Browse Questions

# A and B are two events such that P(A)=0.54,P(B)=0.69 and $P(A\cap B)=0.35$.Find $P(A'\cup B')$

$\begin{array}{1 1}(A)\;0.22\\(B)\;0.12\\(C)\;0.32\\(D)\;0.72\end{array}$

Toolbox:
• $P(A'\cap B')=P(A \cup B)'=1-P(A\cup B)$
• $P(A\cup B)=P(A)+P(B)-P(A \cap B)$
Step 1:
$P(A)=0.54$
$P(B)=0.69$
$P(A \cap B)=0.35$
$P(A\cup B)=P(A)+P(B)-P(A \cap B)$
$\Rightarrow 0.54+0.69-0.35$
$\Rightarrow 0.88$
Step 2:
$P(A'\cap B')=P(A \cup B)'=1-P(A \cup B)$
$\Rightarrow 1-0.88$
$\Rightarrow 0.12$
Hence (B) is the correct answer.