Question

Two fair coins \( S_1 \) and \( S_2 \) are tossed independently once. Let the events \( E, F, \) and \( G \) be defined as follows:
\( E \): Head appears on \( S_1 \)
\( F \): Head appears on \( S_2 \)  
\( G \): The same outcome (head or tail) appears on both \( S_1 \) and \( S_2 \)  
Then which of the following statements is NOT correct?

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

Answer 1

The Correct Option is D

Step 1: Define the sample space and probabilities.

The sample space of the two coin tosses is: \[ \{ (H, H), (H, T), (T, H), (T, T) \}. \]

Each outcome has an equal probability of \( \frac{1}{4} \) since the coins are fair and tossed independently.

The events are defined as:

• \( E = \{ (H, H), (H, T) \} \),
• \( F = \{ (H, H), (T, H) \} \),
• \( G = \{ (H, H), (T, T) \} \).

Step 2: Verify independence between events.

\( E \) and \( F \): The probability of \( E \cap F \) is: \[ P(E \cap F) = P(\{ (H, H) \}) = \frac{1}{4}. \] Since \( P(E) = P(F) = \frac{1}{2} \), we verify: \[ P(E \cap F) = P(E) \cdot P(F) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}. \] Thus, \( E \) and \( F \) are independent.

\( F \) and \( G \): The probability of \( F \cap G \) is: \[ P(F \cap G) = P(\{ (H, H) \}) = \frac{1}{4}. \] Since \( P(F) = P(G) = \frac{1}{2} \), we verify: \[ P(F \cap G) = P(F) \cdot P(G) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}. \] Thus, \( F \) and \( G \) are independent.

\( E \) and \( G^C \): The complement of \( G \) is: \[ G^C = \{ (H, T), (T, H) \}. \] The probability of \( E \cap G^C \) is: \[ P(E \cap G^C) = P(\{ (H, T) \}) = \frac{1}{4}. \] Since \( P(E) = \frac{1}{2} \) and \( P(G^C) = \frac{1}{2} \), we verify: \[ P(E \cap G^C) = P(E) \cdot P(G^C) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}. \] Thus, \( E \) and \( G^C \) are independent.

Step 3: Check mutual independence of \( E, F, \) and \( G \).

For mutual independence, all pairs and triplets of events must be independent. However, the independence of pairs does not guarantee mutual independence.

For example, consider \( P(E \cap F \cap G) \): \[ P(E \cap F \cap G) = P(\{ (H, H) \}) = \frac{1}{4}. \] However: \[ P(E) \cdot P(F) \cdot P(G) = \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{8}. \] Since \( P(E \cap F \cap G) \neq P(E) \cdot P(F) \cdot P(G) \), the events are not mutually independent.