| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
An urn contains a number of balls each of which is either black, white or red. It is given that the urn contains at least one black ball and at least one white ball. k (> 2) balls are drawn from the urn with replacement. Given that the probability that the balls drawn were all black or all white equals the probability that the balls drawn were all red, find k and the number of balls of each colour.
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 1432.1 | marginal problem | HERON::BUCHANAN | Holdfast is the only dog, my duck. | Mon Apr 29 1991 09:40 | 6 |
Ha! A quaint setting. Now, what about the same problem, *without* replacement? Cheers, Andrew. | |||||
| 1432.2 | CADSYS::COOPER | Topher Cooper | Mon Apr 29 1991 13:53 | 4 | |
I have an interesting proof for my answer, but I'm afraid there isn't
room in this note for it.
Topher
| |||||
| 1432.3 | is a countable infinity of answers enough? | CSSE::NEILSEN | Wally Neilsen-Steinhardt | Mon Apr 29 1991 14:10 | 3 |
> Now, what about the same problem, *without* replacement? k > 2, #black < k, #white < k, #red < k | |||||
| 1432.4 | solved for k =< 2 | HERON::BUCHANAN | Holdfast is the only dog, my duck. | Mon Apr 29 1991 14:47 | 34 |
> -< is a countable infinity of answers enough? >- No. Gimme some non-trivial answers. :-) So let's assume that w (& hence b & r) >= k. And k >= 2, otherwise there's another batch of trivial stuff for k = 1, w+b=r. For k = 2, you've got: w(w-1) + b(b-1) = r(r-1) => (2w-1)� + (2b-1)� = (2r-1)� + 1 By a process I went through somewhere else, a�+b�=c�+d� can be solved parametrically. Here, all four summands are odd, and there is no common factor, so we have four +ve integers, f>j,g>h pairwise coprime, with one of them even and with |fh-gj| = 1. Then r = �(fg+hj + 1) w = �(fh+gj + 1) b = �(fg-hj + 1) So, for instance with [f,g,h,j] = [5,7,3,2], you get: r = 21 w = 15 b = 15 Now what about k = 3? Regards, Andrew. | |||||
| 1432.5 | k=3 has many solutions | CLT::TRACE::GILBERT | Ownership Obligates | Tue Apr 30 1991 11:25 | 20 |
You want b!/(b-k)! + w!/(w-k)! = r!/(r-k)! For k=3, and letting b=B+1, w=W+1, r=R+1, this is B�-B + W�-W = R�-R It's easy to numerically find solutions to this. Some are [B,W,R] = [4,4,5], [9,15,16], [21,55,56], [31,56,59], [35,119,120], [40,71,75], [49,139,141], [62,103,110], [85,91,111], [101,291,295], [120,155,176], [137,533,536], [150,430,436], [188,202,246], [220,516,529], [231,904,909], [249,451,475], [253,425,453], [273,689,703], [305,696,715], [315,528,563], [325,664,689], [328,337,419], [350,714,741], [363,416,493], [379,791,819], [400,589,645] [425,449,551], [425,706,754], [435,496,589], [490,870,919] [533,644,748], [609,755,869] These, and their symmetric counterparts (exchanging B and W) are the only solutions with k=3 and R <= 1000. | |||||
| 1432.6 | k=4 | CLT::TRACE::GILBERT | Ownership Obligates | Tue Apr 30 1991 11:43 | 3 |
And b!/(b-4)! + w!/(w-4)! = r!/(r-4)! has solutions: [7,7,8], and [132,190,200] | |||||
| 1432.7 | quick remark | HERON::BUCHANAN | Holdfast is the only dog, my duck. | Wed May 01 1991 09:15 | 6 |
In fact there's at least 1 solution for each k > 0. Let w = b = 2k-1, r = 2k. Cheers, Andrew. | |||||