Question

The firm is requested to send 3 employees who have positive indications of asbestos on to a medical center for further testing. Suppose 40% of the employees have positive indications of asbestos in their lungs. a) Find the probability that exactly 10 employees will be tested in order to find 3 positives

Answer 1

0.0645 = 6.45% probability that exactly 10 employees will be tested in order to find 3 positives

Explanation:

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.

40% of the employees have positive indications of asbestos in their lungs

This means that
`p = 0.4`

a) Find the probability that exactly 10 employees will be tested in order to find 3 positives

2 within the first 9(
`P(X = 2)` when
`n = 9`), and the 10th, with 0.4 probability. So

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

`P(X = 2) = C_(9,2).(0.4)^(2).(0.6)^(7) = 0.1612`

0.4*0.1612 = 0.0645

0.0645 = 6.45% probability that exactly 10 employees will be tested in order to find 3 positives