Question

a student - faculty government committee of 4 people is to be formed from 21 student volunteers and 6 faculty volunteers. find the probability that the committee will consist of the following, assuming the selection is made at random:
part 1 of 4
(a) all faculty members. round your answer to five decimal places.
the probability that the committee will consist of all faculty members is 0.00085.
part 2 of 4
(b) two students and two faculty members. round your answer to five decimal places.
the probability that the committee will consist of two students and two faculty members is 0.17949.
part 3 of 4
(c) all students. round your answer to five decimal places.
the probability that the committee will consist of all students is 0.34103.
part 4 of 4
(d) one faculty member and three students. round your answer to five decimal places.
the probability that the committee will consist of one faculty member and three students is

Explanation:

Step1: Calculate the total number of ways to form the committee

The total number of people is \(21 + 6=27\). We need to form a committee of \(4\) people. Using the combination formula \(C(n,r)=\frac{n!}{r!(n - r)!}\), where \(n = 27\) and \(r = 4\). So, \(C(27,4)=\frac{27!}{4!(27 - 4)!}=\frac{27\times26\times25\times24}{4\times3\times2\times1}=17550\)

Step2: Calculate the number of ways to choose 1 faculty and 3 students

The number of ways to choose \(1\) faculty out of \(6\) is \(C(6,1)=\frac{6!}{1!(6 - 1)!}=6\). The number of ways to choose \(3\) students out of \(21\) is \(C(21,3)=\frac{21!}{3!(21 - 3)!}=\frac{21\times20\times19}{3\times2\times1}=1330\). By the multiplication principle, the number of ways to choose 1 faculty and 3 students is \(C(6,1)\times C(21,3)=6\times1330 = 7980\)

Step3: Calculate the probability

The probability \(P=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\). So, \(P=\frac{7980}{17550}\approx0.45470\)

Answer:

\(0.45470\)