I have the following problem which I am trying to resolve using two methods, but I am getting two different answers.
In a drawer there are $14$ black socks, $11$ blue and $10$ white. You grab $3$ at random, what is the probability of having two of the same color?
Method 1: calculating the probability as: $P(A)=1-P(B)$
Where P(B) is the probability of not having a sock of each color.
I am doing this like this: $$P(B)=\frac{14\times11\times10\times3!}{35\times34\times33}$$
This gives $P(A)=0.765...$
Method 2: calculating $P(A)$ directly:
$$P(A)=\frac{14\times13\times33+11\times10\times33+10\times9\times33}{35\times34\times33}$$
This gives a result of around $0.32$.
So, obviously there is something terribly wrong, but I am not spotting it.
If you want exactly 2 of the same color just subtracting the probability of having all different colors isn't sufficient. The general way of approaching these problems is the hypergeometric. $$\frac{{14 \choose 2}{21 \choose 1}+{11 \choose 2}{24 \choose 1}+{10 \choose 2}{25 \choose 1}}{{35 \choose 3}}$$
Using your complement method: $$1-(\frac{14\times13\times12}{35\times34\times33}+\frac{11\times10\times9}{35\times34\times33}+\frac{10\times9\times8}{35\times34\times33}+\frac{3!(14)(11)(10)}{35\times34\times33})\approx.67$$
Using your direct method: $$\frac{({3 \choose 2}14\times13\times21)+({3 \choose 2}11\times10\times24)+({3 \choose 2}10\times9\times25)}{35\times34\times33}\approx.67$$
Gives the same answer