6$. Find
(i) $P\left(AandB\right)$
(ii) $P\left(AandnotB\right)$
(iii) $P\left(AorB\right)$
(iv) $P\left(neitherAnorB\right)$

If the sum of the above probablities is $m$ enter $100m$

Answer 1

It is given that $P\left(A\right)=0.3$ and $P\left(B\right)=0.6$
Also, $A$ and $B$ are independent events.
(i) $\therefore P\left(AandB\right)=P\left(A\right)\cdot P\left(B\right)$
$\Rightarrow P(A\cap B)=0.3\times 0.6=0.18$
(ii) $P\left(AandnotB\right)=P(A\cap B\prime )$
$=P\left(A\right)-P(A\cap B)$
$=0.3-0.18$
$=0.12$
(iii) $P\left(AorB\right)=P(A\cup B)$
$=P\left(A\right)+P\left(B\right)-P(A\cap B)$
$=0.3+0.6-0.18$
$=0.72$
(iv) $P\left(neitherAnorB\right)=P(A\prime \cap B\prime )$
$=P\left((A\cup B)\prime \right)$
$=1-P(A\cup B)$
$=1-0.72$
$=0.28$