Chapter 4.1, Problem 15E

Rolling Two Dice If two dice are rolled one time, find the probability of getting these results:

a. A sum of 5

b. A sum of 9 or 10

c. Doubles

To determine

To obtain: The probability of getting an outcome as sum of 5.

Answer to Problem 15E

The probability of getting an outcome as sum of 5 is $\frac{1}{9}$.

Explanation of Solution

Given info:

A pair of six-sided dice is rolled.

Calculation:

The possibilities for rolling a pair of six-sided dice is,

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

Thus, the total number of outcomes is 36.

Let event A denote that the outcome as sum of 5. Hence, the possible outcomes to get the sum 5 are ‘ $\left(1,4\right),\left(2,3\right),\left(3,2\right),\left(4,1\right)$’. That is, there are 4 outcomes for event A.

The formula for probability of event A is,

$P\left(A\right)=\frac{\text{Number of outcomes in }A}{\text{Total number of outcomes in the sample space}}$

Substitute 4 for ‘Number of outcomes in A’ and 36 for ‘Total number of outcomes in the sample space’,

$\begin{array}{c}P\left(A\right)=\frac{4}{\text{36}}\\ =\frac{1}{9}\end{array}$

Thus, the probability that the outcome is sum of 5 is $\frac{1}{9}$.

To determine

To obtain: The probability of getting an outcome is sum of 9 or sum 10.

Answer to Problem 15E

The probability of getting an outcome is sum of 9 or sum 10 is $\frac{7}{36}$.

Explanation of Solution

Calculation:

Here, the event B is defined getting sum of 9.

Hence, the possible outcomes are ‘ $\left(3,6\right),\left(4,5\right),\left(5,4\right),\left(6,2\right)$’. That is, there are 4 outcomes for event B.

The probability of event B is,

$P\left(B\right)=\frac{\text{Number of outcomes in }B}{\text{Total number of outcomes in the sample space}}$

Substitute 4 for ‘Number of outcomes in B’ and 36 for ‘Total number of outcomes in the sample space’,

$P\left(B\right)=\frac{4}{36}$

Here, the event C is defined as getting a sum of 10. Hence, the possible outcomes for sum 10 are ‘ $\left(4,6\right),\left(5,5\right),\left(6,4\right)$’. That is, there are 5 outcomes for event C.

The probability of event C is,

$P\left(C\right)=\frac{\text{Number of outcomes in }C}{\text{Total number of outcomes in the sample space}}$

Substitute 3 for ‘Number of outcomes in C’ and 36 for ‘Total number of outcomes in the sample space’,

$P\left(C\right)=\frac{3}{36}$

The formula for probability of event B and C is,

$P\left(B\text{ and }C\right)=\frac{\text{Number of outcomes in }B\text{ and }C}{\text{Total number of outcomes in the sample space}}$

Substitute 0 for ‘Number of outcomes in B and C’ and 36 for ‘Total number of outcomes in the sample space’,

$\begin{array}{c}P\left(B\text{ and }C\right)=\frac{0}{36}\\ =0\end{array}$

Addition Rule:

The formula for probability of getting event B or event C is,

$P\left(B\text{ or }C\right)=P\left(B\right)+P\left(C\right)-P\left(B\text{ and }C\right)$

Substitute $\frac{4}{36}$ for ‘ $P\left(B\right)$’, $\frac{3}{36}$ for ‘ $P\left(C\right)$’ and 0 for ‘ $P\left(B\text{ and }C\right)$’

$\begin{array}{c}P\left(B\text{ or }C\right)=\frac{4}{36}+\frac{3}{36}-0\\ =\frac{7}{36}\end{array}$

Thus, the probability of getting an outcome is sum of 9 or sum 10 is $\frac{7}{36}$.

To determine

To obtain: The probability of getting doubles.

Answer to Problem 15E

The probability of getting doubles is $\frac{1}{6}$.

Explanation of Solution

Calculation:

Let the event D denote getting doubles.

Hence, the possible outcomes are ‘ $\left\{\left(1,1\right),\left(2,2\right),\left(3,3\right),\left(4,4\right),\left(5,5\right),\left(6,6\right)\right\}$’. That is, there are 6 outcomes for event D.

The formula for probability of event D is,

$P\left(D\right)=\frac{\text{Number of outcomes in }D}{\text{Total number of outcomes in the sample space}}$

Substitute 6 for ‘Number of outcomes in D’ and 36 for ‘Total number of outcomes in the sample space’,

$\begin{array}{c}P\left(D\right)=\frac{6}{\text{36}}\\ =\frac{1}{6}\end{array}$

Thus, the probability that the outcome is doubles is $\frac{1}{6}$