Activity for Dylan Callaghanâ€
| Type | On... | Excerpt | Status | Date |
|---|---|---|---|---|
| Comment | Post #295461 |
I think the added term should be $\binom{n-|K|}{n_1-|K|}$, since you are able to choose $n_1-|K|$ out of the $n-|K|$ colors _not already included_ in $K$ to adjoin to $K$. (more) |
— | 7 months ago |
| Comment | Post #295461 |
Thanks for your interest in the question, your insights have been really helpful so far! I really like your use of the inclusion-exclusion principle to model the drawing of exactly $n_1$ different colors. However, If my understanding is correct, I feel that there may be a slight problem in your f... (more) |
— | 7 months ago |
| Comment | Post #295458 |
For the first partial solution, there are two ways of deriving this. Intuitively, there is only one color of interest ($c_1$), accompanied by $c_b$ black balls. You could thus think of lining up all the balls in the order they were drawn, for which on average would see the colored balls equally d... (more) |
— | 7 months ago |
| Edit | Post #295458 | Initial revision | — | 7 months ago |
| Question | — |
Multivariate urn/coupon collector problem without replacement Simply put, the problem I have is: Given an urn containing $m$ total balls of $n$ different colors with $ci$ balls of each color (i.e., $m = c1 + ... + cn$), what is the expected number of balls you would need to draw to see the $k^{\text{th}}$ color when drawing without replacement? This i... (more) |
— | 7 months ago |