Question

A fair coin is tossed three times. Let A be the event of getting exactly two heads and B be the event of getting at most two tails, then \( P(A \cup B) \) is:

• A. \( \frac{1}{2} \)
• B. \( \frac{3}{8} \)
• C. \( \frac{1}{8} \)
• D. \( \frac{7}{8} \)

Hint:

For probability questions involving unions, always remember to subtract the intersection to avoid double-counting outcomes.

Answer 1

The Correct Option is D

Step 1: Define the events. - \( A \) represents the event of obtaining exactly two heads. The set of possible outcomes for \( A \) is \( \{HHT, HTH, THH\} \), therefore \( P(A) = \frac{3}{8} \). - \( B \) represents the event of obtaining at most two tails. The set of possible outcomes for \( B \) includes all outcomes except \( HHH \), thus \( P(B) = \frac{7}{8} \).

Step 2: Apply the formula for the union of events. The formula for the union of two events is: \[P(A \cup B) = P(A) + P(B) - P(A \cap B)\] The intersection of events \( A \) and \( B \), \( P(A \cap B) \), occurs when both conditions are met. The only outcome satisfying both is \( \{HHT, HTH, THH\} \), so \( P(A \cap B) = \frac{3}{8} \). \[P(A \cup B) = \frac{3}{8} + \frac{7}{8} - \frac{3}{8} = \frac{7}{8}\] The calculated probability is \( \frac{7}{8} \). Therefore, the final answer is \( \frac{7}{8} \).