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December 22, 2014

December 22, 2014

Posted by **Amy** on Thursday, December 9, 2010 at 12:31am.

- Math Finite -
**PsyDAG**, Friday, December 10, 2010 at 1:57amWhat you want in the probability of selecting 1, 2 or 3 defective pens.

P of 3 = 4/15 * 3/14 * 2/13

P of 2 = 4/15 * 3/14 * 11/13

P of 1 = 4/15 * 11/14 * 10/13

The answer is the sum of those three products.

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