The mean of the number of heads in a simultaneous toss of three coins is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{3}{2}\)
The number of heads X is binomial with \(n = 3\) and \(p = \frac{1}{2}\), and the mean of a binomial distribution is np:
\( \displaystyle E(X) = np = 3 \times \frac{1}{2} = \frac{3}{2} \)
Directly: \(P(X = 0, 1, 2, 3) = \frac{1}{8}, \frac{3}{8}, \frac{3}{8}, \frac{1}{8}\), so
\( \displaystyle E(X) = \frac{0 + 3 + 6 + 3}{8} = \frac{12}{8} = \frac{3}{2} \)