Question

Consider a random experiment where two fair coins are tossed. Let $A$ be the event that denotes HEAD on both throws, $B$ the event that denotes HEAD on the first throw, and $C$ the event that denotes HEAD on the second throw. Which of the following statements is/are TRUE?

• A. $A$ and $B$ are independent.
• B. $A$ and $C$ are independent.
• C. $B$ and $C$ are independent.
• D. $\Pr(B \mid C) = \Pr(B)$.

Hint:

To test independence: compute $\Pr(X\cap Y)$ and compare with $\Pr(X)\Pr(Y)$. Coin toss events for different coins are independent, but compound events (like $A$) may not be independent of single tosses.

Answer 1

The Correct Option is C

To determine which statements about the events when two fair coins are tossed are true, we begin by analyzing the probabilities of each event.

1. Possible outcomes when two fair coins are tossed are: {HH, HT, TH, TT}.
2. Event A: HEAD on both throws.
   The outcome for this event is {HH}. Thus, the probability of event A occurring is \Pr(A) = \frac{1}{4}.
3. Event B: HEAD on the first throw.
   The outcomes for this event are {HH, HT}. Thus, the probability of event B occurring is \Pr(B) = \frac{2}{4} = \frac{1}{2}.
4. Event C: HEAD on the second throw.
   The outcomes for this event are {HH, TH}. Hence, the probability of event C occurring is \Pr(C) = \frac{2}{4} = \frac{1}{2}.
5. To check the independence of two events, say X and Y, the criterion is whether \Pr(X \cap Y) = \Pr(X) \times \Pr(Y).
6. Checking independence of A and B:
   A \cap B = \{HH\} leading to \Pr(A \cap B) = \frac{1}{4}.
   \Pr(A) \times \Pr(B) = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}.
   Since \Pr(A \cap B) \neq \Pr(A) \times \Pr(B), events A and B are not independent.
7. Checking independence of A and C:
   A \cap C = \{HH\} leading to \Pr(A \cap C) = \frac{1}{4}.
   \Pr(A) \times \Pr(C) = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}.
   Since \Pr(A \cap C) \neq \Pr(A) \times \Pr(C), events A and C are not independent.
8. Checking independence of B and C:
   B \cap C = \{HH\}, leading to \Pr(B \cap C) = \frac{1}{4}.
   \Pr(B) \times \Pr(C) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}.
   Since \Pr(B \cap C) = \Pr(B) \times \Pr(C), events B and C are independent.
9. Checking the condition \Pr(B \mid C) = \Pr(B):
   \Pr(B \mid C) = \frac{\Pr(B \cap C)}{\Pr(C)} = \frac{1/4}{1/2} = \frac{1}{2}.
   \Pr(B) = \frac{1}{2}. Therefore, \Pr(B \mid C) = \Pr(B) is true.

From these computations, we identify that the correct statements are that events B and C are independent, and \Pr(B \mid C) = \Pr(B). The given correct answer states that B and C are independent.