Chapter 10.1, Problem 49E

Solution

(a.)

The probability that a house for sale has neither a garage nor a basement.

It has been determined that the probability that a house for sale has neither a garage nor a basement is $40\%$ .

Given:

A realtor reports that $30\%$ of houses currently listed for sale have garages, $55\%$ have basements and $25\%$ has both a garage and a basement.

Concept used:

If $P\left(E\right)$ is the probability of an event $E$ , then $P\left(A\cup B\right)+P\left(A\cap B\right)=P\left(A\right)+P\left(B\right)$ and $P\left(A^C\right)=100\%-P\left(A\right)$ .

Calculation:

Let $A$ denote the event of a house listed for sale having a garage and let $B$ denote the event of a house listed for sale having a basement.

It is given that a realtor reports that $30\%$ of houses currently listed for sale have garages, $55\%$ have basements and $25\%$ has both a garage and a basement.

Then, $P\left(A\right)=30\%$ , $P\left(B\right)=55\%$ and $P\left(A\cap B\right)=25\%$ .

Now, the probability that a house for sale has neither a garage nor a basement is $P\left({\left(A\cup B\right)}^C\right)$ .

Now, put $P\left(A\right)=30\%$ , $P\left(B\right)=55\%$ and $P\left(A\cap B\right)=25\%$ in $P\left(A\cup B\right)+P\left(A\cap B\right)=P\left(A\right)+P\left(B\right)$ to get,

  $\begin{array}{c}P\left(A\cup B\right)+25\%=30\%+55\%\\ P\left(A\cup B\right)+25\%=85\%\\ P\left(A\cup B\right)=85\%-25\%\\ P\left(A\cup B\right)=60\%\end{array}$

Now, $P\left(A^C\right)=100\%-P\left(A\right)$ .

Then, $P\left({\left(A\cup B\right)}^C\right)=100\%-P\left(A\cup B\right)$ .

Put $P\left(A\cup B\right)=60\%$ in the above equation to get,

  $P\left({\left(A\cup B\right)}^C\right)=40\%$

So, the probability that a house for sale has neither a garage nor a basement is $40\%$ .

Conclusion:

It has been determined that the probability that a house for sale has neither a garage nor a basement is $40\%$ .

(b.)

The probability that a house for sale has a garage or a basement.

It has been determined that the probability that a house for sale has a garage or a basement is $35\%$ .

Given:

A realtor reports that $30\%$ of houses currently listed for sale have garages, $55\%$ have basements and $25\%$ has both a garage and a basement.

Concept used:

If $P\left(E\right)$ is the probability of an event $E$ , then $P\left(A\cup B\right)+P\left(A\cap B\right)=P\left(A\right)+P\left(B\right)$ and $P\left(A^C\right)=100\%-P\left(A\right)$ .

Calculation:

Let $A$ denote the event of a house listed for sale having a garage and let $B$ denote the event of a house listed for sale having a basement.

It is given that a realtor reports that $30\%$ of houses currently listed for sale have garages, $55\%$ have basements and $25\%$ has both a garage and a basement.

Then, $P\left(A\right)=30\%$ , $P\left(B\right)=55\%$ and $P\left(A\cap B\right)=25\%$ .

Now, the probability that a house for sale has a garage or a basement is $P\left(A\cup B\right)-P\left(A\cap B\right)$ .

As determined previously, $P\left(A\cup B\right)=60\%$ .

Put $P\left(A\cup B\right)=60\%$ and $P\left(A\cap B\right)=25\%$ in $P\left(A\cup B\right)-P\left(A\cap B\right)$ to get,

  $P\left(A\cup B\right)-P\left(A\cap B\right)=60\%-25\%$

Simplifying,

  $P\left(A\cup B\right)-P\left(A\cap B\right)=35\%$

So, the probability that a house for sale has a garage or a basement is $35\%$ .

Conclusion:

It has been determined that the probability that a house for sale has a garage or a basement is $35\%$ .

(c.)

The probability that a house for sale has a garage given that it has a basement.

It has been determined that the probability that a house for sale has a garage given that it has a basement is approximately $45.45\%$ .

Given:

A realtor reports that $30\%$ of houses currently listed for sale have garages, $55\%$ have basements and $25\%$ has both a garage and a basement.

Concept used:

If $P\left(E\right)$ is the probability of an event $E$ , then $P\left(A\cup B\right)+P\left(A\cap B\right)=P\left(A\right)+P\left(B\right)$ and $P\left(A^C\right)=100\%-P\left(A\right)$ .

Calculation:

Let $A$ denote the event of a house listed for sale having a garage and let $B$ denote the event of a house listed for sale having a basement.

It is given that a realtor reports that $30\%$ of houses currently listed for sale have garages, $55\%$ have basements and $25\%$ has both a garage and a basement.

