Question

A and B are events with P(A) = $0.5$, P(B) = $0.4$ and $P\left(A\cap B\right)=0.3$. Find the probability that:
(i) A does not occur
(ii) Neither A and nor B occurs

Answer 1

Given:-

$P\left(A\right)=0.5$

$P\left(B\right)=0.4$

$P(A\cap B)=0.3$

To find:-

(i) $P\left(\overline{A}\right)=?$

(ii) $P(\overline{A}+\overline{B})$ = ?$

Solution:-

• (i) Probability of $A$ does not occur-
  

$P\left(\overline{A}\right)=1-P\left(A\right)=1-0.5=0.5$

• (ii) Probability of neither $A$ nor $B$ occurs-
  

$P(\overline{A}\cap \overline{B})=P\left(\overline{A\cup B}\right)=1-P(A\cup B)$

Now as we know that,

$P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$

$\Rightarrow P(A\cup B)=0.5+0.4-0.3=0.6$

Therefore,

$P(\overline{A}\cap \overline{B})=1-P(A\cup B)=1-0.6=0.4$