Probability of choosing 2 co captains AoPS I came across a question on AoPS:

There are 3 math teams in the area, with 5, 7, and 8 students respectively. Each team has two co-captains. If I randomly select a team, and then randomly select two members of that team to give a copy of Introduction to Geometry, what is the probability that both of the people who receive books are co-captains?

The official solution(credit : AoPS) reads
There's a $\dfrac{1}{3}$ chance that I will select each team. Once I have selected a team, let $n$ be the number of students on that team. There are $\dbinom{n}{2}$ ways to choose a pair of those students to give books to, but only one of those pairs will be the two co-captains, which means that once I have selected that team, the probability that I give books to the co-captains is$$\dfrac{1}{\dfrac{n(n-1)}{2}}=\dfrac{2}{n(n-1)}.$$Since the teams have $5,$ $7,$ and $8$ students, this means that the total probability is$$\dfrac{1}{3}\left(\dfrac{2}{5(5-1)}+\dfrac{2}{7(7-1)}+\dfrac{2}{8(8-1)}\right)$$which after a bit of arithmetic simplifies to $\boxed{\dfrac{11}{180}}$.
Doubt 
Although I understood the official solution, during the process of solving on my own, I followed in this direction:
First we decide the size of the Sample Space. Normally, if we considered the size of each team to be the same(say N) then we can choose one team from 3 teams in ${3 \choose 1}$ ways and for each such way, we can choose 2 people from a team in ${N \choose 2}$ ways, this would give the size of sample space to be : ${3 \choose 1} * {N \choose 2}$ ways. Then our event size will be as follows: we can choose one team from 3 teams in ${3 \choose 1}$ ways and for each such way, we can choose 2 people from a 2 co captains in a team in ${2 \choose 2}$ ways. Therefore the probability becomes : $\dfrac{{3 \choose 1} * {2 \choose 2}} {{3 \choose 1} * {N \choose 2}} = \dfrac{1}{{N \choose 2}}$
Now since the size of teams are not the same, instead of applying the multiplication principle in deciding the size of the sample space, i reasoned i should use addition principle. So sample space size is given by : the number of ways of choosing two people from team 1 + the number of ways of choosing two people from team 2 + the number of ways of choosing two people from team 3. It turns out my reasoning is flawed. Although i am failing to understand why. Please help me understand the reason behind the flaw. Thanks in advance.
 A: The way the problem is set up makes the chances of selecting each possible pair unequal (if a smaller team is picked, there is a higher chance the co-captains will be chosen), whereas your approach assumes those chances are equal: hen you declare the sample space to simply pool all three teams together, you implicitly assume equal probabilities.
People sometimes tend to think that when presented multiple possibilities for how something will play out, their chances are all equal – when most of the time they are not.
A: The sample space you have come up with is not valid as it is not an equiprobable space. In other words, each outcome is not equally likely to occur. Depending on which box you choose, the probability of choosing two co-captains is different. Let's take a simpler example.
We have two boxes and we randomly choose one. One has just one blue ball and the other has a blue ball and a red ball. Now let's try and find the probability of picking a red ball.
It should be, $P(R) = \frac{1}{2} \big(0 + \frac{1}{2}\big) = \frac{1}{4}$
Similarly probability of picking a blue ball, $P(B) = \frac{1}{2} \big(1 + \frac{1}{2}\big) = \frac{3}{4}$
And of course, $P(R)+P(B) = 1$
Now try this example using the way you were building sample space and if you can now spot your mistake. Let me know if it clarifies or if you have any further questions.
