Chapter 3.2, Problem 26AQP

a.

Summary Introduction

To determine:

The probability of rolling two six-sided die and obtaining 2 and 3.

Introduction:

The six-sided die is a cube, on which the numbers from $\text{1}\text{to}\text{6}$ are marked once. When the die is rolled one number is obtained as the outcome.

Explanation of Solution

The number 2 is written only once on the die and can be expressed once. The number of chances that two will appear on die is one. The total numbers that can appear on die are six.

The probability for obtaining two on a six-sided die is as follows:

$\begin{array}{c}\text{Probability}=\frac{\text{Number}\text{of}\text{chances}\text{that}\text{2}\text{will}\text{appear}}{\text{Total}\text{numbers}\text{that}\text{can}\text{appear}\text{on}\text{a}\text{die}}\\ =\frac{1}{6}\end{array}$

The number 3 is written only once on the die and can be expressed once. The number of chances that three will appear on die is one. The total numbers that can appear on die are six.

The probability for obtaining three on a six-sided die is as follows:

$\begin{array}{c}\text{Probability}=\frac{\text{Number}\text{of}\text{chances}\text{that}3\text{will}\text{appear}}{\text{Total}\text{numbers}\text{that}\text{can}\text{appear}\text{on}\text{a}\text{die}}\\ =\frac{1}{6}\end{array}$

The probability of obtaining two or three can be calculated using the product rule that is by multiplying individual probabilities of each number. The probability of obtaining two or three is as follows:

$\begin{array}{c}\left(\text{P}\right)\text{of obtaining}\text{two}\text{or}\text{three}=\left(\text{P}\right)\text{ of}\text{obtaining}\text{two}\\ \times \left(\text{P}\right)\text{ of obtaining}\text{three}\\ =\frac{\text{1}}{\text{6}}\times \frac{\text{1}}{\text{6}}\\ =\frac{\text{1}}{\text{36}}\end{array}$

b.

Summary Introduction

To determine:

The probability of rolling two six-sided die and obtaining 6 and 6.

Introduction:

The six-sided die is a cube, on which the numbers from $\text{1}\text{to}\text{6}$ are marked once. When the die is rolled one number is obtained as the outcome.

Explanation of Solution

The number 6 is written only once on the die and can be expressed once. The number of chances that six will appear on die is one. The total numbers that can appear on die are six.

The probability for obtaining six on a six-sided die is as follows:

$\begin{array}{c}\text{Probability}=\frac{\text{Number}\text{of}\text{chances}\text{that}6\text{will}\text{appear}}{\text{Total}\text{numbers}\text{that}\text{can}\text{appear}\text{on}\text{a}\text{die}}\\ =\frac{1}{6}\end{array}$

The probability of obtaining six or six can be calculated using the product rule that is by multiplying individual probabilities of each number. The probability of obtaining six or six is as follows:

$\begin{array}{c}\left(\text{P}\right)\text{of obtaining}\text{six}\text{or}\text{six}=\left(\text{P}\right)\text{ of}\text{obtaining}\text{six}\\ \times \left(\text{P}\right)\text{ of obtaining}\text{six}\\ =\frac{\text{1}}{\text{6}}\times \frac{\text{1}}{\text{6}}\\ =\frac{\text{1}}{\text{36}}\end{array}$

c.

Summary Introduction

To determine:

The probability of rolling two six-sided die and obtaining at least one 6.

Introduction:

The six-sided die is a cube, on which the numbers from $\text{1}\text{to}\text{6}$ are marked once. When the die is rolled one number is obtained as the outcome.

Explanation of Solution

The number 6 is written only once on the die and can be expressed once. The number of chances that six will appear on die is one. The total numbers that can appear on die are six.

The probability for obtaining six on a six-sided die is as follows:

$\begin{array}{c}\text{Probability}=\frac{\text{Number}\text{of}\text{chances}\text{that}6\text{will}\text{appear}}{\text{Total}\text{numbers}\text{that}\text{can}\text{appear}\text{on}\text{a}\text{die}}\\ =\frac{1}{6}\end{array}$

The probability of obtaining any number but a six is as follows:

$\begin{array}{c}\text{Probability}=\frac{\text{Number}\text{of}\text{chances}\text{that}6\text{will}\text{not}\text{appear}}{\text{Total}\text{numbers}\text{that}\text{can}\text{appear}\text{on}\text{a}\text{die}}\\ =\frac{5}{6}\end{array}$

The probability of obtaining at least one six on rolling two six-sided die is as follows:

