Four fair coins are tossed once. Calculate the probability of having equal heads and tails.
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Correct Answer: Option B
Explanation:
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 = \(6p^{2}q^{2} = 6(\frac{1}{2}^{2})(\frac{1}{2}^{2})\)
= \(\frac{6}{16} = \frac{3}{8}\).
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 = \(6p^{2}q^{2} = 6(\frac{1}{2}^{2})(\frac{1}{2}^{2})\)
= \(\frac{6}{16} = \frac{3}{8}\).