Question

A fair coin is tossed four times. Let X denote the number 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, HHTT, HTTT, THHH, ...
For the longest string of heads, X can take the values 0, 1, 2, 3 and 4.
(As when a coin is tossed 4 times, we can get minimum 0 and maximum 4 strings.)
Now,

\[P\left( X = 0 \right) = P\left( 0 \text{ head }\right) = \frac{1}{16}\]
\[P\left( X = 1 \right) = P\left( 1 \text{ head }  \right) = \frac{4}{16}\]
\[P\left( X = 2 \right) = P\left( 2 \text{ heads } \right) = \frac{6}{16}\]
\[P\left( X = 3 \right) = P\left( 3 \text{ heads }  \right) = \frac{4}{16}\]
\[P\left( X = 4 \right) = P\left( 4 \text{ heads } \right) = \frac{1}{16}\]

Thus, the probability distribution of X is given by

x	P(X)
0	\[\frac{1}{16}\]
1	\[\frac{4}{16}\]
2	\[\frac{6}{16}\]
3	\[\frac{4}{16}\]
4	\[\frac{1}{16}\]

Computation of mean and variance

xi	pi	pixi	pixi2
0	\[\frac{1}{16}\]	0	0
1	\[\frac{4}{16}\]	\[\frac{4}{16}\]	\[\frac{4}{16}\]
2	\[\frac{6}{16}\]	\[\frac{12}{16}\]	\[\frac{24}{16}\]
3	\[\frac{4}{16}\]	\[\frac{12}{16}\]	\[\frac{36}{16}\]
4	\[\frac{1}{16}\]	\[\frac{4}{16}\]	1
		`∑`pixi = 2	`∑`pixi2=5

\[\text{ Mean }  = \sum p_i x_i = 2\]
\[\text{ Variance }  = \sum p_i {x_i}2^{}_{} - \left( \text{ Mean }  \right)^2 \]
\[ = 5 - 4\]
\[ = 1\]