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If $P(A \cup B)=0.8$ and $P(A \cap B)=0.3$, then $P(\bar{A})+P(\bar{B})$ is equal to:
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Verified Answer
The correct answer is:
0.9
We have,
$P(A)+P(B)=P(A \cup B)+P(A \cap B)$
$=0.8+0.3=1.1$
$\Rightarrow \quad 1-P(\bar{A})+1-P(\bar{B})=1.1$
$\Rightarrow \quad P(\bar{A})+P(\bar{B})=2-1.1=0.9$
$P(A)+P(B)=P(A \cup B)+P(A \cap B)$
$=0.8+0.3=1.1$
$\Rightarrow \quad 1-P(\bar{A})+1-P(\bar{B})=1.1$
$\Rightarrow \quad P(\bar{A})+P(\bar{B})=2-1.1=0.9$
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