JEE Advanced 2021 — Paper 1 — Question 25

Given $P(E)=\frac{1}{8}, P(F)=\frac{1}{6}, P(G)=\frac{1}{4}, P(E \cap F \cap G)=\frac{1}{10}$. For (A): $P(E \cap F \cap G^c)=P(E \cap F)-P(E \cap F \cap G)$.

Since $E\cap F \subseteq E$, $P(E\cap F)\le P(E)$, so $P(E\cap F\cap G^c)\le P(E)-P(E\cap F\cap G)=\frac{1}{8}-\frac{1}{10}=\frac{1}{40}$.

Hence (A) is true. For (B): $P(E^c\cap F\cap G)=P(F\cap G)-P(E\cap F\cap G)$.

Since $F\cap G \subseteq F$, $P(F\cap G)\le P(F)$, so $P(E^c\cap F\cap G)\le P(F)-P(E\cap F\cap G)=\frac{1}{6}-\frac{1}{10}=\frac{1}{15}$.

Hence (B) is true. For (C): By union bound, $P(E\cup F\cup G)\le P(E)+P(F)+P(G)=\frac{1}{8}+\frac{1}{6}+\frac{1}{4}=\frac{13}{24}$. Hence (C) is true. For (D): By De Morgan's law, $P(E^c\cap F^c\cap G^c)=1-P(E\cup F\cup G)$.

From (C), $P(E\cup F\cup G)\le\frac{13}{24}$, so $P(E^c\cap F^c\cap G^c)\ge 1-\frac{13}{24}=\frac{11}{24}$. Since $\frac{11}{24}>\frac{10}{24}=\frac{5}{12}$, the inequality $\le \frac{5}{12}$ is false. Hence (D) is false. Thus the true statements are (A), (B), and (C).