Question

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

Answer 1

Let X denote the number of tails in three tosses of a coin.
Then, X can take the values 0, 1, 2 and 3.
Now,

\[P\left( X = 0 \right) = P\left( HHH \right) = \frac{1}{8}, \]

\[P\left( X = 1 \right) = P\left( \text{ THH or HHT or HTH }\right) = \frac{3}{8}, \]

\[P\left( X = 2 \right) = P\left(\text{ TTH or THT or HTT }\right) = \frac{3}{8}, \]

\[P\left( X = 3 \right) = P\left( TTT \right) = \frac{1}{8}\]

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	\[\frac{1}{8}\]	\[\frac{3}{8}\]	\[\frac{9}{8}\]
		`∑`pixi = \[\frac{3}{2}\]	`∑`pixi2=3

\[\text{ Mean } = \sum p_i x_i = \frac{3}{2}\]
\[\text{ Variance } = \sum p_i {x_i}^2 - \left( \text{ Mean}  \right)^2 \]
\[ = 3 - \left( \frac{3}{2} \right)^2 \]
\[ = 3 - \frac{9}{4}\]
\[ = \frac{3}{4}\]