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.
a: one bedroom, two bedrooms, three bedrooms, four bedrooms
b: one bathroom, two bathrooms, three bathrooms
c: first floor, second floor, third floor
d: no special view, golf course view, lake view, parking lot view
if there is only one apartment left, what is the probability that it is precisely what a person is looking for, namely three bedrooms, one bathroom, second floor, and a golf course or no special view?

Explanation:

Step1: Calculate total number of combinations

There are 4 options for A, 3 options for B, 3 options for C and 4 options for D. By the counting principle, the total number of combinations is $4\times3\times3\times4 = 144$.

Step2: Determine favorable number of combinations

The person wants three - bedrooms (1 option out of 4 for A), one bathroom (1 option out of 3 for B), second floor (1 option out of 3 for C), and a golf course or no special view (2 options out of 4 for D). So the number of favorable combinations is $1\times1\times1\times2=2$.

Step3: Calculate probability

The probability $P$ is the number of favorable outcomes divided by the total number of outcomes. So $P=\frac{2}{144}=\frac{1}{72}$.

Answer:

$\frac{1}{72}$