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,

\[P\left( X = 0 \right) = P\left( 0 \text{ head }  \right) = \frac{1}{16}\]
\[P\left( X = 1 \right) = P\left( 1 \text{ head }  \right) = \frac{7}{16}\]
\[P\left( X = 2 \right) = P\left( 2 \text{ heads } \right) = \frac{5}{16}\]
\[P\left( X = 3 \right) = P\left( 3 \text{ heads }  \right) = \frac{2}{16}\]
\[P\left( X = 4 \right) = P\left( 4 \text{ heads }  \right) = \frac{1}{16}\]

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
		`∑`pixi = \[\frac{27}{16}\]	`∑`pixi2 =\[\frac{61}{16}\]

\[\text{ Mean }  = \sum p_i x_i = \frac{27}{16} = 1 . 7\]
\[\text{ Variance } = \sum p_i {x_i}2^{}_{} - \left( \text{ Mean }  \right)^2 \]
\[ = \frac{61}{16} - \frac{729}{256}\]
\[ = \frac{247}{256}\]
\[ = 0 . 9\]