Suppose that an employer plans to hire eight people from a group of fourteen equally qualified people, of whom four are minority candidates. If the employer does not know which candidates are minority candidates, and if she selects her employees at random, what is the probability that exactly one minority candidate is hired? Round your answer to four decimal places.

Question

Suppose that an employer plans to hire eight people from a group of fourteen equally qualified people, of whom four are minority candidates. If the employer does not know which candidates are minority candidates, and if she selects her employees at random, what is the probability that exactly one minority candidate is hired? Round your answer to four decimal places. ​

Quizplus · Accepted Answer

We first find the total number of ways to select 8 people from a group of 14 equally qualified people:

${{14}\choose{8}} = 3003$

Next, we find the number of ways to select exactly one minority candidate and seven non-minority candidates:

$({{4}\choose{1}} \cdot {{10}\choose{7}}) = 1200$

We multiply this by the number of ways to arrange these 8 people (since the order in which they are hired doesn't matter) which is $8!$.

So the total number of ways to select exactly one minority candidate and seven non-minority candidates and arrange them is:

$( {{4}\choose{1}} \cdot {{10}\choose{7}} ) \cdot 8! = 1200 \cdot 40,320 = 48,384,000$

Thus, the probability of selecting exactly one minority candidate is:

$\frac{48,384,000}{3003} \approx 0.1603$

Rounding to four decimal places, we get 0.1603, which corresponds to choice C.

Suppose That an Employer Plans to Hire Eight People from a Group