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

Explanation:

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

The total number of people is \(21 + 6=27\). We need to choose \(4\) people. Using the combination formula \(C(n,k)=\frac{n!}{k!(n - k)!}\), where \(n = 27\) and \(k = 4\).

$$C(27,4)=\frac{27!}{4!(27-4)!}=\frac{27\times26\times25\times24}{4\times3\times2\times1}=17550$$

Step2: Calculate the number of ways to form a committee of all students

There are \(21\) students. We need to choose \(4\) students. Using the combination formula with \(n = 21\) and \(k = 4\)

$$C(21,4)=\frac{21!}{4!(21 - 4)!}=\frac{21\times20\times19\times18}{4\times3\times2\times1}=5985$$

Step3: Calculate the probability

The probability \(P\) is the number of favorable outcomes (all - student committees) divided by the number of total outcomes (all possible 4 - person committees)

$$P=\frac{C(21,4)}{C(27,4)}=\frac{5985}{17550}\approx0.34102$$

Answer:

\(0.34102\)