College Algebra, Chapter 10, 10.3, Section 10.3, Problem 18

A pair of dice is rolled, and the numbers showing are observed.

a.) List the sample space of this experiment.

Sample space of rolling two dice

$\left( \begin{array}{cccccc}
(1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6) \\
(2,1) & (2,2) & (2,3) & (2,4) & (2,5) & (2,6) \\
(3,1) & (3,2) & (3,3) & (3,4) & (3,5) & (3,6) \\
(4,1) & (4,2) & (4,3) & (4,4) & (4,5) & (4,6) \\
(5,1) & (5,2) & (5,3) & (5,4) & (5,5) & (5,6) \\
(6,1) & (6,2) & (6,3) & (6,4) & (6,5) & (6,6)
\end{array} \right)$

b.) Find the probability of getting a sum of 7.

There are total of 36 elements in the sample space in which the elements $(1,6), (6,1), (5,2), (2,5), (3,4)$ and $(4,3)$ has the sum of $7$. Thus, the probability of $7$ is

$\displaystyle \frac{6}{36} = \frac{1}{6}$

c.) Find the probability of getting a sum of 9.

Consequently, the elements $(3,6), (6,3), (5,4)$ and $(4,5)$ has the sum of $9$. Thus, the probability in this case is

$\displaystyle \frac{4}{36} = \frac{1}{9}$

d.) Find the probability that the two dice show doubles (the same number).

Moreover, the elements that has the same number are $(1,1), (2,2), (3,3), (4,4), (5,5)$ and $(6,6)$. Thus, we have

$\displaystyle \frac{6}{36} = \frac{1}{6}$

e.) Find the probability that the two dice show different numbers.


In this case, we can use the compliment of the probability in part (d) to get

$\displaystyle 1 - \frac{1}{6} = \frac{5}{6}$

f.) Find the probability of getting a sum of 9 or higher.

Based from the sample space, the probability of getting a sum of $9$ or higher is

$\displaystyle \frac{4}{36} + \frac{3}{36} + \frac{2}{36} + \frac{1}{36} = \frac{10}{36} = \frac{5}{18}$