Question

With one method of a procedure called acceptance sampling, a sample of items is randomly selected without replacement and the entire batch is accepted if every item in the sample is okay. The ABC Electronics Company has just manufactured 2100 write-rewrite CDs, and 60 are defective. If 3 of these CDs are randomly selected for testing, what is the probability that the entire batch will be accepted

Answer 1

Answer:

0.9167 = 91.67% probability that the entire batch will be accepted

Step-by-step explanation:

The CDs are chosen without replacement, which means that the hypergeometric distribution is used.

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.

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

2100 write-rewrite CDs, and 60 are defective.

This means, respectively, that $N = 2100, k = 60$

3 of these CDs are randomly selected for testing

This means that $n = 3$

What is the probability that the entire batch will be accepted?

Probability of none defective, so P(X = 0).

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

$P(X = 0) = h(0,2100,3,60) = \frac{C_{60,0}*C_{2040,3}}{C_{2100,3}} = 0.9167$

0.9167 = 91.67% probability that the entire batch will be accepted