Question

A pair of fair dice is rolled. Let X the random variable representing the sum of the numbers that appear.Find P(3 ≤X≤ 10)

Answer 1

The sample space is,

$S=\begin{Bmatrix} (1,1) & (2,1)&(3,1)&(4,1)&(5,1)&(6,1) \\ (1,2) & (2,2)&(3,2)&(4,2)&(5,2)&(6,2)\\ (1,3)&(2,3)&(3,3)&(4,3)&(5,3)&(6,3)\\ (1,4)&(2,4)&(3,4)&(4,4)&(5,4)&(6,4)\\ (1,5)&(2,5)&(3,5)&(4,5)&(5,5)&(6,5)\\ (1,6)&(2,6)&(3,6)&(4,6)&(5,6)&(6,6) \end{Bmatrix}$

This sample space shows the outcome on the first dice followed by the outcome on the second dice.

Let the random variable $Z$ represent the sum of the outcome on the first dice and the outcome on the second dice.

Taking the sum of the outcome on the first and second dice gives,

$Z=\begin{Bmatrix} 2 & 3&4&5&6&7 \\ 3 & 4&5&6&7&8\\ 4&5&6&7&8&9\\ 5&6&7&8&9&10\\ 6&7&8&9&10&11\\ 7&8&9&10&11&12 \end{Bmatrix}$

Clearly, the random variable $Z$ may take on the values 2,3,4,5,6,7,8,9,10,11,12 with its probability distribution given as,

$z$ 2 3 4 5 6 7 8 9 10 11 12

$p(z)$ $1\over36$ $2\over36$ $3\over36$ $4\over36$ $5\over36$ $6\over36$ $5\over36$ $4\over36$ $3\over36$ $2\over36$ $1\over36$

Now,

$P(3 \le X\le10)={2\over36}+{3\over36}+{4\over36}+{5\over36}+{6\over36}+{5\over36}+{4\over36}+{3\over36}={32\over36}={8\over9}$

The required probability is ${8\over9}$.