Chapter 6.1, Problem 13E

a.

To determine

Obtain the sample space for the event $B^c$.

Answer to Problem 13E

The sample space of possible outcomes is given by:

$\text{Sample space}=\left\{\begin{array}{l}\left(1,1,1\right),\left(1,1,2\right),\left(1,1,3\right),\left(1,2,1\right),\left(1,2,2\right),\left(1,3,1\right),\\ \left(1,3,3\right),\left(2,1,1\right),\left(2,1,2\right),\left(2,2,1\right),\left(2,2,2\right),\left(2,2,3\right),\\ \left(2,3,2\right),\left(2,3,3\right),\left(3,1,1\right),\left(3,1,3\right),\left(3,2,2\right),\left(3,2,3\right),\\ \left(3,3,1\right),\left(3,3,2\right),\left(3,3,3\right)\end{array}\right\}$.

Explanation of Solution

Calculation:

The given information is that there are three people in a family and each person $P_1$, $P_2$ and $P_3$ in the family are assigned to stations 1, 2 and 3.

Sample Space:

A sample space is the set of all possible outcomes of a random experiment.

Here, there are three persons in a family and each person is assigned to three stations.

The 27 possible outcomes are as follows:

$\text{Sample space}=\left\{\begin{array}{l}\left(1,1,1\right),\left(1,1,2\right),\left(1,1,3\right),\left(1,2,1\right),\left(1,2,2\right),\left(1,2,3\right),\left(1,3,1\right),\left(1,3,2\right),\\ \left(1,3,3\right),\left(2,1,1\right),\left(2,1,2\right),\left(2,1,3\right),\left(2,2,1\right),\left(2,2,2\right),\left(2,2,3\right),\left(2,3,1\right),\\ \left(2,3,2\right),\left(2,3,3\right),\left(3,1,1\right),\left(3,1,2\right),\left(3,1,3\right),\left(3,2,1\right),\left(3,2,2\right),\left(3,2,3\right),\\ \left(3,3,1\right),\left(3,3,2\right),\left(3,3,3\right)\end{array}\right\}$.

It is given that the event B is all three people go to different stations.

Therefore, the sample space for the event B is given by:

$\text{Sample space}=\left\{\left(1,2,3\right),\left(1,3,2\right),\left(2,1,3\right),\left(2,3,1\right),\left(3,1,2\right),\left(3,2,1\right)\right\}$.

Complement of an event A is the set of all outcomes that are not in A. It is denoted as $A^c$.

$P\left(A^c\right)=1-P\left(A\right)$

The event $B^c$ denotes the set of all outcomes that are not in B.

Therefore, the sample space for the event $B^c$ is given by:

$\text{Sample space}=\left\{\begin{array}{l}\left(1,1,1\right),\left(1,1,2\right),\left(1,1,3\right),\left(1,2,1\right),\left(1,2,2\right),\left(1,3,1\right),\\ \left(1,3,3\right),\left(2,1,1\right),\left(2,1,2\right),\left(2,2,1\right),\left(2,2,2\right),\left(2,2,3\right),\\ \left(2,3,2\right),\left(2,3,3\right),\left(3,1,1\right),\left(3,1,3\right),\left(3,2,2\right),\left(3,2,3\right),\\ \left(3,3,1\right),\left(3,3,2\right),\left(3,3,3\right)\end{array}\right\}$.

b.

To determine

Obtain the sample space for the event $C^c$.

Answer to Problem 13E

The sample space of possible outcomes is given by:

$\text{Sample space}=\left\{\begin{array}{l}\left(1,1,1\right),\left(1,1,2\right),\left(1,1,3\right),\left(1,2,1\right),\left(1,2,2\right),\left(1,3,1\right),\\ \left(1,3,3\right),\left(2,1,1\right),\left(2,1,2\right),\left(2,2,1\right),\left(2,2,2\right),\left(2,2,3\right),\\ \left(2,3,2\right),\left(2,3,3\right),\left(3,1,1\right),\left(3,1,3\right),\left(3,2,2\right),\left(3,2,3\right),\\ \left(3,3,1\right),\left(3,3,2\right),\left(3,3,3\right)\end{array}\right\}$.

Explanation of Solution

Calculation:

It is given that the event C is that no one goes to station.

Therefore, the sample space for the event C is given by:

$\text{Sample space}=\left\{\begin{array}{l}\left(1,1,1\right),\left(1,1,3\right),\left(1,3,1\right),\left(1,3,3\right),\left(3,1,1\right),\\ \left(3,1,3\right),\left(3,3,1\right),\left(3,3,3\right)\end{array}\right\}$.

Then the event $C^c$ denotes the set of all outcomes that are not in C.

