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If $\mathrm{A}$ and $\mathrm{B}$ are two events of a random experiment such that $P(A \cup B)=0.65$ and $P(A \cap B)=0.15$, then $\mathrm{P}(\overline{\mathrm{A}})+\mathrm{P}(\overline{\mathrm{B}})=$
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2450 Upvotes
Verified Answer
The correct answer is:
$1.2$
Since we know that
$$
\begin{aligned}
& P(A \cup B)=P(A)+P(B)-P(A \cap B) \\
& \Rightarrow \quad P(A)+P(B)=P(A \cup B)+P(A \cap B) \\
& \Rightarrow \quad P(A)+P(B)=0.65+0.15=0.8
\end{aligned}
$$
we also know that
$$
\begin{aligned}
& P(\bar{A})+P(\bar{B})=(1-P(A))+(1-P(B)) \\
& =2-[P(A)+P(B)] \\
& =2-0.8=1.2
\end{aligned}
$$
$$
\begin{aligned}
& P(A \cup B)=P(A)+P(B)-P(A \cap B) \\
& \Rightarrow \quad P(A)+P(B)=P(A \cup B)+P(A \cap B) \\
& \Rightarrow \quad P(A)+P(B)=0.65+0.15=0.8
\end{aligned}
$$
we also know that
$$
\begin{aligned}
& P(\bar{A})+P(\bar{B})=(1-P(A))+(1-P(B)) \\
& =2-[P(A)+P(B)] \\
& =2-0.8=1.2
\end{aligned}
$$
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