Problem Statement
An unbiased coin is tossed $6$ times. Find the probability of getting:
(i) Exactly $4$ heads.
(ii) At least $4$ heads.
Verified Solution & Marking Scheme
Identify Binomial Model Parameters
$n = 6, p = \frac{1}{2}, q = 1 - p = \frac{1}{2}$.
$P(X = r) = \binom{6}{r} \left(\frac{1}{2}\right)^r \left(\frac{1}{2}\right)^{6-r} = \binom{6}{r} \left(\frac{1}{2}\right)^6 = \frac{\binom{6}{r}}{64}$
Part (i): Exactly 4 Heads
$P(X = 4) = \frac{\binom{6}{4}}{64} = \frac{15}{64}$
Part (ii): At Least 4 Heads
$P(X \ge 4) = P(X = 4) + P(X = 5) + P(X = 6) = \frac{\binom{6}{4} + \binom{6}{5} + \binom{6}{6}}{64} = \frac{15 + 6 + 1}{64} = \frac{22}{64} = \frac{11}{32}$