Chapter 5, Problem 64CE

a.

To determine

Find the probability of selecting a family that prepared their own taxes.

Answer to Problem 64CE

The probability of selecting a family that prepared their own taxes is 0.50.

Explanation of Solution

Calculation:

The total number of families is 20 and the number of families prepared their own taxes is 10, the number of families prepared their taxes by a local professional is 7 and the number of families prepared by H&R block is 3.

The probability of selecting a family that prepared their own taxes is obtained as follows:

$\begin{array}{c}\text{Required probability}=\frac{\text{Number of families prepared thier own taxes}}{\text{Total number of families}}\\ =\frac{10}{20}\\ =0.50\end{array}$

Thus, the probability of selecting a family that prepared their own taxes is 0.50.

b.

To determine

Find the probability that selecting two families, both of which prepared their own taxes.

Answer to Problem 64CE

The probability that selecting two families, both of which prepared their own taxes is 0.2368.

Explanation of Solution

Calculation:

The probability that selecting two families, both of which prepared their own taxes is obtained as follows:

$\begin{array}{c}\text{Required probability}=\left(\begin{array}{l}\left(\frac{\left(\begin{array}{l}\text{Number of favorable families for selecting }\\ \text{a first family which prepared their own taxes }\end{array}\right)}{\text{Total number of families}}\right)\times \\ \left(\frac{\left(\begin{array}{l}\text{Number of favorable families for selecting }\\ \text{a second family which prepared their own taxes }\\ \text{after selecting a first family}\end{array}\right)}{\text{Total number of homes after selecting a first family}}\right)\end{array}\right)\\ =\frac{10}{20}\times \frac{9}{19}\\ =\frac{90}{380}\\ =0.2368\end{array}$

Thus, the probability that selecting two families, both of which prepared their own taxes is 0.2368.

c.

To determine

Find the probability of selecting three families, all of which prepared their own taxes.

Answer to Problem 64CE

The probability of selecting three families, all of which prepared their own taxes is 0.1053.

Explanation of Solution

Calculation:

The probability of selecting three families, all of which prepared their own taxes is obtained as follows:

$\begin{array}{c}\text{Required probability}=\left(\begin{array}{l}\left(\frac{\left(\begin{array}{l}\text{Number of favorable families for selecting }\\ \text{a first family which prepared their own taxes }\end{array}\right)}{\text{Total number of families}}\right)\times \\ \left(\frac{\left(\begin{array}{l}\text{Number of favorable families for selecting }\\ \text{a second family which prepared their own taxes }\\ \text{after selecting a first family}\end{array}\right)}{\text{Total number of homes after selecting a first family}}\right)\times \\ \left(\frac{\left(\begin{array}{l}\text{Number of favorable families for selecting }\\ \text{a third family which prepared their own taxes }\\ \text{after selecting a second family}\end{array}\right)}{\text{Total number of homes after selecting a second family}}\right)\end{array}\right)\\ =\frac{10}{20}\times \frac{9}{19}\times \frac{8}{18}\\ =\frac{720}{6,840}\\ =0.1053\end{array}$

Thus, the probability of selecting three families, all of which prepared their own taxes is 0.1053.

d.

To determine

Find the probability that selecting two families, neither of which had their taxes prepared by H&R block.

Answer to Problem 64CE

The probability that selecting two families, neither of which had their taxes prepared by H&R block is 0.7158.

Explanation of Solution

Calculation:

The number of families prepared by H&R block is 3 and the total number of families is 20.

The probability that selecting two families, neither of which had their taxes prepared by H&R block is obtained as follows:

$\begin{array}{c}\text{Required probability}=\left(\begin{array}{l}\left(\frac{\left(\text{Number of not favorable cases for selecting first family }\right)}{\text{Total number of cases}}\right)\times \\ \left(\frac{\left(\text{Number of not favorable cases for selecting second family }\right)}{\text{Total number of cases}}\right)\end{array}\right)\\ =\frac{17}{20}\times \frac{16}{19}\\ =0.7158\end{array}$

Thus, the probability that selecting two families, neither of which had their taxes prepared by H&R block is 0.7158.