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 1500 write-rewrite CDs, and 170 are defective. If 6 of these CDs are randomly selected for testing, what is the probability that the entire batch will be accepted?

Answer 1

Given:

To Determine: The probability that the entire batch will be accepted if 6 CDs were randomly selected

Solution

Please note accepted CDs is the same as non-defective CDs.

If 6 were radomly selected without replacement, let us calculate the probability from the first to the sixth

`\begin{gathered} Recall: \\ P(A)=(n(A))/(n(S)) \\ So, \\ P(accepted)=(n(Accepted))/(n(Total)) \end{gathered}`

For the first accepted CD's

`P(accepted1)=(1330)/(1500)=(133)/(150)`

For the second accepted CD's without replacement

`P(accepted2)=(1330-1)/(1500-1)=(1329)/(1499)=0.886591`

For the third CD's

`P(accepted3)=(1329-1)/(1499-1)=(1328)/(1498)=0.886515`

For the fourth Cd's

`P(accepted4)=(1328-1)/(1498-1)=(1327)/(1497)=0.8864395`

For the fifth Cd's

`P(accepted5)=(1327-1)/(1497-1)=(1326)/(1496)=0.886364`

For the sixth CD's

`P(accepted6)=(1326-1)/(1496-1)=(1325)/(1495)=0.866288`

Therefore, the probability that the entire batch will be accepted would be

`P(accepted)=(1330)/(1500)*(1329)/(1499)*(1328)/(1428)*(1327)/(1427)*(1326)/(1426)*(1325)/(1425)`

`P(accepted)=0.485295\approx0.4853(nearest-4-decimal-place)`

Hence, the probability that the entire batch will be accepted is approximately 0.4853