Question

Find the mean and variance of the number of heads in three tosses of a coin.

Answer 1

Let X denote the number of heads in three tosses of a coin.Then, X can take values 0, 1, 2 and 3.
Now,
$$\begin{gathered} P\left(X=0\right)=P\left(TTT\right)=\frac{1}{8},P\left(X=1\right)=P\left(HTT\mathrm{or}TTH\mathrm{or}THT\right)=\frac{3}{8} \\ P\left(X=2\right)=P\left(HTH\mathrm{or}THH\mathrm{or}HHT\right)=\frac{3}{8},P\left(X=3\right)=P\left(HHH\right)=\frac{1}{8} \end{gathered}$$

Thus, the probability distribution of X is given by

X	P(X)
0	$\frac{1}{8}$
1	$\frac{3}{8}$
2	$\frac{3}{8}$
3	$\frac{1}{8}$

Computation of mean and variance

xi	pi	pixi	pixi2
0	$\frac{1}{8}$	0	0
1	$\frac{3}{8}$	$\frac{3}{8}$	$\frac{3}{8}$
2	$\frac{3}{8}$	$\frac{6}{8}$	$\frac{12}{8}$
3	$\frac{1}{8}$	$\frac{3}{8}$	$\frac{9}{8}$
		${\sum }_{}^{}$pixi = $\frac{3}{2}$	${\sum }_{}^{}$pixi2 = 3

$$\begin{gathered} \mathrm{Mean}=\sum p_i{x}_i=\frac{3}{2} \\ \mathrm{Variance}=\underset{}{\overset{}{\sum p_i{x_i}^2}}-{\left(\mathrm{Mean}\right)}^2 \\ =3-{\left(\frac{3}{2}\right)}^2 \\ =\frac{3}{4} \end{gathered}$$