Question

The probability of a potential employee passing a training course is 86%. If you selected 15 potential employees and gave them the training course, what is the probability that more than 11 will pass the test?

a. 0.648
b.0.900
c. 0.352
d. 0.852

Answer 1

d. 0.852

Explanation:

For each employee, there are only two possible outcomes. Either they will pass the test, or they will not. The probability of an employee passing the test is independent of other employees. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

In which
`C_(n,x)` is the number of different combinations of x objects from a set of n elements, given by the following formula.

`C_(n,x) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

The probability of a potential employee passing a training course is 86%.

This means that
`p = 0.86`

Sample of 15.

This means that
`n = 15`

What is the probability that more than 11 will pass the test?

`P(X > 11) = P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)`

So

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 12) = C_(15,12).(0.86)^(12).(0.14)^(3) = 0.204`

`P(X = 13) = C_(15,13).(0.86)^(13).(0.14)^(2) = 0.29`

`P(X = 14) = C_(15,14).(0.86)^(14).(0.14)^(1) = 0.254`

`P(X = 15) = C_(15,15).(0.86)^(15).(0.14)^(0) = 0.104`

`P(X > 11) = P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) = 0.204 + 0.29 + 0.254 + 0.104 = 0.852`

The correct answer is given by option d.