If a fair coin is tossed four times, what is the probability of obtaining at least one head?
\(\frac{1}{2}\)
\(\frac{1}{4}\)
\(\frac{13}{16}\)
\(\frac{15}{16}\)
Correct answer is D
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) = \(1 - \frac{1}{16} = \frac{15}{16}\)