Probability question above my pay grade.

Greetings all.

I have a small problem that I can't solve without doing it long hand and if possible I'd like to have a formula to solve it. Or the answer would be nice too. (Rofl)

Setup:

I have 6 people, 3 of which are bad guys and 3 are good guys. If I have to choose one out of the group as a bad guy then my odds are 50/50

Take the same 6 people with the same spread... but limit my choice to only one of 4 random people.

What are the odds of my 4 choices including all 3 bad guys

What are the odds of my 4 choices including only 2 bad guys

What are the odd of my 4 choices including only 1 bad guy.

TIA

SloMo

Let me ask the question differently.

Given the same two options as above... choose one of the entire 6 ... or choose one of a random 4.. What is the best decision based on probability.

Re: Probability question above my pay grade.

If you had a really large population, this would be a nice binomial problem. As it is, you may just going to have to count them. I'd start with a tree diagram.

Of course, you could just calculate it.

Just one of 4: $\displaystyle \frac{{3 \choose 1}\cdot {3 \choose 3}}{{6 \choose 4}}$

You do 2 of 4.