I need some help with this exercise, I'm stuck, I'm tryng to resolve
it with the HYPERGEOMETRIC DISTRIBUTION formula but I keep getiing
huge numbers. A population consists of 15 items, 10 of which are
acceptable. In a sample of 4 items, what is the probability that
exactly 3 are acceptable? please explain. thanks

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Clarification of Question by mojomabris-ga on 11 Apr 2006 20:44 PDT

thank you for the explanation guys, mojomabris

You want to choose four acceptable items out of ten, and one
unacceptable item out of five without replacement.  This from five
items out of 15.

If you are choosing 4 out of 10 items 10!/[(4!)(6!)] is written (10C4)
or ten choose four.

Probability = (10C4)(5C1)/(15C5) = 50/143

Ansels explanation is correct however in your question you wanted to
choose 3 acceptable out of ten changing the answer to
(10C3)(5C1)/(15C5)
Which follows the hypergeometric which states suppose a sample size n
chosen w/out replacement from a population N, of which m are
acceptable and N-m are not, then P(X=i)=(mCi)(N-mCn-i)/(NCn)
so in your case "i" (the number acceptable to be drawn) is 3, "m"
(total number of acceptable in the population) is 10, "N" (pop. size)
is 15 making N-m=5, and "n" (sample size) is 4 making n-i=1 we get the
result
(10C3)(5C1)/(15C5) = which I'm I got as 100/1001 but you may want to
recheck the math.

errr maybe 200/1001, definitely recheck...

40/91

The probability of drawing 3 acceptable and then 1 unacceptable is 

(10/15)*(9/14)*(8/13)*(5/12)  =  10/91

There are a total of 4 ways to rearrange the ordering of the
acceptable/unacceptable draws: AAAU, AAUA, AUAA, UAAA.

So the probability of exactly 3 acceptable out of 4 is 40/91.

rracecarr is correct, I actually just woke up realizing I had made a
mistake, the denominator for the prob. should be 15C4 bringing the
answer to 40/91.