Where am I doing it wrong? Trying to develop intuition for probability I have two problems which I know how to solve now, but I am still not quite sure why my initial solutions are incorrect. I would really appreciate a thorough explanation of where I went wrong. Thank you. 
Problem #1: In a pond there are 105 fish, 40 trout, 65 carp. A fisherman catches 8 fish, what is the probability of exactly two of them being trout if at least three of them are not (so they are carp)? I approached this by reducing the sample to 102 (assuming 3 carps), and counting $$\frac{\binom{40}{2} \binom{62}{3}}{\binom{102}{5}}.$$ I thought since 3 carps are already there, we are looking for P that two of the other 5 are trout, and the other 3 are carps. But this was wrong, and the correct answer was $$\frac{\frac{\binom{40}{2} \binom{65}{6}}{\binom{105}{8}}}{\frac{\sum_{x=3}^{8} \binom{40}{8-x} \binom{65}{x}}{\binom{105}{8}}}.$$ This also makes sense, but I don't understand why my original solution was wrong (it over-counted).
Problem #2: A box contains 18 tennis balls, 8 new 10 old. 3 balls are picked randomly and played with (so if any of them were new, they become 'old'), and returned to the box. If we pick 3 balls for the second time (after this condition), what is P that they are all new? I broke this down into 4 pieces: P(3 new second round|3 new first round)P(3 new first round) + P(3 new second round|2 new 1 old first round)P(2 new 1 old first round) + P(3 new second round|1 new 2 old first round)P(1 new 2 old first round) + P(3 new second round|3 old first round)(3 old first round). However, I was supposed to used binomials to count this. Instead I had a feeling that I should just multiply probabilities this way: $$\begin{align*}
\frac{5\times4\times3}{18\times17\times16} &\times \frac{8\times7\times 6}{18\times 17\times 16} + \frac{6\times5\times 4}{18\times17\times16} \times \frac{8\times7\times10}{18\times17\times16}\\
&\quad + \frac{7\times6\times5}{18\times17\times16} \times \frac{8\times10\times9}{18\times17\times16} + \frac{8\times7\times6}{18\times17\times16} \times \frac{10\times9\times8}{18\times17\times16}.
\end{align*}$$ I get the correct answer with binomials, but this equation that I constructed undercounts the possibilities. Could you tell me what I am missing? ty!
 A: Let's reduce problem 1 to see where you are going wrong. Let's say that there are 7 fishes, 4 trout and 3 carp, and you want to count how many ways there are of catching 2 fishes, at least one of them a carp.
You can catch two carp in $\binom{3}{2}=3$ ways; you can catch a carp and a trout in $\binom{4}{1}\binom{3}{1}=12$ ways; that's a total of $15$ ways.
You are trying to count them by saying "put one carp away; then I just need to select one more fish from the remaining 6; that's $\binom{6}{1} = 6$ ways. And the total number of catches is just one fish from the remaining ones, so that's also $6$". This is what you did in your fraction; but it doesn't give the right count: it undercounts the total number of possible catches, and the number of catches of the type you want. So you are not quite so much "overcounting", as you are undercounting the "universe" and also undercounting your desired outcomes; in the end, the two errors don't quite compensate and you end up with a smaller denominator than you should and a somewhat smaller numerator than you should, and the quotient ends up too big. (In my example, the probability would have come out to $1$, when it should be $12/21$).
In your second one, there is a problem with "hidden denominators". Your first product is okay, because the number of ways of selecting the 5 new balls is $\frac{5\times4\times 3}{3\times 2\times 1}$ (order does not matter), and the way of selecting 3 balls from 18 is $\frac{18\times 17\times 16}{3\times 2\times 1}$, so those denominators cancel; likewise with your second factor.
But in your second summand, this does not work out: when you are selecting two old and one new ball, you "really" have $\frac{8\times 7\times 10}{2}$, because the order which you select the new balls does not matter; but the denominator is still $\binom{18\times 17\times 16}{3\times 2\times 1}$; so you are short by a factor of $3$, since you have
$$\frac{8\times 7\times 10}{18\times 17\times 16}$$
instead of
$$\frac{\quad\frac{8\times 7\times 10}{2}\quad}{\quad\frac{18\times 17\times 16}{3\times 2\times 1}\quad} = \frac{(8\times 7\times 10)\times(3\times 2\times 1)}{2\times(18\times 17\times 16)} = \frac{8\times 7\times 10\times 3}{18\times 17\times 16}.$$
Likewise, your third summand is incorrect by a factor of $3$ for the same reason. You are taking into account order when you shouldn't, and because of the cancellation it shows up as a fraction that is too small (but not really an "undercount"; in fact, you are overcounting both the "desired outcome" and the "possible outcomes", but overcounting the latter more). 