Therefore, the sample space for the event $C^c$ is given by:

$\text{Sample space}=\left\{\begin{array}{l}\left(1,1,2\right),\left(1,2,1\right),\left(1,2,2\right),\left(1,2,3\right),\left(1,3,2\right),\left(2,1,1\right),\left(2,1,2\right),\\ \left(2,1,3\right),\left(2,2,1\right),\left(2,2,2\right),\left(2,2,3\right),\left(2,3,1\right),\left(2,3,2\right),\left(2,3,3\right),\\ \left(3,1,2\right),\left(3,2,1\right),\left(3,2,2\right),\left(3,2,3\right),\left(3,3,2\right)\end{array}\right\}$.

c.

To determine

Obtain the sample space for the event $A\cup B$.

Answer to Problem 13E

The sample space of possible outcomes is given by:

$\text{Sample space}=\left\{\begin{array}{l}\left(1,1,1\right),\left(2,2,2\right),\left(3,3,3\right),\left(1,2,3\right),\left(1,3,2\right),\left(2,1,3\right),\\ \left(2,3,1\right),\left(3,1,2\right),\left(3,2,1\right)\end{array}\right\}$.

Explanation of Solution

Calculation:

It is given that the event A is that all three people go to the same station.

Then the possibilities for event A are either three persons are assigned to station 1 or three persons are assigned to station 2 or three persons are assigned to station 3.

Therefore the sample space for A is given by:

$\text{Sample space}=\left\{\left(1,1,1\right),\left(2,2,2\right),\left(3,3,3\right)\right\}$.

It is given that the event B is that all three people go to the different stations.

Therefore, the sample space for the event B is given by:

$\text{Sample space}=\left\{\left(1,2,3\right),\left(1,3,2\right),\left(2,1,3\right),\left(2,3,1\right),\left(3,1,2\right),\left(3,2,1\right)\right\}$.

The sample space for the event $A\cup B$ contains all the outcomes in A and B.

Therefore, the sample space for the event $A\cup B$ is given by:

$\text{Sample space}=\left\{\begin{array}{l}\left(1,1,2\right),\left(1,2,1\right),\left(1,2,2\right),\left(1,2,3\right),\left(1,3,2\right),\left(2,1,1\right),\left(2,1,2\right),\\ \left(2,1,3\right),\left(2,2,1\right),\left(2,2,2\right),\left(2,2,3\right),\left(2,3,1\right),\left(2,3,2\right),\left(2,3,3\right),\\ \left(3,1,2\right),\left(3,2,1\right),\left(3,2,2\right),\left(3,2,3\right),\left(3,3,2\right)\end{array}\right\}$.

d.

To determine

Obtain the sample space for the event $A\cap B$.

Answer to Problem 13E

The sample space of possible outcomes is given by:

$\text{Sample space}=\varphi$.

Explanation of Solution

Calculation:

It is given that the event A is all three people go to the same station.

Then the possibilities for event A are either three persons are assigned to station 1 or three persons are assigned to station 2 or three persons are assigned to station 3.

Therefore, the sample space for A is given by:

$\text{Sample space}=\left\{\left(1,1,1\right),\left(2,2,2\right),\left(3,3,3\right)\right\}$.

It is given that the event B is that all three people go to the different station.

Therefore, the sample space for the event B is given by:

$\text{Sample space}=\left\{\left(1,2,3\right),\left(1,3,2\right),\left(2,1,3\right),\left(2,3,1\right),\left(3,1,2\right),\left(3,2,1\right)\right\}$.

The sample space for the event $A\cap B$ contains the outcomes that are common in both A and B. There is no outcomes that are common in both A and B.

Therefore, the sample space for the event $A\cap B$ is given by:

$\text{Sample space}=\varphi$.

e.

To determine

Obtain the sample space for the event $A\cap C$.

Answer to Problem 13E

The sample space of possible outcomes is given by:

$\text{Sample space}=\left\{\left(1,1,1\right),\left(3,3,3\right)\right\}$.

Explanation of Solution

Calculation:

It is given that the event A is that all three people go to the same station.

Then the possibilities for event A are either three person are assigned to station 1 or three person are assigned to station 2 or three person are assigned to station 3.

Therefore, the sample space for A is given by:

$\text{Sample space}=\left\{\left(1,1,1\right),\left(2,2,2\right),\left(3,3,3\right)\right\}$.

It is given that the event C is that no one goes to station.

Therefore, the sample space for the event C is given by:

$\text{Sample space}=\left\{\begin{array}{l}\left(1,1,1\right),\left(1,1,3\right),\left(1,3,1\right),\left(1,3,3\right),\left(3,1,1\right),\\ \left(3,1,3\right),\left(3,3,1\right),\left(3,3,3\right)\end{array}\right\}$.

The sample space for the event $A\cap C$ contains the outcomes that are common in both A and B. The outcomes that are common in both A and B are $\left(1,1,1\right)$ and. $\left(3,3,3\right)$.

Therefore, the sample space for the event $A\cap C$ is given by:

$\text{Sample space}=\left\{\left(1,1,1\right),\left(3,3,3\right)\right\}$.