Question

an engineering company has four openings and the applicant pool consists of seven database administrators and nine network engineers. if the hiring is done without regard for the specific qualifications of the applicants, find the probability that the four hired will be
part 1 of 5
(a) all network engineers. round your answer to five decimal places.
the probability that the four people hired will be all network engineers is 0.06923.
part 2 of 5
(b) two database administrators and two network engineers. round your answer to five decimal places.
the probability that two database administrators and two network engineers will be hired is 0.41538.
part 3 of 5
(c) all database administrators. round your answer to five decimal places.
the probability that the four people hired will be all database administrators is 0.01923.
part 4 of 5
(a) three database administrators and one network engineer. round your answer to five decimal places.
the probability that three database administrators and one network engineer will be hired is 0.17308.
part 4 / 5
part 5 of 5
(a) one database administrator and three network engineers. round your answer to five decimal places.
the probability that one database administrator and three network engineers will be hired is □

Explanation:

Step1: Calculate the total number of applicants

The applicant pool consists of \(7\) database administrators and \(9\) network engineers. So the total number of applicants \(N=7 + 9=16\)

Step2: Use the combination formula

The combination formula is \(C(n,k)=\frac{n!}{k!(n - k)!}\), where \(n\) is the total number of items, and \(k\) is the number of items to be chosen.
We want to choose \(1\) database administrator out of \(7\) (\(C(7,1)\)) and \(3\) network engineers out of \(9\) (\(C(9,3)\)), and the total number of ways to choose \(4\) people out of \(16\) is \(C(16,4)\)

Step3: Calculate each combination

• \(C(7,1)=\frac{7!}{1!(7 - 1)!}=\frac{7!}{1!6!}=7\)
• \(C(9,3)=\frac{9!}{3!(9 - 3)!}=\frac{9\times8\times7\times6!}{3\times2\times1\times6!}=84\)
• \(C(16,4)=\frac{16!}{4!(16 - 4)!}=\frac{16\times15\times14\times13\times12!}{4\times3\times2\times1\times12!}=1820\)

Step4: Calculate the probability

The probability \(P=\frac{C(7,1)\times C(9,3)}{C(16,4)}\)
Substitute the values: \(P=\frac{7\times84}{1820}=\frac{588}{1820}\approx0.323\)

Answer:

\(0.323\)