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: Direct sum:
$\binom95 + \binom96 + \binom97 + \binom98 + \binom99 = 126 + 84 + 36 + 9 + 1 = 256$.

Step 2: Divide by total:
$P = \frac{256}{2^9} = \frac{256}{512} = \frac12$.

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