Question

If $P(E)=0.15,P(F)=0.09$ and $P(E\cup F)=0.19$, then the value of $P\left(E^C\cap F\right)$ is

• A. 0.07
• B. 0.04
• C. 0.02
• D. 0.05

Answer 1

Answer: (B)

$P\left(E^C\cap F\right)=P(F)-P(E\cap F)\dots (1)$
$\text{ and }P(E\cap F)=P(E)+P(F)-P(E\cup F)$
$P(E\cap F)=0.15+0.09-0.19=0.05$
$\therefore P\left(E^C\cap F\right)=0.09-0.05=0.04$