Four fair coins are tossed once. Calculate the probability of having equal heads and tails.
\(\frac{1}{4}\)
\(\frac{3}{8}\)
\(\frac{1}{2}\)
\(\frac{15}{16}\)
Correct answer is B
Let \(p(head) = p = \frac{1}{2}\) and \(p(tail) = q = \frac{1}{2}\)
\((p + q)^{4} = p^{4} + 4p^{3}q + 6p^{2}q^{2} + 4pq^{3} + q^{4}\)
The probability of equal heads and tails = \(6p^{2}q^{2} = 6(\frac{1}{2}^{2})(\frac{1}{2}^{2})\)
= \(\frac{6}{16} = \frac{3}{8}\).