Probability - An urn with twenty balls (10 are black/ other 10 are red) pick three. Urn with ten red balls and ten black balls, what is the probability that you will get balls of both colors if you draw three times without replacement?
I created a tree with all the possible combinations (BBR, BRB, and etc.). There are eight variables and only two of them would not work BBB/RRR. So wouldn't be the answer be 75%?
 A: Since you are drawing without replacement, the $8$ events you are thinking about are not all equally likely.  If, for example, you have drawn $2$ balls, and gotten BB, the probability the next one is a B is slightly lower than if your first two balls were AA, or AB. Your reasoning would be perfectly correct if the drawing were done with replacement. 
I would probably attack the problem like this. The probability of BBB is $\frac{10}{20}\cdot\frac{9}{19}\cdot\frac{8}{18}$.  
By symmetry the probability of RRR is the same. So the probability you will get a monochrome set of balls is $q=2\cdot\frac{10}{20}\cdot\frac{9}{19}\cdot\frac{8}{18}$.
The probability there will be some colour variation is $1-q$.   
A: The key here is 'without replacement'.  To pick the three balls, the scheme will always be two of one color and one of the other.  Thus, we have probability
$$\frac{10\cdot{9}\cdot{10}}{20\cdot{19}\cdot{18}}$$
There are 6 out of 8 ways to do this so the answer is
$$6\times\frac{10\cdot{9}\cdot{10}}{20\cdot{19}\cdot{18}}=\frac{60}{76}$$
