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Let $A$ and $B$ are two mutually exclusive events with $\mathrm{P}(\mathrm{A})=$ $\frac{1}{3} \mathrm{P}(\mathrm{B})=\frac{1}{4} .$ What is the value of $\mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}}) ?$
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The correct answer is:
$\frac{5}{12}$
$A \& B$ are mutualy exclusive events. i.e., $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=0$
We know, $\mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}})=\mathrm{P}(\overline{\mathrm{A} \cup \mathrm{B}})$
$=1-\mathrm{P}(\mathrm{A} \cup \mathrm{B})$
$=1-[\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})]$
$=1-\left[\frac{1}{3}+\frac{1}{4}-0\right]$
$=1-\left[\frac{7}{12}\right]=\frac{5}{12} .$
We know, $\mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}})=\mathrm{P}(\overline{\mathrm{A} \cup \mathrm{B}})$
$=1-\mathrm{P}(\mathrm{A} \cup \mathrm{B})$
$=1-[\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})]$
$=1-\left[\frac{1}{3}+\frac{1}{4}-0\right]$
$=1-\left[\frac{7}{12}\right]=\frac{5}{12} .$
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