What am I missing here? (Probability) An organization consists of 10 married couples. A lottery without replacement will decide a chairman and secretary. What's the probability that a married couple gets chosen?
My solution attempt:
There are 2 ways a couple can get the positions, so {(secretary, chairman),(chairman, secretary)}, but one is redundant so there is 1 combination. Now we need to determine how many ways of choosing 2 out of 2, that is $^{20}C_2$. Let our event be $A$, then $P(A) = \frac{1}{20C2}=\frac{1}{45}.$
The actual solution is $\frac{1}{19}$. I'm guessing that I have to multiply the ways of choosing 2 out of 2 by 10 to account for every couple? That would give the right solution, but I can't seem to convince myself of why that is true.

Suppose we first choose chairman and then secretary.
There are 20 candidates.
Suppose a chairman is picked. $19$ people left. Only one person is the new chairman's husband/wife. Thus, with probability $1/19$ they get the position of secretary.
