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.
- \( A \) is the event of getting exactly two heads. The possible outcomes are \( HHT, HTH, THH \), so \( P(A) = \frac{3}{8} \). - \( B \) is the event of getting at most two tails. The possible outcomes are all three outcomes except \( HHH \), so \( P(B) = \frac{7}{8} \).

Step 2: Use 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 only outcome in which both events occur is \( HHT, HTH, THH \), so \( P(A \cap B) = \frac{3}{8} \). \[ P(A \cup B) = \frac{3}{8} + \frac{7}{8} - \frac{3}{8} = \frac{7}{8} \] Thus, the correct answer is \( \frac{7}{8} \). Therefore, the correct answer is 4. \( \frac{7}{8} \).