Chapter 4.1, Problem 16E

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

a. A sum less than 9

b. A sum greater than or equal to 10

c. A 3 on one die or on both dice.

To determine

To obtain: The probability of getting a sum less than 9.

Answer to Problem 16E

The probability of getting a sum less than 9 is $\frac{13}{18}$.

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 getting a sum less than 9.

Hence, the possible outcomes some less than 9 are,

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

That is, there are 26 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 26 for ‘Number of outcomes in A’ and 36 for ‘Total number of outcomes in the sample space’,

$\begin{array}{c}P\left(A\right)=\frac{26}{\text{36}}\\ =\frac{13}{18}\end{array}$

Thus, the probability that the outcome is sum less than 9 is $\frac{13}{18}$.

To determine

To obtain: The probability of getting a number greater than or equal to 10.

Answer to Problem 16E

The probability of getting a number greater than or equal to 10 is $\frac{1}{6}$.

Explanation of Solution

Calculation:

Here, the event B is defined as rolling a number greater than or equal to 10. Hence, the possible outcomes greater than or equal to 10 are ‘ $\left(4,6\right),\left(5,5\right),\left(5,6\right),\left(6,4\right),\left(6,5\right),\left(6,6\right)$’. That is, there are 6 outcomes greater than or equal to 10.

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

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

Thus, the probability of getting a number greater than or equal to 10 is $\frac{1}{6}$.

To determine

To obtain: The probability of getting an outcome 3 on one die or 3 on both dice.

Answer to Problem 16E

The probability of getting an outcome 3 on one die or 3 on both dice is $\frac{11}{36}$.

Explanation of Solution

Calculation:

Here event C is defined as rolling a number 3 on one die.

Hence, the possible outcomes for 3 on one die are $\left\{\left(3,1\right),\left(3,2\right),\left(3,4\right),\left(3,5\right),\left(3,6\right),\left(1,3\right),\left(2,3\right),\left(4,3\right),\left(5,3\right),\left(6,3\right)\right\}$.

That is, there are 10 outcomes for event B.

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

$P\left(C\right)=\frac{10}{36}$

Here, the event D is defined as rolling a number 3 on both dice.

Hence, the possible outcome for 3 on both dice is $\left(3,3\right)$. The 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 1 for ‘Number of outcomes in D’ and 36 for ‘Total number of outcomes in the sample space’,

$P\left(C\right)=\frac{1}{36}$

The formula for probability of event C and D is,

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

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

$P\left(C\text{ and }D\right)=\frac{0}{36}$

Addition Rule:

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

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

Substitute $\frac{10}{36}$ for ‘ $P\left(C\right)$’, $\frac{1}{36}$ for ‘ $P\left(D\right)$’ and 0 for ‘ $P\left(C\text{ and }D\right)$’

$\begin{array}{c}P\left(C\text{ or }D\right)=\frac{10}{36}+\frac{1}{36}-\frac{0}{36}\\ =\frac{11}{36}\end{array}$

Thus, the probability of getting an outcome 3 on one die or 3 on both dice is $\frac{11}{36}$