if 5 of these CD's are randomly selected for testing, what is the probability that the entire batch will be accepted?

Question

with one method of a proced 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 ABcD Electronics Company has just manufactured 6000 write-rewrite CD's, and 170 are defective.

Answer 1

What if the question was about a sample of just one item? Then the probability would be (6000-170)/6000, or 1 - 170/6000 that we get a non-defective sample item, and therefore 0.9716... . This is called a Bernoulli trial. We will accept the entire batch with probability approximately 0.97.

 

Now, first we consider a series of Bernoulli trials. This is if we had REPLACEMENT (in this problem we don't). A coin toss is another example of a Bernoulli trial, with probability 1/2 instead of .97. The probability of getting two heads is 1/2 x 1/2 = 1/4, three heads 1/2 x 1/2 x 1/2 = 1/8, etc. Similarly here, the probability of getting five non-defective items in a row is 0.971666⁵, that is, the probability of getting one then raised to the fifth power. This is equal to 

 

(1-170/6000)⁵ = 0.8661...

 

Going back to the first paragraph, if we draw a non-defective item, and DON'T REPLACE IT, there are now 5999 total and 170 defective. The Bernoulli trial the second time around is (1 - 170/5999) etc.

 

Without replacement, the problem becomes (1 - 170/6000) x (1 - 170/5999) x (1 - 170/5998) x (1 - 170/5997) x (1 - 170/5996). Just like in the case with replacement, we are multiplying to get the probability of two statistically independent events happening. I hope that helps.