\( \quad[3 \mathrm{mks}] \) A fair coin is tossed 10 times. What is the probability of observing exactly 8 heads? \( [3 \mathrm{mks}] \)

To find the probability of observing exactly 8 heads when tossing a fair coin 10 times, we can use the binomial probability formula: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] where: - \( n \) is the number of trials (10 tosses), - \( k \) is the number of successes (8 heads), - \( p \) is the probability of success on a single trial (0.5 for a fair coin). Calculating, we get: \[ P(X = 8) = \binom{10}{8} (0.5)^8 (0.5)^{2} = \binom{10}{8} (0.5)^{10} \] The binomial coefficient \( \binom{10}{8} \) can also be calculated as \( \binom{10}{2} = \frac{10 \times 9}{2 \times 1} = 45 \). Thus: \[ P(X = 8) = 45 \times (0.5)^{10} = 45 \times \frac{1}{1024} = \frac{45}{1024} \] Therefore, the probability of observing exactly 8 heads is \( \frac{45}{1024} \).