Question

Home modifications for wheelchair users. The American Journal of Public Health (January 2002) reported on a study of elderly wheelchair users who live at home. A sample of 306 wheelchair users, age 65 or older, were surveyed about whether they had an injurious fall during the year and whether their home features any one of five structural modifications: bathroom modifications, widened doorways/hallways, kitchen modifications, installed railings, and easy-open doors. The responses are summarized the accompanying table. Suppose we select, at random, one of the 306 surveyed wheelchair users.

a. Find the probability that the wheelchair user had an injurious fall.

b. Find the probability that the wheelchair user had all five features installed in the home.

c. Find the probability that the wheelchair user had no falls and none of the features installed in the home.

d. Given the wheelchair user had all five features installed, what is the probability that the user had an injurious fall?

e. Given the wheelchair user had none of the features installed, what is the probability that the user had an injurious fall?

Short answer

1. The probability is 0.157.
2. The probability is 0.029.
3. The probability is 0.291.
4. The probability is 0.222.
5. The probability is 0.183

Step-by-step solution

Important formula

The formula for probability are $\mathbf{P}\mathbf{=}\frac{\mathrm{Favourable}\mathrm{out}\mathrm{comes}}{\mathrm{Total}\mathrm{out}\mathrm{comes}}$

(a) The probability that the wheelchair user had an injurious fall

$\begin{array}{c}\mathbf{P}\mathbf{(}\mathbf{injuries}\mathbf{}\mathbf{fall}\mathbf{)}\mathbf{=}\frac{48}{306}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{157}\end{array}$

Hence, the probability is 0.157.

(b) The probability that the wheelchair user had all five features installed in the home

$\begin{array}{l}\mathbf{P}\mathbf{(}\mathbf{all}\mathbf{}\mathbf{5}\mathbf{}\mathbf{at}\mathbf{}\mathbf{home}\mathbf{)}\mathbf{=}\frac{9}{306}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{029}\end{array}$

Hence, the probability is 0.029.

(c) The probability that the wheelchair user had no falls and none of the features installed in the home

$\begin{array}{l}\mathbf{P}\mathbf{\left(}\mathbf{NF}\mathbf{\right)}\mathbf{=}\frac{89}{306}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{291}\end{array}$

Hence, the probability is 0.291.

(d) The probability that the user had an injurious fall

$\begin{array}{l}\mathbf{P}\mathbf{\left(}\mathbf{I}\mathbf{\right|}\mathbf{FI}\mathbf{)}\mathbf{=}\frac{2}{9}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{222}\end{array}$

Hence, the probability is 0.222.

(e) The probability that the user use none had an injurious fall

$\begin{array}{c}\mathbf{P}\mathbf{\left(}\mathbf{I}\mathbf{\right|}\mathbf{NF}\mathbf{)}\mathbf{=}\frac{20}{109}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{183}\end{array}$

Hence, the probability is 0.183.