# Combined Probability

#### cflo1214

##### New Member
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

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.

#### cflo1214

##### New Member
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)