91Ó°ÊÓ

Suppose that two balanced dice are rolled, and letXdenote the absolute value of the difference between thetwo numbers that appear.

When two balanced dices are rolled, the total possible outcomes is 36

Xdenotes the absolute value of the difference between thetwo numbers that appear.

With X=0, the pairs are (1,1), (2,2), (3,3), (3,2), (4,4), (5,5) and (6,6), so

\(\begin{aligned}{}\Pr \left( {X = 0} \right) &= \frac{6}{{36}}\\& = \frac{1}{6}\end{aligned}\)

With X=1, the pairs are (1,2), (2,1), (2,3), (3,2), (3,4), (4,3), (4,5), (5,4), (5,6) and (6,5), so

\(\begin{aligned}{}\Pr \left( {X = 1} \right)& = \frac{{10}}{{36}}\\ &= \frac{5}{{18}}\end{aligned}\)

With X=2, the pairs are (1,3), (3,1), (2,4), (4,2), (3,5), (5,3), (4,6) and (6,4), so

\(\begin{aligned}{}\Pr \left( {X = 2} \right) &= \frac{8}{{36}}\\ &= \frac{2}{9}\end{aligned}\)

With X=3, the pairs are (1,4), (4,1), (2,5), (5,2), (3,6) and (6,3), so

\(\begin{aligned}{}\Pr \left( {X = 3} \right)& = \frac{6}{{36}}\\ &= \frac{1}{6}\end{aligned}\)

With X=4, the pairs are (1,5), (5,1), (2,6) and (6,2), so

p(4) = 4/36 = 1/9

\(\begin{aligned}{}\Pr \left( {X = 4} \right) &= \frac{4}{{36}}\\ &= \frac{1}{9}\end{aligned}\)

Finally, with X=5, the pairs are (1,6) and (6,1), so

\(\begin{aligned}{}\Pr \left( {X = 5} \right)& = \frac{2}{{36}}\\& = \frac{1}{{18}}\end{aligned}\)

Therefore the probability function is

\(f\left( x \right) = \left\{ \begin{array}{l}\frac{1}{6}\,\,\,\,\,\,\,for\,\,\,x = 0\\\frac{5}{{18}}\,\,\,\,for\,\,\,x = 1\\\frac{2}{9}\,\,\,\,\,\,\,for\,\,\,x = 2\\\frac{1}{6}\,\,\,\,\,\,\,for\,\,\,x = 3\\\frac{1}{9}\,\,\,\,\,\,\,for\,\,\,x = 4\\\frac{1}{{18}}\,\,\,\,for\,\,\,x = 5\end{array} \right.\)