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


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

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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

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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