Can you solve this Halloween raffle probability problem? Five raffle tickets are sold for a Halloween raffle. Marty buys three of the five raffle tickets. Two of the tickets are to be drawn as winning tickets. What is the probability of Marty not having a winning ticket? What is the probability of Mart have one winning ticket? What is the probability of Marty having two winning tickets? 
 A: Given Marty's choice of tickets, how many ways are there to select 2 winning tickets that Marty has not chosen, how many ways are there to select 2 winning tickets one of which Marty has chosen, and how many ways are there to select 2 winning tickets both of which Marty has chosen? Compare each of these quantities with the total number of ways of selecting 2 winning tickets.
The probabilities can easily be calculated from this information as follows. The probability of a certain event, such as the selection of 2 winning tickets that Marty has not chosen, is the number of choices of tickets where this event occurs divided by the total number of choices of tickets.
I'll answer one of your questions as an example. I'll calculate the probability that Marty has chosen neither of the selected winning tickets. First of all, there are 2 tickets that Marty hasn't chosen. There is one possible choice of 2 winning tickets among these 2 tickets: namely, both of them. There are 5 choose 2, or $\frac{5!}{2! 3!}=\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5}{(1 \cdot 2) \cdot (1 \cdot 2 \cdot 3)} = 10$, overall choices of 2 winning tickets among the 5 tickets. Therefore there is $\frac{1}{10}$ probability that Marty has chosen neither winning ticket.
You can apply the same idea to the other two questions.