$\begin{array}{c}\left(\text{P}\right)\text{of obtaining}\text{at}\text{least}\text{one}\text{six}=\left(\begin{array}{l}\left(\text{P}\right)\text{ of}\text{obtaining}\text{six}\text{on}\text{a}\text{first}\text{die}+\\ \text{[((P))}\text{of}\text{obtaining}\text{any}\text{number}\text{but}\text{six}\times \\ \text{(P)}\text{of}\text{obtaining}\text{six}\text{on}\text{a}\text{second}\text{die]}\end{array}\right)\\ =\frac{\text{1}}{\text{6}}+\frac{\text{5}}{\text{6}}\times \frac{1}{6}\\ =\frac{6+5}{\text{36}}\\ =\frac{11}{36}\end{array}$

d.

Summary Introduction

To determine:

The probability of rolling two six-sided die and obtaining two of the same number (two 1 s, or two 2s, or two 3s, etc.).

Introduction:

The six-sided die is a cube, on which the numbers from $\text{1}\text{to}\text{6}$ are marked once. When the die is rolled one number is obtained as the outcome.

Explanation of Solution

The number 1 is written only once on the die and can be expressed once. The number of chances that one will appear on die is one. The total numbers that can appear on die are six.

The probability for obtaining one on a six-sided die is as follows:

$\begin{array}{c}\text{Probability}=\frac{\text{Number}\text{of}\text{chances}\text{that}1\text{will}\text{appear}}{\text{Total}\text{numbers}\text{that}\text{can}\text{appear}\text{on}\text{a}\text{die}}\\ =\frac{1}{6}\end{array}$

The probability of obtaining one or one can be calculated using the product rule that is by multiplying individual probabilities of each number. The probability of obtaining one or one is as follows:

$\begin{array}{c}\left(\text{P}\right)\text{of obtaining}\text{one}\text{or}\text{one}=\left(\text{P}\right)\text{ of}\text{obtaining}\text{one}\\ \times \left(\text{P}\right)\text{ of obtaining}\text{one}\\ =\frac{\text{1}}{\text{6}}\times \frac{\text{1}}{\text{6}}\\ =\frac{\text{1}}{\text{36}}\end{array}$

Similarly, the probability of obtaining two or two will be $\frac{1}{36}$ and so on.

The probability of obtaining two 1s, two 2s, two 3s, two 4s, two 5s, and two 6s is as follows:

$\text{Probability}=\left(\begin{array}{l}\text{Probability}\text{of}\text{obtaining}\text{two}\text{1s}+\\ \text{Probabilty}\text{of}\text{obtaining}\text{two}\text{2s}+\\ \text{Probabilty}\text{of}\text{obtaining}\text{two}\text{3s}+\\ \text{Probabilty}\text{of}\text{obtaining}\text{two}\text{4s}+\\ \text{Probabilty}\text{of}\text{obtaining}\text{two}\text{5s}+\\ \text{Probabilty}\text{of}\text{obtaining}\text{two}\text{6s}+\end{array}\right)$

$\begin{array}{c}\text{Probability}=\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}\\ =\frac{1+1+1+1+1+1}{36}\\ =\frac{6}{36}\\ =\frac{1}{6}\end{array}$

e.

Summary Introduction

To determine:

The probability of rolling two six-sided die and obtaining an even number on both dice.

Introduction:

The six-sided die is a cube, on which the numbers from $\text{1}\text{to}\text{6}$ are marked once. When the die is rolled one number is obtained as the outcome.

Explanation of Solution

A die has three even numbers. The probability of obtaining an even number is as follows:

$\begin{array}{c}\text{Probability}=\frac{\text{Number}\text{of}\text{chances}\text{that}\text{an}\text{even}\text{will}\text{appear}}{\text{Total}\text{numbers}\text{that}\text{can}\text{appear}\text{on}\text{a}\text{die}}\\ =\frac{3}{6}\end{array}$

The probability of obtaining an even number on both die can be calculated using the product rule that is by multiplying individual probabilities of each number. The probability of obtaining an even number is as follows:

$\begin{array}{c}\left(\text{P}\right)=\left(\text{P}\right)\text{ of}\text{obtaining}\text{an}\text{even}\text{number}\\ \times \left(\text{P}\right)\text{ of obtaining}\text{an}\text{even}\text{number}\\ =\frac{\text{3}}{\text{6}}\times \frac{\text{3}}{\text{6}}\\ =\frac{\text{9}}{\text{36}}\\ =\frac{1}{4}\end{array}$

f.

