there are $N$ couples in a party. $2K$ people from the party are invited to play a game $(2K\leq N)$.
What is the probability that out of the $2K$ people that were invited to play the game:
$\mathbf A.$ exactly $K$ couples were invited to play
$\mathbf B.$ exactly $one$ couple was invited to play
$$$$
this is what I did:
first I calculated the Omega: there is no order (it doesn't matter if one couple was chosen before the other) and after we chose a couple we can't choose them again, so:
$$|\Omega| = \frac{N!}{(N-2K)!}$$
now we can look at a couple as an individual- so I have to chose $K$ couples from $N$ couples, so:
$$P(A) = \frac {|A|}{|\Omega|} = \frac {\frac {N!}{(N-K)!}} {\frac {N!}{(N-2K)!}} = \mathbf{\frac{(N-2K)!}{(N-K)!}}$$
Am I right?
for $\mathbf B$ I am a little confused. I know I need to choose one couple so: ${N \choose 1}$ , but now I'm not sure how to continue.
Thank you!
Best Answer
The sample space consists of all subsets of $2K$ people that can be selected from the $2N$ people at the party. Hence, $$|\Omega| = \binom{2N}{2K} = \frac{(2N)!}{(2K)!(2N - 2K)!}$$ In your sample space, you counted ordered selections of $2K$ people from the $N$ couples at the party. While you could use ordered selections if you also use ordered selections for the favorable cases, you should have selected $2K$ people from the $2N$ people at the party.
Observe that if exactly $K$ couples are selected, then all $2K$ of the people selected to play the game must be members of those $K$ couples. Hence, the number of favorable cases is the number of ways we can select exactly $K$ of the $N$ couples at the party, which is $$\binom{N}{K}$$ Hence, $$\Pr(\text{exactly $K$ couples are selected}) = \frac{\dbinom{N}{K}}{\dbinom{2N}{2K}}$$
Strategy: Use the same sample space as above. For the favorable cases: