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Earlier discussion
1 older comment kept for context.
Victor Titus
P(at least one head) = 1 - P(4 tails)
Let \(p = \frac{1}{2}\) = probability of head and \(q = \frac{1}{2}\) = probability of tail.
\((p + q)^{4} = p^{4} + 4p^{3}q + 6p^{2}q^{2} + 4pq^{3} + q^{4}\)
P(4 tails) = \(q^{4} = (\frac{1}{2})^{4} = \frac{1}{16}\)
P(at least one head) = \(1 - \frac{1}{16} = \frac{15}{16}\)