probability question with couple if 5 couples attend an event with all ten people entering the raffle, 
and all 10 people enter a raffle. Four of the 10 people are selected at random to win, what is probability that both the husband and wife will win the raffle?
 A: As I understand the question, you have a specific husband-wife pair we care about, say.. Mr. and Mrs. Jones.  The question I think is, what is the chance that both Mr. and Mrs. Jones win simultaneously.
As given in the problem statement you have five couples, so there are ten people total.  Four people will be selected to win (also assuming someone can win only once).
As it doesn't matter to us in what order the winners are selected, and we only select a person a single time, we can treat this as a combinations question.  The total number of outcomes is then $\binom{10}{4}$.
The number of outcomes in which both Mr. and Mrs. Jones win could be thought of as $\binom{8}{2}$  (as it is deciding who the $\textit{other}$ two winners are, again ignoring order).
So, the probability is the number of events with the situation we want divided by the number of events possible and would be $\frac{\binom{8}{2}}{\binom{10}{4}}$

In the event that we are curious about at least one married couple simultaneously winning, but not worrying about which couple it was, again the total number of possibilities is $\binom{10}{4}$
The number of possibilities of at least one married couple simultaneously being winners is equal to the number of possibilities of exactly one married couple plus the number of possibilities of exactly two married couples.
For exactly one married couple., first pick which couple it is, then pick the two remaining winners, taking care to account for symmetry in selecting them.  So you have $5\cdot 8\cdot 6 \cdot \frac{1}{2}$ number of possibilities  (pick the couple, pick the next winner out of the remaining eights, pick the next winner out of the remaining six [we excluded the spouse of the previous choice], and divide by 2 to account for the possibility of picking the same two people in a different order)
For picking exactly two couples you have $\binom{5}{2}$ number of possibilities.
So.. $\frac{\binom{5}{2} + 5\cdot8\cdot6\cdot\frac{1}{2}}{\binom{10}{4}}$ would be the final probability.

As you can see, the interpretation of the question can lead to very different answers, so it helps to be clear what the question is actually asking.
