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?
First, restating the facts:
For every 4chambered peanut, there are 2 3chambered peanuts, 4 2chambered peanuts, and 8 1chambered peanuts. That means that for every 15 pods, there are 26 kernels.
The odds that a given pod will have only one kernel is 8:7 (the probability is 8/15), so the probability that a bag of 1000 pods will have exactly 1000 kernels is (8/15) ^ 1000, an infitesimal number. Because of this, I believe that the question intends that a bag of "1000 peanuts" has just enough pods so that the number of kernels tends toward 1000.
1000 kernels is 38 sets of 26 kernels and 12 "odd" kernels, so I'm going to assume that a bag of "1000 peanuts" is actually a bag of 577 ([38 * 15] + 7) pods. If the 38 complete sets of pods average out to 988 kernels, and the 7 "odd" ones to 9  15 kernels, the bag will have 997  1003 kernels. Close enough for the peanut company and the government inspectors.
But the problem asks what are the actual odds that the bag will contain exactly 1000 kernels. Since each pod can be one of four types, there are (577 ^ 4) combinations of pods. One of these adds up to 577 kernels (all singlechambered); another adds up to 2308 kernels (all 4chambered), but with the the largest number in the vacinity of 1000. Once it is determined what the number of combinations is that add up to 1000 kernels (call it "C") we can determine both the probabily (C / 2308) and the odds (C : [2308  C])

Posted by TomM
on 20030218 02:15:21 