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

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- $P(A \cup B)=P(A)+P(B)-P(A \cap B)$

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.

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