A local fraternity is conducting a raffle where 50 tickets are to be sold – one per customer. There are three prizes to be awarded. If the four organizers of the raffle each buy one ticket, what is the probability that the four organizers win exactly two prizes?
I want to do this using combinations. I know that the probability of them winning all three of the prizes equals $\dfrac{4 \choose 3}{50 \choose 3} = \dfrac{1}{4900}$. So intuitively, I'd guess that the probability of them winning two prizes is $\dfrac{4 \choose 2}{\binom{50}{2} \times \binom{3}{2}}$, the $\binom{3}{2}$ being because there are 3 tickets to choose from in total. But this is apparently wrong. Why?
Best Answer
You have calculated the probability of selecting two from four winning tickets, out of all ways to select two from fifty tickets and two from other three things. This is not ... like anything that you want to find.
You want the probability of selecting two from four winning tickets and one from forty-six losing tickets, out of all the ways to select any three from all fifty tickets.