Question

If $A$ and $B$ are mutually exclusive events, such that $P(A)=0.25,P(B)=0.4,$ then $P(A^c\cap B^c)$ is equal to

• A. $0.45$
• B. $0.55$
• C. $0.9$
• D. $0.35$

Answer 1

Answer: (D)

Given, A and B are mutually exclusive events.
$\Rightarrow$ $P(A\cap B)=0$
By addition theorem of probability,
$P(A\cup B)=P(A)+P(B)-P(A\cap B)$
$\Rightarrow$ $P(A\cup B)=P(A)+P(B)$
$\Rightarrow$ $P(A\cup B)=0.25+0.4$
$[\because P(A)=0.25,P(B)=0.4(given)]$
$P(A\cup B)=0.65$ ..(i)
Now, we have $P(A^c\cap B^c)=P{(A\cup B)}^c$
$\Rightarrow$ $P(A^c\cap B^c)=1-P(A\cup B)$
$=1-0.65=0.35$