Combination - Probability. Probability within a set of 2 I am stuck on a homework question. The question is

If there are 6 couples. (12 individuals) and 6 prizes are to be given out to these 12 individuals. What is the probability of a couple receiving a price.

I know that there are $^{12}C_6$ possible ways to give out a prize. So that is the total number of outcomes.

Whats confusing me is how should I find the probability of the set consisting of 2 people i.e a couple
any hints to help me out?

Edit any one can receiv a price, Even two people who are coupled
 A: Assuming you are looking at a particular couple and you want the probability that at least one wins a prize, you could approach it by considering:


*

*What is the probability the first member of the couple wins a prize?

*If the first member of the couple does not win a prize, what is the conditional probability the second wins a prize? Multiply this by the probability the first member of the couple does not win a prize.

*Add the two previous results together 


Another approach giving the same result:


*

*What is the probability the first member of the couple does not win a prize? 

*If the first member of the couple does not win a prize, what is the conditional probability the second does not win a prize?

*Multiply these together to get the probability the couple does not win a prize.

*Find the probability the couple wins at least one prize

A: We interpret the question as asking for the probability that there is at least one couple such that each member of the couple receives a prize. 
We first find the probability of the complementary event, that there is no couple such that each member of the couple receives a prize.
Since conveniently there are $6$ prizes and $6$ couples, one member of each couple must receive a prize. Which member? There are $2^6$ choices. 
So now we can find the probability that no two coupled people get a prize, and then the probability that there is at least one couple both of whose members get a prize.  
A: Interpreting the question as asking for the probability, for one specific couple, of receiving at least one prize, one can argue as follows. There are $N=\binom{12}6$ ways to distribute the prizes, (supposedly) all equally likely. Among these there are $M=\binom{10}6$ ways for the prizes to be distributed only among the remaining couples (our couple loses). Then the probability asked for is $\frac{N-M}N$.
