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 3 employees who have positive indications of asbestos on to a medical center for further testing. If 60% of the employees have positive indications of asbestos in their lungs, find the probability that 11 employees must be tested in order to find 3 positives?

Answer 1

Answer:

$P(X=11) = (11-1 C 3-1)0.6^3 (1-0.6)^{11-3}= (10C2) 0.6^3 0.4^8 =0.00637$

So then the probability that 11 employees must be tested in order to find 3 positives is 0.00637

Step-by-step explanation:

Previous concepts

A negative binomial random variable "is the number X of repeated trials to produce r successes in a negative binomial experiment. The probability distribution of a negative binomial random variable is called a negative binomial distribution, this distribution is known as the Pascal distribution".

And the probability mass function is given by:

$P(X=x) = (x-1 C r-1)p^r (1-p)^{x-r}$

Where r-1 represent the number successes after the x-1 trials and p is the probability of a success on any given trial.

Solution to the problem

Let X the random variable that represent the number of trials on which the third employee has founded with a positive result of asbestos.

For this case the random variable Y follows a negative binomial distribution given by:

$X \sim Neg Bin(r=3, p=0.6)$

And we want to find the probability P(X=11) and if we replace in the mass function we got:

$P(X=11) = (11-1 C 3-1)0.6^3 (1-0.6)^{11-3}= (10C2) 0.6^3 0.4^8 =0.00637$

So then the probability that 11 employees must be tested in order to find 3 positives is 0.00637