<PRE><B>Question 872700</B>
 Out of each batch of 15 items, 4 are faulty. Items are examined one by one, and items checked are not replaced. 
a) what is the probability that there are exactly 3 defectives in the first 8 examined
This is a Binomial Problem with n = 8 and p(defect) = 4/15
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There are 8C3 = 56 ways to have 3 of 8 defective items
The probability of eachset of 3 defect and 5 not defect i (4/15)^3*(11/15)^8
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Answer to your Problem::
P(x = 3 defect in 8) = 56(4/15)^3*(11/15)^8 = 0.2252
If you use a TI calculator you get binompdf(8,4/15,3) = 0.2252
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b) probability that the 9th item examined is the 4th defective one found
Patterm:: nnnnnnnd = (11/15)^7*(4/15) = 0.000000017168..
Cheers,
Stan H.
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