Hey, I don't think I am doing this question correctly, my solutions just seem to simple.

Question:

There are 8 opposite-sex married couples at a party. Two people are chosen at random to win a door prize.

a) What is the probability that the 2 people will be married to each other?

b) What is the probability that the 2 people will be of the same sex?

c) If 6 people are chosen, what is the probability that they are 3 married couples?

Solution:

a) $\displaystyle n(s) = 8C2 = 120 $

$\displaystyle n(married couples) = 8 $

So, that means that it's 8/120, or 1/15.

b) $\displaystyle (\frac {8}{16})(\frac {7}{15}) $

$\displaystyle = \frac {7}{30} $

c) $\displaystyle n(s) = 16C6 = 8008 $

There are 8 married couples, so 8C3.

$\displaystyle \frac {56}{8008} $

$\displaystyle \frac {1}{143} $

I am unsure if any of these are correct, as I am not used to doing these sorts of problems with things like married couples.