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
(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
undo

Explanation:

Step1: Calculate the total number of ways to choose 4 people from the applicant pool

The applicant pool has \(7 + 9=16\) people. The number of ways to choose 4 people from 16 is given by the combination formula \(C(n,r)=\frac{n!}{r!(n - r)!}\), where \(n = 16\) and \(r=4\).

$$C(16,4)=\frac{16!}{4!(16 - 4)!}=\frac{16\times15\times14\times13}{4\times3\times2\times1}=1820$$

Step2: Calculate the number of ways to choose 4 database administrators

There are 7 database administrators. The number of ways to choose 4 from 7 is \(C(7,4)=\frac{7!}{4!(7 - 4)!}=\frac{7\times6\times5}{3\times2\times1} = 35\)

Step3: Calculate the probability

The probability \(P\) that the four people hired are all database administrators is the number of favorable outcomes (choosing 4 database administrators) divided by the number of total outcomes (choosing 4 people from 16).

$$P=\frac{C(7,4)}{C(16,4)}=\frac{35}{1820}\approx0.01923$$

Answer:

\(0.01923\)