I'm having an incredible amount of trouble with this stats homework... I'm sure it's something simple, but it just isn't clicking. Could someone help me out a bit? Not necessarily looking for answers, but someone to explain it to me.

There are 11 (4 men and 7 women) of us wanting to go to a concert. However, we only have 4 tickets.

a. How many ways can we assign tickets? -- I did 11C4 which I'm pretty sure is correct? Got 330.

b. What is the probability that 2 men and 2 women go to the concert? -- This is where it starts to confuse me. I know that I need to find my "m" value for P = m/n, and that m is the # of ways we can get 2 men and 2 women, but I don't know how to find m....
This is what I did, pretty sure it's incorrect: m = (4C2)(7C2) = 126
P = .3818

c. What is the probability that at least one man goes to the concert? -- Again, no idea. I feel like if I could figure out b. then I might be able to get this one.
This is what I did: m = (4C1)(4C2)(4C3)(4C4) = 96
P = .2909

I know for a fact that at least one of them is wrong, because part c. should be greater than b.

I'd really appreciate any help, thanks.

A and b are right

Part c
at least one man goes
so find
p(1man3girl)+p(2man2girl)+p(3man1girl)+p(4man)

Or 1- p(4girl)

a. is right as stated above.
b. is right as soon as you assume that tickets are randomly (uniformly) assigned.
c. it seems easier to find the probability that no man goes to the concert : (7C4)*(4C0) = (7C4) = ... over 11C4

This post was edited by feanur on Aug 21 2013 04:25pm

Yes, that last part is what I did finally xD
Thank you.

I will probably need more help before this assignment is done, but until then, thanks.