The mean number of heads in three tosses of a fair coin is :

Let X be no. of heads in three tosses of a fair coin

Using binomial probability distribution

For X = 0 we get  P(X = 0) = ${ }^3 C_0\left(\frac{1}{2}\right)^3$

For X = 1 we get  P(X = 1) = ${ }^3 C_1\left(\frac{1}{2}\right)^3$

For X = 2 we get  P(X = 2) = ${ }^3 C_2\left(\frac{1}{2}\right)^3$

For X = 3 we get  P(X = 3) = ${}^3 C_3\left(\frac{1}{2}\right)^3$

So

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

for Mean y = $\sum P(X) X$

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

$=\frac{3}{8}+\frac{6}{8}+\frac{3}{8}=\frac{12}{8}=\frac{3}{2}$