Although some of the things in this problem aren't true in real life just assume they are in the question.

When growing peanuts the following happens:

for every 2 single chambered peanuts there will be one double chambered peanut

for every 2 double chambered peanuts there will be one triple chambered peanut

for every 2 triple chambered peanuts there will be one quadruple chambered peanut.

When a company packages 1000 peanuts in one bag they take peanuts randomly from a giant bin that contins all the peanuts grown. What are the odds that there will be 1000 individual nuts?

(In reply to Ground Rules and First Thoughts by TomM)

TomM's interpretation (ground rules) is the only viable one: the company actually packages 577 peanuts in the hopes of averaging 1000 chambers in the package. However there are not 577^4 combinations of pods. That number would represent all the combinations of 0-576 quads, 0-576 triples, 0-576 doubles and 0-576 singles. Most by far of the totals wouldn't be 577 and only 4 combinations that would total 577 are excluded (i.e., 577+0+0+0, etc.). The partition of 577 into 4 numbers n(1), n(2), n(3) and n(4) can be accomplished in (577+3)C3 ways. Think of 577 peanuts mixed with 3 markers laid out in a row. Up to the first marker are the quads, between that and the next marker are the triples, then come the doubles, and after the third marker come the singles. There are 580 objects of which 3 are markers so there are 580C3 ways of doing this. This is 32,350,660. Of these, 14,700 have a total of 1000 chambers.

However, the probability is not 14,700/32,350,660, as not all the combinations are equally likely. The least likely is all-quads, at (1/15)^577, with 678 zeros after the decimal before the first non-zero. Not much higher is the probability of all-singles, at (8/15)^577, still with 157 zero-digits after the decimal point. These are at the extremes of distance away from the desired 1000 chambers. Those combinations closer to producing 1000 chambers are more common, such as 38 quads, 77 triples, 155 doubles, 307 singles, which has probability 0.000127732 of occurring. This probability is 577C38 * (577-38)C77 * (577-38-77)C155 * (1/15)^38 * (2/15)^77 * (4/15)^155 * (8/15)^307.

Posted by Charlie on 2003-02-19 03:16:03