The probability that fewer than 4 of them say job applicants should follow up within two weeks is: 0.03426
How to solve Binomial probability distribution?
To find the probability that fewer than 4 out of 11 randomly selected human resource managers say job applicants should follow up within two weeks, we can use the binomial probability formula.
The formula is:
P(X = x) = (ⁿCₓ)*(pˣ)*(1 - p)⁽ⁿ ⁻ ˣ⁾
where:
n is the number of trials (in this case, the number of selected human resource managers, which is 11)
k is the number of successful trials (in this case, the number of managers who say applicants should follow up within two weeks, which is 0, 1, 2, or 3)
p is the probability of success in a single trial (in this case, the probability that a manager says applicants should follow up within two weeks, which is 0.59)
(1 - p) is the probability of failure in a single trial
Thus, the probability that fewer than 4 of them say job applicants should follow up within two weeks is:
P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
P(X < 4) = 0.00006 + 0.00087 + 0.00627 + 0.02706
P(X < 4) = 0.03426
Complete question is:
Assume that when human resource managers are randomly selected, 59% say job applicants should follow up within two weeks. If 11 human resource managers are randomly selected, find the probability that fewer than 4 of them say job applicants should follow up within two weeks.