Browse Questions

# The probability that at least one of the events A and B occurs is 0.6.If A and B occur simultaneously with probability 0.2.Then $P(\bar{A})+P(\bar{B})$ is

$\begin{array}{1 1}(A)\;0.4\\(B)\;0.8\\(C)\;1.2\\(D)\;1.6\end{array}$

Toolbox:
• $P(A \cup B)=P(A)+P(B)-P(A \cap B)$
Step 1:
Given :
$P(A \cup B)=0.6,P(A \cap B)=0.2$
$\therefore P(A \cup B)=P(A)+P(B)-P(A \cap B)$
$0.6=P(A)+P(B)-0.2$
$\Rightarrow 0.6+0.2=P(A)+P(B)$
$\Rightarrow P(A)+P(B)=0.8$
Step 2:
$P(\bar{A})+P(\bar{B})=(1-P(A))+1-P(B))$
$\Rightarrow 2-(P(A)+P(B))$
$\Rightarrow 2-0.8=1.2$
Hence (C) is the correct answer.