Suppose a committee is to be formed with 5 engineers and 3 consultants selected randomly from a group of 8 engineers and 3 consultants.

Question

Suppose a committee is to be formed with 5 engineers and 3 consultants selected randomly from a group of 8 engineers and 3 consultants.

(a) What is the probability that engineer Elizabeth and consultant Carl are on the committee?

(b) What is the conditional probability that engineer Eddie is on the committee given that engineer Elizabeth is on the committee?

Answer 1

a) Since all 3 consultants are chosen, we know Carl will be chosen. Now we only need to think about Elizabeth. There are C(8,5) total ways to choose 5 engineers from 8. This is the denominator. For the numerator there are C(1,1) ways to choose Elizabeth and C(7,4) to choose the other 4 members. [C(1,1) = 1]

So P = C(7,4)/C(8,5) = 5/8

b) The three consultants and Eddie are chosen leaving a total of C(7,4) for choosing 4 engineers from 7. In similar logic to a) the numerator is now C(6,3)

So P = C(6,3)/C(7,4) = 4/7