Then, $P\left(A\right)=30\%$ , $P\left(B\right)=55\%$ and $P\left(A\cap B\right)=25\%$ .

The probability that a house for sale has a garage given that it has a basement is $\frac{P\left(A\cap B\right)}{P\left(B\right)}$ .

Put $P\left(A\cap B\right)=25\%$ and $P\left(B\right)=55\%$ in $\frac{P\left(A\cap B\right)}{P\left(B\right)}$ to get,

  $\frac{P\left(A\cap B\right)}{P\left(B\right)}=\frac{25\%}{55\%}$

Simplifying,

  $\frac{P\left(A\cap B\right)}{P\left(B\right)}=\frac{5}{11}$

Converting to percentage,

  $\frac{P\left(A\cap B\right)}{P\left(B\right)}\approx 45.45\%$

So, the probability that a house for sale has a garage given that it has a basement is approximately $45.45\%$ .

Conclusion:

It has been determined that the probability that a house for sale has a garage given that it has a basement is approximately $45.45\%$ .

(d.)

The probability that a house for sale has a basement given that it has a garage.

It has been determined that the probability that a house for sale has a basement given that it has a garage is approximately $83.33\%$ .

Given:

A realtor reports that $30\%$ of houses currently listed for sale have garages, $55\%$ have basements and $25\%$ has both a garage and a basement.

Concept used:

If $P\left(E\right)$ is the probability of an event $E$ , then $P\left(A\cup B\right)+P\left(A\cap B\right)=P\left(A\right)+P\left(B\right)$ and $P\left(A^C\right)=100\%-P\left(A\right)$ .

Calculation:

Let $A$ denote the event of a house listed for sale having a garage and let $B$ denote the event of a house listed for sale having a basement.

It is given that a realtor reports that $30\%$ of houses currently listed for sale have garages, $55\%$ have basements and $25\%$ has both a garage and a basement.

Then, $P\left(A\right)=30\%$ , $P\left(B\right)=55\%$ and $P\left(A\cap B\right)=25\%$ .

The probability that a house for sale has a basement given that it has a garage is $\frac{P\left(A\cap B\right)}{P\left(A\right)}$ .

Put $P\left(A\cap B\right)=25\%$ and $P\left(A\right)=30\%$ in $\frac{P\left(A\cap B\right)}{P\left(A\right)}$ to get,

  $\frac{P\left(A\cap B\right)}{P\left(A\right)}=\frac{25\%}{30\%}$

Simplifying,

  $\frac{P\left(A\cap B\right)}{P\left(A\right)}=\frac{5}{6}$

Converting to percentage,

  $\frac{P\left(A\cap B\right)}{P\left(A\right)}\approx 83.33\%$

So, the probability that a house for sale has a basement given that it has a garage is approximately $83.33\%$ .

Conclusion:

It has been determined that the probability that a house for sale has a basement given that it has a garage is approximately $83.33\%$ .

(e.)

If having a garage and having a basement are independent events.

It has been determined that having a garage and having a basement are not independent events.

Given:

A realtor reports that $30\%$ of houses currently listed for sale have garages, $55\%$ have basements and $25\%$ has both a garage and a basement.

Concept used:

If $P\left(E\right)$ is the probability of an event $E$ , then $P\left(A\cup B\right)+P\left(A\cap B\right)=P\left(A\right)+P\left(B\right)$ and $P\left(A^C\right)=100\%-P\left(A\right)$ .

Calculation:

Let $A$ denote the event of a house listed for sale having a garage and let $B$ denote the event of a house listed for sale having a basement.

It is given that a realtor reports that $30\%$ of houses currently listed for sale have garages, $55\%$ have basements and $25\%$ has both a garage and a basement.

Then, $P\left(A\right)=30\%$ , $P\left(B\right)=55\%$ and $P\left(A\cap B\right)=25\%$ .

Now, if $A$ and $B$ are independent events, then $P\left(A\cap B\right)=P\left(A\right)P\left(B\right)$ .

Put $P\left(A\right)=30\%$ and $P\left(B\right)=55\%$ in $P\left(A\right)P\left(B\right)$ to get,

  $\begin{array}{c}P\left(A\right)P\left(B\right)=\left(30\%\right)\left(55\%\right)\\ =\left(\frac{30}{100}\right)\left(\frac{55}{100}\right)\\ =\left(\frac{30}{100}\right)\left(\frac{55}{100}\right)\times 100\%\\ =\left(\frac{30}{100}\right)\left(55\right)\%\end{array}$

Solving,

  $P\left(A\right)P\left(B\right)=16.5\%$

It is given that $P\left(A\cap B\right)=25\%$ .

Therefore, clearly $P\left(A\cap B\right)\ne P\left(A\right)P\left(B\right)$ .

This implies that $A$ and $B$ are not independent events.

Thus, having a garage and having a basement are not independent events.

Conclusion:

It has been determined that having a garage and having a basement are not independent events.