Question

Let \( P \) be a probability function that assigns the same weight to each of the points of the sample space \( \Omega = \{1, 2, 3, 4\} \). Consider the events \( E = \{1, 2\}, F = \{1, 3\}, G = \{3, 4\}. \) Then which of the following statement(s) is (are) true? 
 

• A. \( E \) and \( F \) are independent
• B. \( E \) and \( G \) are independent
• C. \( F \) and \( G \) are independent
• D. \( E \), \( F \), and \( G \) are independent

Hint:

To check if events are independent, calculate the probability of their intersections and compare with the product of their individual probabilities.

Answer 1

The Correct Option is A, C

Step 1: Calculate probabilities

$$P(E) = P({1,2}) = \frac{2}{4} = \frac{1}{2}$$ $$P(F) = P({1,3}) = \frac{2}{4} = \frac{1}{2}$$ $$P(G) = P({3,4}) = \frac{2}{4} = \frac{1}{2}$$

Step 2: Calculate intersections

$$E \cap F = {1,2} \cap {1,3} = {1}, \quad P(E \cap F) = \frac{1}{4}$$ $$E \cap G = {1,2} \cap {3,4} = \emptyset, \quad P(E \cap G) = 0$$ $$F \cap G = {1,3} \cap {3,4} = {3}, \quad P(F \cap G) = \frac{1}{4}$$ $$E \cap F \cap G = {1} \cap {3,4} = \emptyset, \quad P(E \cap F \cap G) = 0$$

Option (A): $E$ and $F$ are independent

Check if $P(E \cap F) = P(E) \cdot P(F)$: $$P(E \cap F) = \frac{1}{4}$$ $$P(E) \cdot P(F) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$$

Since $P(E \cap F) = P(E) \cdot P(F)$, events $E$ and $F$ are independent.

Option (A) is TRUE 

Option (B): $E$ and $G$ are independent

Check if $P(E \cap G) = P(E) \cdot P(G)$: $$P(E \cap G) = 0$$ $$P(E) \cdot P(G) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$$

Since $P(E \cap G) \neq P(E) \cdot P(G)$, events $E$ and $G$ are NOT independent.

Option (B) is FALSE 

Option (C): $F$ and $G$ are independent

Check if $P(F \cap G) = P(F) \cdot P(G)$: $$P(F \cap G) = \frac{1}{4}$$ $$P(F) \cdot P(G) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$$

Since $P(F \cap G) = P(F) \cdot P(G)$, events $F$ and $G$ are independent.

Option (C) is TRUE 

Option (D): $E$, $F$ and $G$ are independent

For three events to be mutually independent, we need:

1. Pairwise independence: $P(E \cap F) = P(E)P(F)$, $P(E \cap G) = P(E)P(G)$, $P(F \cap G) = P(F)P(G)$
2. Three-way independence: $P(E \cap F \cap G) = P(E)P(F)P(G)$

From our checks:

• $E$ and $F$ are independent ✓
• $E$ and $G$ are NOT independent ✗
• $F$ and $G$ are independent ✓

Since $E$ and $G$ are not pairwise independent, the three events are NOT mutually independent.

Option (D) is FALSE 

Answer: (A) and (C)