Question

A public interest group hires students to solicit donations by telephone. After a brief training period students make calls to potential donors and are paid on a commission basis. Experience indicates that early​ on, these students tend to have only modest success and that 60​% of them give up their jobs in their first two weeks of employment. The group hires 5 students, which can be viewed as a random sample.

What is the probability that at least 2 of the 5 will give up in the first two weeks?

What is the probability that at least 2 of the 5 will not give up in the first two weeks?

Answer 1

a)
Here, n = 5, p = 0.6, (1 - p) = 0.4 and x = 2
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X >= 2).
P(X >= 2) = (5C2 * 0.6^2 * 0.4^3) + (5C3 * 0.6^3 * 0.4^2) + (5C4 * 0.6^4 * 0.4^1) + (5C5 * 0.6^5 * 0.4^0)
P(X >= 2) = 0.2304 + 0.3456 + 0.2592 + 0.0778
P(X >= 2) = 0.9130

b)
Here, n = 5, p = 0.4, (1 - p) = 0.6 and x = 2
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X >= 2).
P(X >= 2) = (5C2 * 0.4^2 * 0.6^3) + (5C3 * 0.4^3 * 0.6^2) + (5C4 * 0.4^4 * 0.6^1) + (5C5 * 0.4^5 * 0.6^0)
P(X >= 2) = 0.3456 + 0.2304 + 0.0768 + 0.0102
P(X >= 2) = 0.6630