Answer to Question #308407 in Statistics and Probability for Zel

Question

Using the sample space for rolling two dice, construct a probability distribution for the random variable X representing the sum of the numbers that appear.

Answer 1

 

The sample space $\varOmega$ is

 $\left( 1,1 \right) ,\left( 1,2 \right) ,\left( 1,3 \right) ,\left( 1,4 \right) ,\left( 1,5 \right) ,\left( 1,6 \right) \\\left( 2,1 \right) ,\left( 2,2 \right) ,\left( 2,3 \right) ,\left( 2,4 \right) ,\left( 2,5 \right) ,\left( 2,6 \right) \\\left( 3,1 \right) ,\left( 3,2 \right) ,\left( 3,3 \right) ,\left( 3,4 \right) ,\left( 3,5 \right) ,\left( 3,6 \right) \\\left( 4,1 \right) ,\left( 4,2 \right) ,\left( 4,3 \right) ,\left( 4,4 \right) ,\left( 4,5 \right) ,\left( 4,6 \right) \\\left( 5,1 \right) ,\left( 5,2 \right) ,\left( 5,3 \right) ,\left( 5,4 \right) ,\left( 5,5 \right) ,\left( 5,6 \right) \\\left( 6,1 \right) ,\left( 6,2 \right) ,\left( 6,3 \right) ,\left( 6,4 \right) ,\left( 6,5 \right) ,\left( 6,6 \right)$

The sum for each $\omega \in \varOmega$ is respectively

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

Hence for X - sum of rolls:

$P\left( X=2 \right) =\frac{1}{36}\\P\left( X=3 \right) =\frac{2}{36}=\frac{1}{18}\\P\left( X=4 \right) =\frac{3}{36}=\frac{1}{12}\\P\left( X=5 \right) =\frac{4}{36}=\frac{1}{9}\\P\left( X=6 \right) =\frac{5}{36}\\P\left( X=7 \right) =\frac{6}{36}=\frac{1}{6}\\P\left( X=8 \right) =\frac{5}{36}\\P\left( X=9 \right) =\frac{4}{36}=\frac{1}{9}\\P\left( X=10 \right) =\frac{3}{36}=\frac{1}{12}\\P\left( X=11 \right) =\frac{2}{36}=\frac{1}{18}\\P\left( X=12 \right) =\frac{1}{36}$