Question

A fair coin is tossed 9 times. On each toss, a man predicts that the outcome will be heads. The probability that the number of successful predictions is strictly greater than the number of unsuccessful predictions is...

• A. \(\frac{3}{4}\)
• B. \(\frac{1}{6}\)
• C. \(\frac{1}{2}\)
• D. \(-\frac{1}{2}\)

Hint:

With 9 tosses, successes beat failures when there are at least 5 heads, and the distribution is symmetric.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The number of heads \(X\) in \(9\) tosses of a fair coin follows \(B(9, \frac12)\). A prediction of "heads" is successful when the toss is heads.

Step 2: Key Formula or Approach:
Successes exceed failures when \(X > 9 - X\), i.e. \(X \geq 5\).

Step 3: Detailed Explanation:
\(P(X = r) = \binom9r\left(\frac12\right)^9\), which is symmetric: \(P(X = r) = P(X = 9 - r)\).
So \(P(X \geq 5) = P(X \leq 4)\). Since there are no ties (9 is odd), these two events cover everything, and each has probability \(\frac12\).
\[ P = \frac12 \]
Option D is negative, so it cannot be a probability.

Final Answer:
The probability is \(\frac{1}{2}\), option (C). \[ \boxed{\frac{1}{2}} \]