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

The correct answer is Option (3) → 1

Let X be the number of heads. The sample space is {HH, HT, TH, TT}, so

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

The mean (expectation) is

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

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

Shortcut: X is binomial with n = 2 and p = 1/2, so the mean is \(np=2\times\frac{1}{2}=1\).