1. The problem statement, all variables and given/known data a box contains 4 black balls and 6 white balls. After randomly selecting 2 balls with out replacement, what is the probability that both are the same color. 2. Relevant equations C(n,k)=(n!)/(k!(n-k)!) 3. The attempt at a solution I'm just guessing here but, P=(C(4,2)C(6,2))/C(10,2)= (6x5x4x3)/(2x10x9) The result makes sense I guess. I mean, the probability of picking 2 black balls is (6x5)/(10X9) because the first pick generates a probability of 6/10 and the second pick is out of 9 balls and the possible events is the first 6 times the remaining 5. The same goes for the white balls. And using that logic, it makes sense that the probability is just all the possible events multiplied divided by the twice the overall outcomes( twice because there are 9x10 outcomes for picking 2 of either white or black so twice when the whole picture is taken into account? IDK.) Is my logic correct or did I screw this up?