Chapter 4, Problem 25CQ

Selecting a Sweater At a men’s clothing store, 12 men purchased blue golf sweaters, 8 purchased green sweaters, 4 purchased gray sweaters, and 7 bought black sweaters. If a customer is selected at random, find the probability that he purchased

a. A blue sweater

b. A green or gray sweater

c. A green or black or blue sweater

d. A sweater that was not black

To determine

To obtain: The probability of the person purchased a blue sweater.

Answer to Problem 25CQ

The probability of the person purchased a blue sweateris $\frac{12}{31}$.

Explanation of Solution

Given info:

12 men purchased blue golf sweaters, 8 purchased green sweaters, 4 purchased gray sweaters and 7 bought black sweaters at men’s clothing store.

Calculation:

From the given data, it can be observed that total number of outcomes is 31.

Let event A denote the person purchased a blue sweater.

The possible number of outcomes for the person purchased a blue sweateris 12.

That is, there are 12 outcomes for getting an 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 12 for ‘Number of outcomes in A’ and 31 for ‘Total number of outcomes in the sample space’,

$P\left(A\right)=\frac{12}{\text{31}}$

Thus, the probability of the person purchased a blue sweateris $\frac{12}{31}$.

To determine

To obtain: The probability of the person purchased a green or gray sweater.

Answer to Problem 25CQ

The probability of the person purchased a green or gray sweater is $\frac{12}{31}$.

Explanation of Solution

Calculation:

Mutually exclusive:

If the two events cannot occur at the same time then the events are mutually exclusive or disjoint events.

Let event B denote the person purchased a green sweater.

The possible number of outcomes for the person purchased a green sweater is 8.

That is, there are 8 outcomes for getting an event B.

The formula for 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 8 for ‘Number of outcomes in B’ and 31 for ‘Total number of outcomes in the sample space’,

$P\left(B\right)=\frac{8}{\text{31}}$

Let event C denote the person purchased a gray sweater.

The possible number of outcomes for the person purchased a gray sweater is 4.

That is, there are 4 outcomes for getting an event C.

The formula for probability of event B is,

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

Substitute 4 for ‘Number of outcomes in C’ and 31 for ‘Total number of outcomes in the sample space’,

$P\left(C\right)=\frac{4}{\text{31}}$

Here, the event B and C cannot occur at the same time. So the events are mutually exclusive or disjoint events.

Addition Rule:

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

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

Substitute $\frac{8}{\text{31}}$ for $P\left(A\right)$ and $\frac{4}{\text{31}}$ for $P\left(B\right)$

$\begin{array}{c}P\left(B\text{ or }C\right)=\frac{8}{31}+\frac{4}{31}\\ =\frac{8+4}{31}\\ =\frac{12}{31}\end{array}$

Thus, the probability of the person purchased a green or gray sweater is $\frac{12}{31}$.

To determine

To obtain: The probability of the person purchased a green or black or blue sweater.

Answer to Problem 25CQ

The probability of the person purchased a green or black or blue sweater is $\frac{27}{31}$.

Explanation of Solution

Calculation:

Let event D denote the person purchased a green sweater.

The possible number of outcomes for the person purchased a green sweater is 8.

That is, there are 8 outcomes for getting an 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 8 for ‘Number of outcomes in D’ and 31 for ‘Total number of outcomes in the sample space’,

$P\left(D\right)=\frac{8}{\text{31}}$

Let event E denote the person purchased a black sweater.

The possible number of outcomes for the person purchased a black sweater is 7.

That is, there are 7 outcomes for getting an event E.

The formula for probability of event B is,

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

Substitute 7 for ‘Number of outcomes in E’ and 31 for ‘Total number of outcomes in the sample space’,

$P\left(E\right)=\frac{7}{\text{31}}$

Let event F denote the person purchased a blue sweater.

The possible number of outcomes for the person purchased a blue sweater is 12.

That is, there are 12 outcomes for getting an event E.

The formula for probability of event E is,

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

Substitute 12 for ‘Number of outcomes in E’ and 31 for ‘Total number of outcomes in the sample space’,

$P\left(E\right)=\frac{12}{\text{31}}$

Let event F denote the person purchased a blue sweater.

The possible number of outcomes for the person purchased a blue sweater is 12.

That is, there are 12 outcomes for getting an event F.

The formula for probability of event F is,

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

Substitute 12 for ‘Number of outcomes in F’ and 31 for ‘Total number of outcomes in the sample space’,

$P\left(F\right)=\frac{12}{\text{31}}$

Here, the event D, E and F cannot occur at the same time. So the events are mutually exclusive or disjoint events.

Addition Rule:

The formula for probability of getting event D or event E or F is,

$P\left(D\text{ or }E\text{ or }F\right)=P\left(D\right)+P\left(E\right)+P\left(F\right)$

Substitute $\frac{8}{\text{31}}$ for $P\left(D\right)$, $\frac{7}{\text{31}}$ for $P\left(E\right)$ and $\frac{12}{\text{31}}$ for $P\left(F\right)$

$\begin{array}{c}P\left(D\text{ or }E\text{ or }F\right)=\frac{8}{31}+\frac{7}{31}+\frac{12}{31}\\ =\frac{8+7+12}{31}\\ =\frac{27}{31}\end{array}$

Thus, the probability of the person purchased a green or black or blue sweater is $\frac{27}{31}$.

To determine

To obtain: The probability of the person purchased a sweater that was not black.

Answer to Problem 25CQ

The probability of the person purchased a sweater that was not black is $\frac{24}{31}$.

Explanation of Solution

Calculation:

Let event G denote the person purchased a sweater that was blackand the event $\overline{G}$ denote the person purchased a sweater that was not black

The possible number of outcomes for the person purchased a black sweater is 7.

That is, there are 7 outcomes for getting an event G.

The formula for probability of event G is,

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

Substitute 7 for ‘Number of outcomes in G’ and 31 for ‘Total number of outcomes in the sample space’,

$P\left(G\right)=\frac{7}{\text{31}}$

The formula for obtaining the probability for the person purchased a sweater that was not black is,

$P\left(\overline{G}\right)=1-P\left(G\right)$

Substitute $\frac{7}{\text{31}}$ for $P\left(G\right)$

$\begin{array}{c}P\left(\overline{G}\right)=1-\frac{7}{31}\\ =\frac{31-7}{31}\\ =\frac{24}{31}\end{array}$

Thus, the probability of the person purchased a sweater that was not black is $\frac{24}{31}$.