Question

A pair of fair dice is thrown. Let X be the random variable which denotes the minimum of the two numbers which appear. Find the probability distribution, mean and variance of X.

 

Answer 1

Let X denote the event of getting twice the number. Then, X can take the values 1, 2, 3, 4, 5 and 6.
Thus, the probability distribution of X is given by

x	P(X)
1	\[\frac{11}{36}\]
2	\[\frac{9}{36}\]
3	\[\frac{7}{36}\]
4	\[\frac{5}{36}\]
5	\[\frac{3}{36}\]
6	\[\frac{1}{36}\]

Computation of mean and variance

xi	pi	pixi	pixi2
1	\[\frac{11}{36}\]	\[\frac{11}{36}\]	\[\frac{11}{36}\]
2	\[\frac{9}{36}\]	\[\frac{18}{36}\]	1
3	\[\frac{7}{36}\]	\[\frac{21}{36}\]	\[\frac{63}{36}\]
4	\[\frac{5}{36}\]	\[\frac{20}{36}\]	\[\frac{80}{36}\]
5	\[\frac{3}{36}\]	\[\frac{15}{36}\]	\[\frac{75}{36}\]
6	\[\frac{1}{36}\]	\[\frac{6}{36}\]	1
		`∑`pixi =\[\frac{91}{36} = 2 . 5\]	`∑`pixi2=\[\frac{301}{36} = 8 . 4\]

\[\text{ Mean }  = \sum p_i x_i = 2 . 5\]
\[\text{ Variance }  = \sum p_i {x_i}2^{}_{} - \left( \text{ Mean }  \right)^2 = 8 . 4 - 6 . 25 = 2 . 15\]