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 3
(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 3 / 5
part 4 of 5
(d) 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

Explanation:

Step1: Calculate the combination of total applicants

The total number of applicants is \(7 + 9=16\). The number of ways to choose \(4\) people out of \(16\) is given by the combination formula \(C(n,r)=\frac{n!}{r!(n - r)!}\), where \(n = 16\) and \(r=4\). So \(C(16,4)=\frac{16!}{4!(16 - 4)!}=\frac{16\times15\times14\times13}{4\times3\times2\times1}=1820\).

Step2: Calculate the combination for part (d)

We need to choose \(3\) database administrators out of \(7\) and \(1\) network engineer out of \(9\). The number of ways to choose \(3\) database administrators out of \(7\) is \(C(7,3)=\frac{7!}{3!(7 - 3)!}=\frac{7\times6\times5}{3\times2\times1} = 35\). The number of ways to choose \(1\) network engineer out of \(9\) is \(C(9,1)=\frac{9!}{1!(9 - 1)!}=9\). By the multiplication principle, the number of favorable cases is \(C(7,3)\times C(9,1)=35\times9 = 315\).

Step3: Calculate the probability for part (d)

The probability \(P=\frac{\text{Number of favorable cases}}{\text{Total number of cases}}\). So \(P=\frac{315}{1820}\approx0.17308\)

Answer:

\(0.17308\)