Inequalities for Probabilities of Intersections and Unions of Three Events

To determine which of the given statements are true, we analyze each option step-by-step using the basic axioms and properties of probability.

We are given: $P(E) = \frac{1}{8}, \quad P(F) = \frac{1}{6}, \quad P(G) = \frac{1}{4}, \quad P(E \cap F \cap G) = \frac{1}{10}$

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Analysis of Option (A):

The set $E \cap F$ can be partitioned into two mutually disjoint sets: $E \cap F = (E \cap F \cap G) \cup (E \cap F \cap G^c)$

Taking the probability on both sides: $P(E \cap F) = P(E \cap F \cap G) + P(E \cap F \cap G^c)$

Since $E \cap F \subseteq E$, we know that $P(E \cap F) \le P(E)$. Therefore: $P(E \cap F \cap G) + P(E \cap F \cap G^c) \le P(E)$

Rearranging for $P(E \cap F \cap G^c)$: $P(E \cap F \cap G^c) \le P(E) - P(E \cap F \cap G)$

Substitute the given values: $P(E \cap F \cap G^c) \le \frac{1}{8} - \frac{1}{10} = \frac{5 - 4}{40} = \frac{1}{40}$

Thus, Option (A) is TRUE.

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Analysis of Option (B):

Similarly, the set $F \cap G$ can be partitioned into two mutually disjoint sets: $F \cap G = (E \cap F \cap G) \cup (E^c \cap F \cap G)$

Taking the probability on both sides: $P(F \cap G) = P(E \cap F \cap G) + P(E^c \cap F \cap G)$

Since $F \cap G \subseteq F$, we have $P(F \cap G) \le P(F)$. Therefore: $P(E \cap F \cap G) + P(E^c \cap F \cap G) \le P(F)$

Rearranging for $P(E^c \cap F \cap G)$: $P(E^c \cap F \cap G) \le P(F) - P(E \cap F \cap G)$

Substitute the given values: $P(E^c \cap F \cap G) \le \frac{1}{6} - \frac{1}{10} = \frac{5 - 3}{30} = \frac{2}{30} = \frac{1}{15}$

Thus, Option (B) is TRUE.

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Analysis of Option (C):

By Boole's Inequality (the Union Bound): $P(E \cup F \cup G) \le P(E) + P(F) + P(G)$

Substitute the given values: $P(E \cup F \cup G) \le \frac{1}{8} + \frac{1}{6} + \frac{1}{4} = \frac{3 + 4 + 6}{24} = \frac{13}{24}$

Thus, Option (C) is TRUE.

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Analysis of Option (D):

By De Morgan's Law, the complement of the union of events is the intersection of their complements: $E^c \cap F^c \cap G^c = (E \cup F \cup G)^c$

Taking the probability: $P(E^c \cap F^c \cap G^c) = 1 - P(E \cup F \cup G)$

Using the upper bound derived in Option (C), $P(E \cup F \cup G) \le \frac{13}{24}$: $P(E^c \cap F^c \cap G^c) \ge 1 - \frac{13}{24} = \frac{11}{24}$

Since $\frac{11}{24} > \frac{10}{24} = \frac{5}{12}$, it follows that: $P(E^c \cap F^c \cap G^c) \ge \frac{11}{24} > \frac{5}{12}$

This proves that $P(E^c \cap F^c \cap G^c)$ cannot be less than or equal to $\frac{5}{12}$.

Thus, Option (D) is FALSE.

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Conclusion:

The correct statements are A, B, and C.