Double urn probability and multiple instances I've got a probability problem based on the following problem.
I provided some scenarios which I need to solve but I'm thankful for any hints or links on how to solve just one of them or how to described this scenario (and/or solutions) in proper 'mathematical notaion'
The Problem:
An urn contains 10 balls numbered from 1 to 10, every number occurs excactly once.
Person A picks balls from the urne and distributes the balls randomly among 4 other persons, say persons B, C, D and E (it is not neccessary for every person to get the same amount of balls). 
After the distribution process, person C picks $N$ ball(s) from its own urn, which holds another 10 balls, again numbered from 1 to 10.
What is the probability that person C received one or more balls from person A which have the same number as the ball(s) person C just picked from its own urn?
Some scenarios:
Scenario 1: Person C received one ball with nr.7. What is the probability that person C picks the ball with the same number (7) from its own urn?
Scenario 2: Person C received two balls with nr.7 and nr.8. What is the probability that person C picks the same two balls (7 and 8) from its own urn?
Scenario 3: Person C received two balls with nr.7 and nr.8. What is the probability that person C picks one ball from its own urn which number is either 7 or 8?
Scenario 4: Person C received one ball with nr.7. What is the probability that person C picks two balls of its own urn out of which one is ball nr.7?
Scenario 5: Person C received one ball with nr.7 and person D received three other balls, say balls 4, 5 and 6. What is the probability that person C picks one ball from its own urn which number is either 4, 5, 6 or 7? (Person C and person D are merged, they count as one)
 A: For the general question, assuming picks are without replacement:


*

*for each numbered ball picked by C from their own urn, the probability C did not get that numbered ball from A is $\displaystyle \frac34$ (i.e. it went to B, D or E), 

*so the probability of no matches is $\displaystyle \left(\frac34\right)^N$, 

*the probability of at least one match is $\displaystyle 1-\left(\frac34\right)^N$ and 

*the probability of $n$ matches is $\displaystyle {N \choose n}\frac{3^{N-n}}{4^N}$. 
For the scenarios the numbered balls received by C from A are determined by the scenario, so:


*

*if C picks one from ten balls, the probability it is number $7$ is $\dfrac{1}{10}$

*if C picks two from ten balls, the probability they are numbers $7$  and $8$ is $\dfrac{1}{45}$

*if C picks two from ten balls, the probability exactly one is number $7$  or $8$ is $\dfrac{16}{45}$, and the probability at least one is number $7$  or $8$ is $\dfrac{17}{45}$ 

*if C picks two from ten balls, the probability one is number $7$ is $\dfrac{1}{5}$ 

*if C picks one from ten balls, the probability it is numbers $4$, $5$, $6$ or $7$ is $\dfrac{2}{5}$