Summary Introduction

To determine:

The probability of rolling two six-sided die and obtaining an even number on at least one die.

Introduction:

The six-sided die is a cube, on which the numbers from $\text{1}\text{to}\text{6}$ are marked once. When the die is rolled one number is obtained as the outcome.

Explanation of Solution

A die has three even numbers. The probability of obtaining an even number is as follows:

$\begin{array}{c}\text{Probability}=\frac{\text{Number}\text{of}\text{chances}\text{that}\text{an}\text{even}\text{will}\text{appear}}{\text{Total}\text{numbers}\text{that}\text{can}\text{appear}\text{on}\text{a}\text{die}}\\ =\frac{3}{6}\end{array}$

A die has three even numbers. The probability of not obtaining an even number is as follows:

$\begin{array}{c}\text{Probability}=\frac{\text{Number}\text{of}\text{chances}\text{that}\text{an}\text{even}\text{will}\text{not}\text{appear}}{\text{Total}\text{numbers}\text{that}\text{can}\text{appear}\text{on}\text{a}\text{die}}\\ =\frac{3}{6}\end{array}$

An even number can be an outcome on first die but not on second die.

The probability of obtaining an even number on first die and not obtaining on second die can be calculated using the product rule that is by multiplying individual probabilities of each number. The probability of obtaining an even number on first die and not obtaining on second die is as follows:

$\begin{array}{c}\left(\text{P}\right)=\left(\begin{array}{c}\left(\text{P}\right)\text{of}\text{obtaining}\text{an}\text{even}\text{number}\text{on}\text{first}\text{die}\times \\ \left(\text{P}\right)\text{of}\text{not}\text{obtaining}\text{an}\text{even}\text{number}\text{on}\text{second}\text{die}\end{array}\right)\\ =\frac{\text{3}}{\text{6}}\times \frac{\text{3}}{\text{6}}\\ =\frac{\text{9}}{\text{36}}\end{array}$

An even number cannot be an outcome on first die but can be on second die.

The probability of obtaining an even number on second die and not obtaining on first die can be calculated using the product rule that is by multiplying individual probabilities of each number. The probability of obtaining an even number on second die and not obtaining on first die is as follows:

$\begin{array}{c}\left(\text{P}\right)=\left(\begin{array}{c}\left(\text{P}\right)\text{of}\text{obtaining}\text{an}\text{even}\text{number}\text{on}\text{second}\text{die}\times \\ \left(\text{P}\right)\text{of}\text{not}\text{obtaining}\text{an}\text{even}\text{number}\text{on}\text{first}\text{die}\end{array}\right)\\ =\frac{\text{3}}{\text{6}}\times \frac{\text{3}}{\text{6}}\\ =\frac{\text{9}}{\text{36}}\end{array}$

An even number can be an outcome on both die.

The probability of obtaining an even number on both die can be calculated using the product rule that is by multiplying individual probabilities of each number. The probability of obtaining an even number on both die is as follows:

$\begin{array}{c}\left(\text{P}\right)=\left(\begin{array}{c}\left(\text{P}\right)\text{of}\text{obtaining}\text{an}\text{even}\text{number}\text{on}\text{first}\text{die}\times \\ \left(\text{P}\right)\text{of}\text{not}\text{obtaining}\text{an}\text{even}\text{number}\text{on}\text{second}\text{die}\end{array}\right)\\ =\frac{\text{3}}{\text{6}}\times \frac{\text{3}}{\text{6}}\\ =\frac{\text{9}}{\text{36}}\end{array}$

The probability of obtaining an even number on at least one die is as follows:

$\text{Probability}=\left(\begin{array}{l}\text{Probability}\text{of}\text{obtaining}\text{an}\text{even}\text{number}\text{on}\text{first}\text{die}\text{but}\text{not}\text{on}\text{a}\text{second}\text{die}+\\ \text{Probability}\text{of}\text{obtaining}\text{an}\text{even}\text{number}\text{on}\text{second}\text{die}\text{but}\text{not}\text{on}\text{a}\text{first}\text{die}+\\ \text{Probability}\text{of}\text{obtaining}\text{an}\text{even}\text{number}\text{on}\text{both}\text{die}\end{array}\right)$

$\begin{array}{c}\text{Probability}=\frac{9}{36}+\frac{9}{36}+\frac{9}{36}\\ =\frac{9+9+9}{36}\\ =\frac{27}{36}\\ =\frac{3}{4}\end{array}$