Chapter 4, Problem 46SE

The Wall Street Journal/Harris Personal Finance poll asked 2082 adults if they owned a home (All Business website, January 23, 2008). A total of 1249 survey respondents answered Yes. Of the 450 respondents in the 18–34 age group, 117 responded Yes.

1. a. What is the probability that a respondent to the poll owned a home?
2. b. What is the probability that a respondent in the 18–34 age group owned a home?
3. c. What is the probability that a respondent to the poll did not own a home?
4. d. What is the probability that a respondent in the 18–34 age group did not own a home?

To determine

Compute the probability that respondent to the poll owned a home.

Answer to Problem 46SE

The probability that respondent to the poll owned a home is 0.60.

Explanation of Solution

Calculation:

The data shows that there are a total of 2,082 adults. Total of 1,249 survey respondent answered as yes. Among the 450 respondent of age 18-34, 117 responded yes.

Event A denotes respondent to the poll owned a home.

The probability of the event can be obtained as follows:

$P\left(A\right)=\frac{\text{Number of ways event occurs}}{\text{Total frequency}}$

From the data, there were a total of 2,082 adults. Among the respondent 1,249 of them answer yes. Substitute these values in the above equation.

Therefore,

$\begin{array}{c}P\left(A\right)=\frac{1,249}{2,082}\\ =0.5999\\ =0.60\end{array}$

Thus, the probability that respondent to the poll owned a home is 0.60.

To determine

Compute the probability that respondent in the 18-34 age group owned a home.

Answer to Problem 46SE

The probability that respondent in the 18-34 age group owned a home is  0.26.

Explanation of Solution

Calculation:

Event B denotes respondent in the 18-34 age group owned a home.

The probability of the event can be obtained as follows:

$P\left(B\right)=\frac{\text{Number of ways event occurs}}{\text{Total frequency}}$

From the data, it is clear that there are total 450 respondent of age 18-34, and 117 of them responded yes. Substitute tis values in the above equation.

Therefore,

$\begin{array}{c}P\left(B\right)=\frac{117}{450}\\ =0.26\end{array}$

Thus, the probability that respondent in the 18-34 age group owned a home is 0.26.

To determine

Compute the probability that respondent to the poll did not own a home.

Answer to Problem 46SE

The probability that respondent to the poll did not own a home is  0.40.

Explanation of Solution

Calculation:

Event $A^c$  denotes respondent to the poll did not own a home.

Complementation Rule:

For any event A the complementation rule states that,

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

From part (a), $P\left(A\right)=0.60$ . Substitute tis values in the above equation.

Therefore,

$\begin{array}{c}P\left(A^c\right)=1-0.60\\ =0.40\end{array}$

Thus, probability that respondent to the poll did not own a home is 0.40.

To determine

Compute the probability that respondent in the 18-34 age group did not own a home.

Answer to Problem 46SE

The probability that respondent in the 18-34 age group did not own a home is 0.74.

Explanation of Solution

Calculation:

Event $B^c$ denotes respondent in the 18-34 age group did not own a home.

Complementation Rule:

For any event B the complementation rule states that,

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

From part (b), $P\left(B\right)=0.26$ . Substitute tis values in the above equation.

Therefore,

$\begin{array}{c}P\left(B^c\right)=1-0.26\\ =0.74\end{array}$

Thus, the probability that respondent in the 18-34 age group did not own a home is 0.74.