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 re: Ground Rules and First Thoughts by Charlie)

To get equally likely outcomes we'd need to do the permutations, which would amount to 4^577 (4 choices for each peanut iterated 577 times), an unwieldy number. It's best to consider the combinations and total their respective probabilities. When the probabilities of all 14,700 combinations of 577 peanuts that add to 1000 chambers are calculated as in the last paragraph of my preceding post and added together, the total probability is 0.0178809942162995744.... Calling this p, the odds against the occurrence are (1-p)/p = 54.92530191... to 1 odds, or about 55-to-1, or 1 in 56.

(Each probability is 577Cquads * (577-quads)Ctriples * (577-quads-triples)Cdoubles * (1/15)^quads * (2/15)^triples * (4/15)^doubles * (8/15)^singles)

Posted by Charlie on 2003-02-19 03:21:09