Question

question 12 4 pts of the 18 residents in an apartment complex, 6 own the units and the rest are renters. 6 residents are randomly interviewed regarding the living conditions at the apartment complex. what is the probability that of the 6 residents who were interviewed, 3 of them own the unit in which they live? note: if your answer has up to four decimal places, please enter it below without rounding the number. but if your answer has more than four decimal places, round the number to four decimal places.

Explanation:

Step1: Calculate the number of renters

The number of renters is \(18 - 6=12\).

Step2: Use the combination formula \(C(n,k)=\frac{n!}{k!(n - k)!}\)

The total number of ways to choose \(6\) residents out of \(18\) is \(C(18,6)=\frac{18!}{6!(18 - 6)!}=\frac{18\times17\times16\times15\times14\times13}{6\times5\times4\times3\times2\times1}=18564\)
The number of ways to choose \(3\) owners out of \(6\) is \(C(6,3)=\frac{6!}{3!(6 - 3)!}=\frac{6\times5\times4}{3\times2\times1}=20\)
The number of ways to choose \(6-3 = 3\) renters out of \(12\) is \(C(12,3)=\frac{12!}{3!(12 - 3)!}=\frac{12\times11\times10}{3\times2\times1}=220\)

Step3: Calculate the probability

By the multiplication principle, the number of ways to choose \(3\) owners and \(3\) renters is \(C(6,3)\times C(12,3)=20\times220 = 4400\)
The probability \(P=\frac{C(6,3)\times C(12,3)}{C(18,6)}=\frac{4400}{18564}\approx0.2370\)

Answer:

\(0.2370\)