JEE Advance - Mathematics (2007 - Paper 2 Offline - No. 1)
Let $${E^c}$$ denote the complement of an event $$E.$$ Let $$E, F, G$$ be pairwise independent events with $$P\left( G \right) > 0$$ and $$P\left( {E \cap F \cap G} \right) = 0.$$ Then $$P\left( {{E^c} \cap {F^c}|G} \right)$$ equals
$$P\left( {{E^c}} \right) + P\left( {{F^c}} \right)$$
$$P\left( {{E^c}} \right) - P\left( {{F^c}} \right)$$
$$P\left( {{E^c}} \right) - P\left( F \right)$$
$$P\left( E \right) - P\left( {{F^c}} \right)$$
Explanation
$$\begin{aligned}
& \mathrm{E}_{1} f_{1} G \text { are pairwise independent events. } \\\\
& \therefore \quad \mathrm{P}(\mathrm{E} \cap \mathrm{F})=\mathrm{P}(\mathrm{E}) \cdot \mathrm{P}(\mathrm{F}) \\\\
& \mathrm{P}(\mathrm{F} \cap \mathrm{G})=\mathrm{P}(\mathrm{F}) \cdot \mathrm{P}(\mathrm{G}) \\\\
& \mathrm{P}(\mathrm{G} \cap \mathrm{E})=\mathrm{P}(\mathrm{G}) \cdot \mathrm{P}(\mathrm{E}) \\\\
& \mathrm{P}\left(\frac{\mathrm{E}^{\mathrm{c}} \cap \mathrm{F}^{\mathrm{c}}}{\mathrm{G}}\right)=\mathrm{P}\left(\frac{\left(E^{\mathrm{c}} \cap \mathrm{F}^{\mathrm{c}}\right) \cap G}{\mathrm{P}(\mathrm{G})}\right) \\\\
&=\frac{\mathrm{P}(\mathrm{G})-\mathrm{P}(\mathrm{G} \cap \mathrm{E})-\mathrm{P}(\mathrm{G} \cap \mathrm{F})}{\mathrm{P}(\mathrm{G})} \\\\
&=1-\mathrm{P}(\mathrm{E})-\mathrm{P}(\mathrm{F}) \\\\
&=\mathrm{P}\left(\mathrm{E}^{\mathrm{c}}\right)-\mathrm{P}(\mathrm{F}) .
\end{aligned}$$
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