Question 1078083: say that there are three defective items in a bag of 50 items. a sample of 10 items is taken at random. let a denote the number of defective items in the sample. find the probability that the sample contains
a) exactly one defective items
b) at least three good items;
c) at most one defective items.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! The first is 3C1*47C9*10
divided by 50C10
This is the same as multiplying 3/50*47/49*46/48*...*39/41*10=0.3980
Once you know there are 3 ways to choose the first, make the top and bottom of the combination add to the 50C10, which is the denominator.
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At least 3 good items. In a sample of 10, with only three defective items, there is no way there can be fewer than 7 good items. The probability is 1.
At most one defective item: We know the probability of 1.
We need the probability of 0 (or subtract the probability of 2 or 3).
probability of 0 is
47C10/50C10=0.5041
The sum of those two is 0.9021.
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