Question

With one method of a procedure called acceptance sampling, a sample of items is randomly selected without replacement and the entire batch is rejected if there is at least one defective unit. The ABCD Electronics Company has just manufactured 4500 write-rewrite CDs, and 150 are defective. If 4 of these CDs are randomly selected and tested, what is the probability that the entire batch will be rejected

Answer 1

0.1269 = 12.69% probability that the entire batch will be rejected.

Explanation:

CD's are chosen without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

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

`P(X = x) = h(x,N,n,k) = (C_(k,x)*C_(N-k,n-x))/(C_(N,n))`

In which:

x is the number of successes.

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) = (n!)/(x!(n-x)!)`

In this question:

4500 CDs, which means that
`N = 5000`

150 defective, which means that
`k = 150`

4 are selected, which means that
`n = 4`

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

At least 1 defective, which is:

`P(X \geq 1) = 1 - P(X = 0)`

In which

`P(X = x) = h(x,N,n,k) = (C_(k,x)*C_(N-k,n-x))/(C_(N,n))`

`P(X = 0) = h(0,4500,4,150) = (C_(150,0)*C_(4350,4))/(C_(4500,4)) = 0.8731`

So

`P(X \geq 1) = 1 - P(X = 0) = 1 - 0.8731 = 0.1269`

0.1269 = 12.69% probability that the entire batch will be rejected.