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

The correct answer is Option (3) → 1

X → no of heads in two tosses

$X=0⇒P(X=0)={^2C}_0(\frac{1}{2})^0(\frac{1}{2})^2=\frac{1}{4}$

$X=1⇒P(X=1)={^2C}_1(\frac{1}{2})(\frac{1}{2})=\frac{2}{4}=\frac{1}{2}$

$X=2⇒P(X=2)={^2C}_2(\frac{1}{2})^2(\frac{1}{2})^0=\frac{1}{4}$

Mean = $∑P(X_i)=1$