Question

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

Short answer

Therefore, the probability function is:

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

Step-by-step solution

Given information

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

Calculating the probability function

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

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

The sketch of the random variable is