GATE CS 2023 Q43 — Probability: Random Variables and Distributions

Let the sample space for tossing two fair coins be $S = \{HH, HT, TH, TT\}$, where each outcome is equally likely with probability $\frac{1}{4}$.

The events are defined as:

• $A = \text{HEAD on both throws} = \{HH\} \implies P(A) = \frac{1}{4}$
• $B = \text{HEAD on the first throw} = \{HH, HT\} \implies P(B) = \frac{2}{4} = \frac{1}{2}$
• $C = \text{HEAD on the second throw} = \{HH, TH\} \implies P(C) = \frac{2}{4} = \frac{1}{2}$

Now, let us evaluate the options using the definition of independent events ($P(X \cap Y) = P(X)P(Y)$):

1. For Option A: $A \cap B = \{HH\} \implies P(A \cap B) = \frac{1}{4}$.
   Since $P(A)P(B) = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8} \neq P(A \cap B)$, $A$ and $B$ are not independent.
2. For Option B: $A \cap C = \{HH\} \implies P(A \cap C) = \frac{1}{4}$.
   Since $P(A)P(C) = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8} \neq P(A \cap C)$, $A$ and $C$ are not independent.
3. For Option C: $B \cap C = \{HH\} \implies P(B \cap C) = \frac{1}{4}$.
   Since $P(B)P(C) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = P(B \cap C)$, $B$ and $C$ are independent.
4. For Option D: Since $B$ and $C$ are independent, we have:
   $\operatorname{Prob}(B \mid C) = \frac{P(B \cap C)}{P(C)} = \frac{1/4}{1/2} = \frac{1}{2} = P(B)$.
   Thus, this statement is true.