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 1800 write-rewrite CDs, and 100 are defective. If 5 of these CDs are randomly selected for testing, what is the probability that the entire batch will be accepted?
Rounded to four decimal place accuracy.
To find the probability that the entire batch will be accepted, we need to calculate the probability that all 5 CDs randomly selected for testing are not defective.
First, let's calculate the probability of selecting a not defective CD from the total population of CDs.
The total number of CDs in the batch is 1800, and the number of defective CDs is 100. Therefore, the number of non-defective CDs in the batch is 1800 - 100 = 1700.
The probability of selecting a not defective CD on the first selection is 1700/1800.
Next, the probability of selecting a not defective CD on the second selection, without replacement, is (1700 - 1)/(1800 - 1).
Following the same logic, we can calculate the probabilities for the subsequent selections:
- Probability of not defective CD on third selection = (1700 - 2)/(1800 - 2)
- Probability of not defective CD on fourth selection = (1700 - 3)/(1800 - 3)
- Probability of not defective CD on fifth selection = (1700 - 4)/(1800 - 4)
To find the probability that all 5 CDs selected are not defective, we need to multiply the probabilities of each individual selection:
P(all 5 CDs are not defective) = (1700/1800) * ((1700 - 1)/(1800 - 1)) * ((1700 - 2)/(1800 - 2)) * ((1700 - 3)/(1800 - 3)) * ((1700 - 4)/(1800 - 4))
Now, let's calculate this probability:
P(all 5 CDs are not defective) ≈ (1700/1800) * (1699/1799) * (1698/1798) * (1697/1797) * (1696/1796)
P(all 5 CDs are not defective) ≈ 0.8779
Therefore, the probability that the entire batch will be accepted is approximately 0.8779, rounded to four decimal places.