Question

If a fair coin is tossed 10 times. Find the probability of getting at most six heads.

Hint:

For binomial probabilities, sum \( \binom{n}{k} p^k (1-p)^{n-k} \); use symmetry for fair coin.

Answer 1

Binomial distribution: \( X \sim B(10, 0.5) \).
\[ P(X \leq 6) = \sum_{k=0}^{6} \binom{10}{k} \left( \frac{1}{2} \right)^k \left( \frac{1}{2} \right)^{10-k} = \sum_{k=0}^{6} \binom{10}{k} \left( \frac{1}{2} \right)^{10}. \] \[ = \frac{1}{2^{10}} \sum_{k=0}^{6} \binom{10}{k} = \frac{1}{1024} (1 + 10 + 45 + 120 + 210 + 252 + 210) = \frac{848}{1024} = \frac{53}{64}. \] Answer: \( \frac{53}{64} \).