The mean of the number of heads in the two tosses of a coin is

The correct answer is Option (1) → 1

Let X be the number of heads in two tosses of a fair coin. The outcomes HH, HT, TH, TT are equally likely, so the probability distribution of X is

    \( \displaystyle P(X = 0) = \frac{1}{4},\quad P(X = 1) = \frac{2}{4},\quad P(X = 2) = \frac{1}{4} \)

The mean is \(E(X) = \sum x\,P(x)\):

    \( \displaystyle E(X) = 0 \cdot \frac{1}{4} + 1 \cdot \frac{2}{4} + 2 \cdot \frac{1}{4} \)

    \( \displaystyle = \frac{2}{4} + \frac{2}{4} = 1 \)

This agrees with the binomial mean \(np = 2 \times \frac{1}{2} = 1\).

CBSE Applied Mathematics (Class XII):
Mathematical Expectation of Discrete Probability Distribution