Probability: boxes and Discrete probability distribution 1) We have n white balls and 2n black balls. In how many ways we could put it into n boxes if in every box it have to be at least one black ball have to be in every box.
My answer is:
First of all let put one black ball into each box. So now we have a problem how to put n white and n black balls into n boxes which we can be done in $\binom{N+N-1}{N}\binom{N+N-1}{N}$  ways.
2) In the lake we have got k - A-fishes, 2k - B-fishes, 4-k - C-fishes Jack caught 7 fishes. Calculate the probability that caught at least one A-fish.
I'm not sure which Discrete probability distribution I could use...
 A: Your analysis of the first problem is fine, and the answer $\binom{n+n-1}{n}\binom{n+n-1}{n}$ is correct.
For the second, we will assume that the number of Type A fish caught has Hypergeometric Distribution. It is easier to find the probability of catching no Type A fish.
There are $\binom{7k}{7}$ ways to choose $7$ fish. There are $\binom{6k}{7}$ ways to choose $7$ fish none of Type A. So the probability of catching no Type A is $\frac{\binom{6k}{7}}{\binom{7k}{7}}$. The probability of at least one Type A is $1-\frac{\binom{6k}{7}}{\binom{7k}{7}}$.
Remarks: In order to use the hypergeometric distribution, we needed to  assume first that we do not replace the fish we catch. We also needed to assume that all $\binom{7k}{7}$ possible outcomes are equally likely. This is unreasonable, some types of fish are more easily caught. Also, we do not have independence, fish often gather in schools.
If $k$ is at all large, the no replacement and replacement models give very similar answers, and the probability of at least one Type A fish is approximately $1-\left(\frac{6}{7}\right)^7$. 
