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If $\mathrm{A}$ and $\mathrm{B}$ are mutually exclusive events, then
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1892 Upvotes
Verified Answer
The correct answer is:
$\mathrm{P}(\mathrm{A}) \leq \mathrm{P}(\overline{\mathrm{B}})$
$\mathrm{P}(\mathrm{A}) \leq \mathrm{P}(\overline{\mathrm{B}})$
Given that A and B are two mutually exclusively events Then,
$$
\begin{aligned}
&\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B}) \quad[\because(\mathrm{A} \cap \mathrm{B})=\phi] \\
&\text { since, } \mathrm{P}(\mathrm{A} \cup \mathrm{B}) \leq 1 \therefore \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B}) \leq 1 \\
&\Rightarrow \mathrm{P}(\mathrm{A})+1-\mathrm{P}(\overline{\mathrm{B}}) \leq 1 \Rightarrow \mathrm{P}(\mathrm{A}) \leq \mathrm{P}(\overline{\mathrm{B}})
\end{aligned}
$$
$$
\begin{aligned}
&\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B}) \quad[\because(\mathrm{A} \cap \mathrm{B})=\phi] \\
&\text { since, } \mathrm{P}(\mathrm{A} \cup \mathrm{B}) \leq 1 \therefore \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B}) \leq 1 \\
&\Rightarrow \mathrm{P}(\mathrm{A})+1-\mathrm{P}(\overline{\mathrm{B}}) \leq 1 \Rightarrow \mathrm{P}(\mathrm{A}) \leq \mathrm{P}(\overline{\mathrm{B}})
\end{aligned}
$$
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