A local home improvement store mailed surveys to people living within a five-mile radius of the store. One question asked which type of saw the homeowner is most likely to buy in the near future. The survey responses for this question are shown in the table.

Question

Home Improvement Store Survey
Type of
Saw Number of
People
Chain 14
Circular 8
Jig 4
Miter 10
Reciprocating 2
Table 12

The store randomly chose two surveys, one at a time, where the first survey is replaced before the second survey is chosen. What is the probability that both surveys chosen are in favor of buying a miter saw for this question?
A.

B.

C.

D.

Answer 1

To find the probability that both surveys chosen are in favor of buying a miter saw, we first need to determine the total number of surveys and the probability of selecting a miter saw preference.

1. Calculate the total number of surveys: \[ \text{Total Number of People} = 14 + 8 + 4 + 10 + 2 + 12 = 50 \]

2. Determine the probability of selecting a miter saw preference: The number who prefer a miter saw is 10.

   \[ P(\text{miter saw}) = \frac{\text{Number of people who prefer a miter saw}}{\text{Total number of people}} = \frac{10}{50} = \frac{1}{5} \]

3. Since the first survey is replaced before choosing the second survey, the probability remains the same for both selections. Thus, the probability of choosing a miter saw preference twice is: \[ P(\text{both miter saw}) = P(\text{miter saw}) \times P(\text{miter saw}) = \left(\frac{1}{5}\right) \times \left(\frac{1}{5}\right) = \frac{1}{25} \]

So, the probability that both surveys chosen are in favor of buying a miter saw is \( \frac{1}{25} \).

To match this with the choices that are likely available (A, B, C, D), you would look for an option that corresponds to \( \frac{1}{25} \).