Chapter 15, Problem 1E

(a)

To determine

To find: the probability that a home for sale has a pool or a garage.

Answer to Problem 1E

0.68

Explanation of Solution

Given:

  $\begin{array}{l}P\left(O\right)=0.21\\ P\left(G\right)=0.64\\ P\left(O\cap G\right)=0.17\end{array}$

Formula used:

Addition rule of probability

  $P\left(O\cup G\right)=P\left(O\right)+P\left(G\right)+P\left(O\cap G\right)$

Calculation:

Let O = home is having swimming pool

Let G = home is having garage

By addition rule of probability

  $\begin{array}{c}P\left(O\cup G\right)=P\left(O\right)+P\left(G\right)+P\left(O\cap G\right)\\ =0.21+0.64-0.17\\ =0.68\end{array}$

There is 68% possibility that chosen home has garage or swimming pool.

(b)

To determine

To find: the probability that a home for sale has neither a pool nor a garage.

Answer to Problem 1E

0.32

Explanation of Solution

Given:

  $\begin{array}{l}P\left(O\right)=0.21\\ P\left(G\right)=0.64\\ P\left(O\cap G\right)=0.17\end{array}$

Formula used:

  $P\left(O^{\text{'}}\cap G^{\text{'}}\right)=1-P\left(O\cup G\right)$

Calculation:

Calculating this using complement

  $\begin{array}{c}P\left(O^{\text{'}}\cap G^{\text{'}}\right)=1-P\left(O\cup G\right)\\ =1-0.68\\ =0.32\end{array}$

There is 32% possibility that chosen home has neither pool nor garage

(c)

To determine

To find: the probability that a home for sale has a pool but no garage.

Answer to Problem 1E

0.04

Explanation of Solution

Given:

  $\begin{array}{l}P\left(O\right)=0.21\\ P\left(G\right)=0.64\\ P\left(O\cap G\right)=0.17\end{array}$

Formula used:

  $P\left(O\cap G^{\text{'}}\right)=P\left(O\right)-P\left(O\cap G\right)$

Calculation:

Probability that chosen home has a pool but not garage is given by

  $\begin{array}{c}P\left(O\cap G^{\text{'}}\right)=P\left(O\right)-P\left(O\cap G\right)\\ =0.21-0.17\\ =0.04\end{array}$

There is 4% possibility that chosen home has swimming pool but not garage