The variance of number of heads in three tosses of a coin is :

In three tosses of a coin let X be number of heads, probability distribution

X =	0	1	2	3
P(X) =	${ }^3 C_0 \frac{1}{2^3}$	${ }^3 C_1 \frac{1}{2^3}$	${ }^3 C_2 \frac{1}{2^3}$	${ }^3 C_3 \frac{1}{2^3}$

So mean = µ = $\sum x_i P\left(x_i\right)$

$=0 \times {}^3 C_0 \frac{1}{2^3}+1 \times {}^3 C_1 \frac{1}{2^3}+2 \times {}^3 C_2 \frac{1}{2^3}+3 \times {}^3 C_3 \frac{1}{2^3}$

$µ = E(X) = 0 + \frac{3}{8} + \frac{6}{8} + \frac{3}{8} = \frac{12}{8} = \frac{3}{2} = E(X)$

$E(X^2) = 0^2 × {}^3 C_0 \frac{1}{2^3} + 1^2 × {}^3 C_1 \frac{1}{2^3} + 2^2 × {}^3 C_2 \frac{1}{2^3} + 3^2 × {}^3 C_3 \frac{1}{2^3}$

$E(X^2) = \frac{3}{8} + \frac{12}{8} + \frac{9}{8} = \frac{24}{8} = \frac{6}{2} = 3$

So  σ² (variance) = $E(X^2)-[E(X)]^2$

$= 3 - \left(\frac{3}{2}\right)^2 = 3 - \frac{9}{4} = \frac{12-9}{4} = \frac{3}{4}$

$σ^2 = \frac{3}{4}$