Question

A fair coin is tossed four times. Let X denote the longest string of heads occurring. Find the probability distribution, mean and variance of X.

Answer 1

If a coin is tossed 4 times, then the possible outcomes are:
HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH and TTTT .

For the longest string of heads, X can take the values 0, 1, 2, 3 and 4.
(As when the coin is tossed 4 times, we can get maximum 4 and minimum 0 strings.)

Now,
$$\begin{gathered} P\left(X=0\right)=P\left(0\mathrm{head}\right)=\frac{1}{16} \\ P\left(X=1\right)=P\left(1\mathrm{head}\right)=\frac{7}{16} \\ P\left(X=2\right)=P\left(2\mathrm{heads}\right)=\frac{5}{16} \\ P\left(X=3\right)=P\left(3\mathrm{heads}\right)=\frac{2}{16} \\ P\left(X=4\right)=P\left(4\mathrm{heads}\right)=\frac{1}{16} \end{gathered}$$

Thus, the probability distribution of X is given by

X	P(X)
0	$\frac{1}{16}$
1	$\frac{7}{16}$
2	$\frac{5}{16}$
3	$\frac{2}{16}$
4	$\frac{1}{16}$

Computation of mean and variance

xi	pi	pixi	pixi2
0	$\frac{1}{16}$	0	0
1	$\frac{7}{16}$	$\frac{7}{16}$	$\frac{7}{16}$
2	$\frac{5}{16}$	$\frac{10}{16}$	$\frac{20}{16}$
3	$\frac{2}{16}$	$\frac{6}{16}$	$\frac{18}{16}$
4	$\frac{1}{16}$	$\frac{4}{16}$	1
		${\sum }_{}^{}$pixi = $\frac{27}{16}$	${\sum }_{}^{}$pixi2 = $\frac{61}{16}$

$$\begin{gathered} \mathrm{Mean}=\sum p_i{x}_i=\frac{27}{16}=1.7 \\ \mathrm{Variance}=\underset{}{\overset{}{\sum p_i{x_i}^2}}-{\left(\mathrm{Mean}\right)}^2 \\ =\frac{61}{16}-\frac{729}{256} \\ =\frac{247}{256} \\ =0.9 \end{gathered}$$