Combined Probability

#1
How do I solve this problem?

Four departments: PE (6 team members), QA (8 team members), EE (5 team members), and AS (28 team members). If 8 team members are randomly selected, what is the probability that you will:

1) Have at least one team member from each department
2) All AS team members and none from the other groups
3) No AS team members
 

Dason

Ambassador to the humans
#2
Hi! :welcome: We are glad that you posted here! This looks like a homework question though. Our homework help policy can be found here. We mainly just want to see what you have tried so far and that you have put some effort into the problem. I would also suggest checking out this thread for some guidelines on smart posting behavior that can help you get answers that are better much more quickly.
 
#3
For part B, this is what I was trying...

((28/8)(6/0)(8/0)(5/0))/((47/8)) = (((28x27x…x10x9x8x…x1)/(8x…x1))(1)(1)(1))/(((47x46x…x10x9x8x…x1)/(8x…x1))) = 1/(47x46x…30x29)

For part A, this is what I was trying...

1 minus the probability of 0 for each group. But the numbers were coming back so crazy, it doesn't seem like that's the way to go.

[(28/0)(6/0)(8/0)(5/0)]/(47/8)
 

Dason

Ambassador to the humans
#4
For the first question your approach doesn't quite work because there is probably overlap in what you're calculating. Like if PE doesn't contain any representation that doesn't necessarily exclude the possibility that QA also doesn't contain representation. The typical way to do this type of problem is to force one person from each group into the committee and count how many ways there are to do that. Then once you do that you can just take the remaining people and select how many more people you need from there not worrying about which team they're from.