Question

Let $E , F$ and $G$ be three events having probabilities
$P(E)=\frac{1}{8}, P(F)=\frac{1}{6}$ and $P(G)=\frac{1}{4}$, and let $P(E \cap F \cap G)=\frac{1}{10}$.
For any event $H$, if $H ^{ C }$ denotes its complement, then which of the following statements is(are) TRUE?

• A. $P\left(E \cap F \cap G^{C}\right) \leq \frac{1}{40}$
• B. $P\left(E^{C} \cap F \cap G\right) \leq \frac{1}{15}$
• C. $P(E \cup F \cup G) \leq \frac{13}{24}$
• D. $P\left(E^{C} \cap F^{C} \cap G^{C}\right) \leq \frac{5}{12}$

Answer 1

The Correct Option is A, B, C

Step 1: Given Information
We are given the following probabilities for events \( E \), \( F \), and \( G \):
- \( P(E) = \frac{1}{8} \), \( P(F) = \frac{1}{6} \), and \( P(G) = \frac{1}{4} \).
- \( P(E \cap F \cap G) = \frac{1}{10} \).

Step 2: Understanding the Question
We are asked to verify the truth of the following statements:
- (A) \( P(E \cap F \cap G^{C}) \leq \frac{1}{40} \)
- (B) \( P(E^{C} \cap F \cap G) \leq \frac{1}{15} \)
- (C) \( P(E \cup F \cup G) \leq \frac{13}{24} \).

Step 3: Analyzing Statement A
We are asked to find \( P(E \cap F \cap G^{C}) \). Using the inclusion-exclusion principle, we know that:
\[ P(E \cap F \cap G^{C}) = P(E \cap F) - P(E \cap F \cap G) \] From the given information, \( P(E \cap F \cap G) = \frac{1}{10} \), and \( P(E \cap F) \) can be calculated as the intersection of the two events \( E \) and \( F \), which is at most \( P(E) \) or \( P(F) \). Hence, we have:
\[ P(E \cap F \cap G^{C}) \leq \frac{1}{40} \] Therefore, statement (A) is true.

Step 4: Analyzing Statement B
We are asked to find \( P(E^{C} \cap F \cap G) \). Again, using the inclusion-exclusion principle, we have:
\[ P(E^{C} \cap F \cap G) = P(F \cap G) - P(E \cap F \cap G) \] From the given information, \( P(E \cap F \cap G) = \frac{1}{10} \). We know that \( P(F \cap G) \) is at most \( P(F) \) or \( P(G) \), so we have:
\[ P(E^{C} \cap F \cap G) \leq \frac{1}{15} \] Therefore, statement (B) is true.

Step 5: Analyzing Statement C
We are asked to find \( P(E \cup F \cup G) \). Using the inclusion-exclusion principle, we have:
\[ P(E \cup F \cup G) = P(E) + P(F) + P(G) - P(E \cap F) - P(E \cap G) - P(F \cap G) + P(E \cap F \cap G) \] Substituting the given probabilities, we find that:
\[ P(E \cup F \cup G) \leq \frac{13}{24} \] Therefore, statement (C) is true.

Final Answer:
The correct answers are: 
- (A) \( P(E \cap F \cap G^{C}) \leq \frac{1}{40} \) 
- (B) \( P(E^{C} \cap F \cap G) \leq \frac{1}{15} \) 
- (C) \( P(E \cup F \cup G) \leq \frac{13}{24} \)