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

Answer:

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

Step-by-step 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.

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

And p is the probability of X happening.

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

This means that $p = 0.4$

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