Question

10. A random sample of 10 fibers is taken from a collection of 92 fibers that consists of 43 fibers of polymer A, 17 fibers of polymer B, and 32 fibers of polymer C. a. What is the probability that the sample does not contain any fibers of polymer B

Answer 1

Answer:

0.1150 = 11.50% probability that the sample does not contain any fibers of polymer B

Step-by-step explanation:

Hypergeometric distribution:

The probability of x sucesses is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of sucesses.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

In this question:

Fibers are chosen without replacement, which means that we use the hypergeometric distribution.

92 fibers means that $N = 92$

Sample of 10 means that $n = 10$

No polymeter B which means that $x = 0$

17 fibers of polymeter B which means that $k = 17$

a. What is the probability that the sample does not contain any fibers of polymer B

This is P(X = 0).

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 0) = h(0,92,10,17) = \frac{C_{17,0}*C_{75,10}}{C_{92,10}} = 0.1150$

0.1150 = 11.50% probability that the sample does not contain any fibers of polymer B