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The probability that atleast one of the events A and $\mathrm{B}$ occurs is $0.6$. If $\mathrm{A}$ and $\mathrm{B}$ occur simultaneously with probability $0.2$, then $\mathrm{P}(\overline{\mathrm{A}})+\mathrm{P}(\overline{\mathrm{B}})$ is
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The correct answer is:
$1.2$
$\mathrm{P}(\mathrm{A} \cup \mathrm{B})=0.6$ and $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=0.2$ we know that
$$
\begin{array}{l}
\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \\
\Rightarrow 0.6=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-0.2 \\
\Rightarrow \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})=0.8 \\
\Rightarrow 1-\mathrm{P}(\overline{\mathrm{A}})+1-\mathrm{P}(\overline{\mathrm{B}})=0.8 \\
\Rightarrow-[\mathrm{P}(\overline{\mathrm{A}})+\mathrm{P}(\overline{\mathrm{B}})]=0.8-2 \\
\Rightarrow \mathrm{P}(\overline{\mathrm{A}})+\mathrm{P}(\overline{\mathrm{B}})=1.2
\end{array}
$$
$$
\begin{array}{l}
\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \\
\Rightarrow 0.6=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-0.2 \\
\Rightarrow \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})=0.8 \\
\Rightarrow 1-\mathrm{P}(\overline{\mathrm{A}})+1-\mathrm{P}(\overline{\mathrm{B}})=0.8 \\
\Rightarrow-[\mathrm{P}(\overline{\mathrm{A}})+\mathrm{P}(\overline{\mathrm{B}})]=0.8-2 \\
\Rightarrow \mathrm{P}(\overline{\mathrm{A}})+\mathrm{P}(\overline{\mathrm{B}})=1.2
\end{array}
$$
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