Question 11 Mark
If A and B are mutually exclusive events, P(A) = 0.35 and P(B) = 0.45, find:
$\text{P}(\text{A}'\cap\text{B}')$
$\text{P}(\text{A}'\cap\text{B}')$
Answer
View full question & answer→Since, it is given that, A and B are matually exclusive events.
$\therefore\ \text{P}(\text{A}\cap\text{B})=0$ $[\because\ \text{A}\cap\text{B}=\phi]$
$\text{P}(\text{A}'\cap\text{B}')=\text{P}(\text{A}\cup\text{B}')$
$=1-\text{P}(\text{A}\cup\text{B})$
$=1-0.8$
$=0.2$
$\therefore\ \text{P}(\text{A}\cap\text{B})=0$ $[\because\ \text{A}\cap\text{B}=\phi]$
$\text{P}(\text{A}'\cap\text{B}')=\text{P}(\text{A}\cup\text{B}')$
$=1-\text{P}(\text{A}\cup\text{B})$
$=1-0.8$
$=0.2$