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!