Question

an apartment complex offers apartments with four different options, designated by a through d. there are an equal number of apartments with each combination of options. if there is only one apartment left, what is the probability that it is precisely what a person is looking for, namely two bedrooms, two bathrooms, first floor, and a golf course or no special view? a one bedroom two bedrooms three bedrooms b one bathroom two bathrooms c first floor second floor d no special view golf course view lake view question 11, 11.5.28 hw score: 61.9%, 8.67 of 14 points o points: 0 of 1

Explanation:

Step1: Count total combinations

There are 4 different options (A - D) for each category and an equal number of apartments with each combination. So the total number of combinations is \(4\times4\times4\times4 = 256\) (assuming each of the 4 - category (bedrooms, bathrooms, floor, view) choices are independent). But since we are only interested in the proportion of combinations that match the criteria, we can think of it in terms of the relative frequencies of each option within each category.

Step2: Determine favorable combinations

For bedrooms, the favorable option (two - bedrooms) is 1 out of 3. For bathrooms, the favorable option (two - bathrooms) is 1 out of 2. For floor, the favorable option (first - floor) is 1 out of 2. For view, the favorable options (golf course view or no special view) are 2 out of 3.

Step3: Calculate probability

Using the multiplication rule for independent events in probability \(P = \frac{1}{3}\times\frac{1}{2}\times\frac{1}{2}\times\frac{2}{3}\)

$$P=\frac{1\times1\times1\times2}{3\times2\times2\times3}=\frac{2}{36}=\frac{1}{18}$$

Answer:

\(\frac{1}{18}\)