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The question is as follows: In a sample of shall we say 1000 items, 65 come from source A and the rest 935 from source B. If a random sample is drawn from this population of 100 how do I work out the probability of NO samples from source A being drawn?

The formula for how to work this out would be appreciated

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closed as unclear what you're asking by Matthew Drury, mdewey, gung, whuber Dec 12 '16 at 14:50

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Seems it would be best to add a self-study tag. Also, is the 1000 the population? it is a tad unclear to me $\endgroup$ – Yuval Spiegler Dec 11 '16 at 21:52
  • $\begingroup$ I think the OP is treating the 1000 to be a finite population and the sample is taken without replacement. The probability of not drawing A on the first draw is 935/1000 and on the second draw (934/999) since A was not drawn on the first draw. Continue this way untill you get to 835/900. Then multiply these probabilities together. It should be a pretty small number. $\endgroup$ – Michael Chernick Dec 11 '16 at 22:29
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Based upon the axiom of conditional probability, if you were to randomly draw $k$ items from an urn of $N$ (where $x$ items are from source B, $N-x$ from other source), then the probability that all items drawn will be from source B is given by: $$\prod_{i=0}^{k-1}\frac{x-i}{N-i}$$

More specifically, if you were to draw $10$ items from an urn of $100$, where $35$ are from source B, i.e. $k = 10$, $N = 100$ and $x = 35$, then probability can be calculated by

$$\frac{35}{100} \times \frac{34}{99} \times \frac{33}{98} \times \frac{32}{97} \times \frac{31}{96} \times \frac{30}{95} \times \frac{29}{94} \times \frac{28}{93} \times \frac{27}{92} \times \frac{26}{91}$$

Upon the first draw there are 35 source B items out of 100 in the urn, and you are assuming the item will come from those 35. Upon the second draw there are 34 source B items out of the 99 remaining, again assume you get one of the 34, and so on.

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