Question

The employees of a firm that manufactures insulation are being tested for indications of asbestos in their lungs. The firm is requested to send three employees who have positive indications of asbestos to a medical center for further testing. If 40% of the employees have positive indications of asbestos in their lungs, find the probability that fifteen employees must be tested in order to find three positives. (Round your answer to three decimal places.)

Answer 1

Answer:

0.013 = 1.3% probability that fifteen employees must be tested in order to find three positives.

Step-by-step explanation:

For each employee, there are only two possible outcomes. Either they test positive, or they do not. The probability of an employee testing positive is independent of any other employee, which means that the binomial probability distribution is used 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.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

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

This means that $p = 0.4$

Find the probability that fifteen employees must be tested in order to find three positives.

2 during the first 14(given by P(X = 2) when n = 14).

The 15th is positive, with 0.4 probability. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{14,2}.(0.4)^{2}.(0.6)^{12} = 0.0317$

0.0317*0.4 = 0.013.

0.013 = 1.3% probability that fifteen employees must be tested in order to find three positives.