Verified Solution • 2025–26 Session moderate QJEEAdv07Ma32e3: JEE Advanced 2007 Mathematics - Class 12 JEEAdv07Ma32e3 Solution: Let denote the complement of an event Let be pairwise… Let Ec{E^c}Ec denote the complement of an event E.E.E. Let E,F,GE, F, GE,F,G be pairwise independent events with P\left( G \right) > 0 and P(E∩F∩G)=0.P\left( {E \cap F \cap G} \right) = 0.P(E∩F∩G)=0. Then P(Ec∩Fc∣G)P\left( {{E^c} \cap {F^c}|G} \right)P(Ec∩Fc∣G) equals Multiple Choice Options A P(Ec)+P(Fc)P\left( {{E^c}} \right) + P\left( {{F^c}} \right)P(Ec)+P(Fc) B P(Ec)−P(Fc)P\left( {{E^c}} \right) - P\left( {{F^c}} \right)P(Ec)−P(Fc) C P(Ec)−P(F)P\left( {{E^c}} \right) - P\left( F \right)P(Ec)−P(F) D P(E)−P(Fc)P\left( E \right) - P\left( {{F^c}} \right)P(E)−P(Fc) Step-by-Step Verified Solution amp;E1f1G are pairwise independent events. amp;∴P(E∩F)=P(E)⋅P(F)amp;P(F∩G)=P(F)⋅P(G)amp;P(G∩E)=P(G)⋅P(E)amp;P(Ec∩FcG)=P((Ec∩Fc)∩GP(G))amp;=P(G)−P(G∩E)−P(G∩F)P(G)amp;=1−P(E)−P(F)amp;=P(Ec)−P(F).\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}amp;E1f1G are pairwise independent events. amp;∴P(E∩F)=P(E)⋅P(F)amp;P(F∩G)=P(F)⋅P(G)amp;P(G∩E)=P(G)⋅P(E)amp;P(GEc∩Fc)=P(P(G)(Ec∩Fc)∩G)amp;=P(G)P(G)−P(G∩E)−P(G∩F)amp;=1−P(E)−P(F)amp;=P(Ec)−P(F). Related Learning Resources & Exam Preparation JEE Advanced PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring