Chapter 4.3, Problem 4.1E

Tossing a Die An experiment involves tossing a single die. These are some events:

A: Observe a 2

B: Observe an even number

C: Observe a number greater than 2

D: Observe both A and B

E: Observe A or B or both

F: Observe both A and C

a. List the simple events in the sample space.

b. List the simple events in each of the events A through F.

c. What probabilities should you assign to the simple events?

d. Calculate the probabilities of the six events A through F by adding the appropriate simple-event probabilities.

To determine

To list:the simple events in sample events.

Answer to Problem 4.1E

  $\left\{E_1,E_2,E_3,E_4,E_5,E_6\right\}$

Explanation of Solution

Given:

An experiment involves tossinga single die. The following events are given below:

  $\begin{array}{l}A:\text{Observe a }2\\ B:\text{Observe an even number }2\\ C:\text{Observe a number greater than }2\\ D:\text{Observe both }A\text{ and }B\\ E:\text{Observe }A\text{ or }B\text{ or both}\\ F:\text{Observe both }A\text{ and }C\end{array}$

Calculation:

Let $E_1$ be the event that $i$ is observed on die.

Therefore, the simple events in the sample are $\left\{E_1,E_2,E_3,E_4,E_5,E_6\right\}$ .

To determine

To list: the simple events in each of the events $A$ through $F$ .

Answer to Problem 4.1E

  $\begin{array}{l}A:E_2;B:E_2,E_4,E_6;\\ C:E_3,E_4,E_5,E_6;D:E_2;\\ E:E_2,E_4,E_6;F:\text{None}\text{.}\end{array}$

Explanation of Solution

Given:

An experiment involves tossing a single die. The following events are given below:

  $\begin{array}{l}A:\text{Observe a }2\\ B:\text{Observe an even number }2\\ C:\text{Observe a number greater than }2\\ D:\text{Observe both }A\text{ and }B\\ E:\text{Observe }A\text{ or }B\text{ or both}\\ F:\text{Observe both }A\text{ and }C\end{array}$

Calculation:

Let $E_1$ be the event that $i$ is observed on die.

Therefore, the simple events in each of the events $A$ through $F$ are:

  $\begin{array}{l}A:E_2\\ B:E_2,E_4,E_6\\ C:E_3,E_4,E_5,E_6\\ D:E_2\\ E:E_2,E_4,E_6\\ F:\text{None}\end{array}$

Hence,the listis

  $\begin{array}{l}A:E_2;B:E_2,E_4,E_6;\\ C:E_3,E_4,E_5,E_6;D:E_2;\\ E:E_2,E_4,E_6;F:\text{None}\text{.}\end{array}$ .

To determine

To find: the probability to assign the simple event.

Answer to Problem 4.1E

  $P\left(E_i\right)=\frac{1}{6}$

Explanation of Solution

Given:

An experiment involves tossing a single die. The following events are given below:

  $\begin{array}{l}A:\text{Observe a }2\\ B:\text{Observe an even number }2\\ C:\text{Observe a number greater than }2\\ D:\text{Observe both }A\text{ and }B\\ E:\text{Observe }A\text{ or }B\text{ or both}\\ F:\text{Observe both }A\text{ and }C\end{array}$

Calculation:

Let $E_1$ be the event that $i$ is observed on die.

Therefore, the simple events in the sample are $\left\{E_1,E_2,E_3,E_4,E_5,E_6\right\}$ .

As each event $E_i$ is equally likely,

Therefore, the probability getting to assign the simple event is $P\left(E_i\right)=\frac{1}{6}$ .

To determine

To find: the probability to assign the simple events $A$ through $F$ by adding the appropriate simple-event probabilities.

Answer to Problem 4.1E

  $P\left(A\right)=\frac{1}{6};P\left(B\right)=\frac{1}{2};P\left(C\right)=\frac{2}{3};P\left(D\right)=\frac{1}{6};P\left(E\right)=\frac{1}{2};P\left(F\right)=0$

Explanation of Solution

Given:

An experiment involves tossing a single die. The following events are given below:

  $\begin{array}{l}A:\text{Observe a }2\\ B:\text{Observe an even number }2\\ C:\text{Observe a number greater than }2\\ D:\text{Observe both }A\text{ and }B\\ E:\text{Observe }A\text{ or }B\text{ or both}\\ F:\text{Observe both }A\text{ and }C\end{array}$

Calculation:

Let $E_1$ be the event that $i$ is observed on die.

  $\begin{array}{l}P\left(A\right)=P\left(E_2\right)=\frac{1}{6}\\ P\left(B\right)=P\left(E_2\right)+P\left(E_4\right)+P\left(E_6\right)\\ P\left(B\right)=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}\\ P\left(B\right)=\frac{1}{2}\end{array}$

  $\begin{array}{l}P\left(C\right)=P\left(E_3\right)+P\left(E_4\right)+P\left(E_5\right)+P\left(E_6\right)\\ P\left(C\right)=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}\\ P\left(C\right)=\frac{2}{3}\\ P\left(D\right)=P\left(E_2\right)=\frac{1}{6}\end{array}$

  $\begin{array}{l}P\left(E\right)=P\left(E_2\right)+P\left(E_4\right)+P\left(E_6\right)\\ P\left(E\right)=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}\\ P\left(E\right)=\frac{1}{2}\\ P\left(F\right)=0\end{array}$

Therefore, the probability to assign the simple events $A$ through $F$ is $P\left(A\right)=\frac{1}{6};P\left(B\right)=\frac{1}{2};P\left(C\right)=\frac{2}{3};P\left(D\right)=\frac{1}{6};P\left(E\right)=\frac{1}{2};P\left(F\right)=0$ .