Probability of balls being same color There are X red balls and Y white balls. I randomly choose Z balls (without replacement, so I can choose the same ball twice). What is the probability that all Z balls have same color?
i am getting
$$\left ( \frac{X}{X+Y} \right )^{Z} + \left ( \frac{Y}{X+Y} \right )^{Z} $$
Please verify.
 A: If you are choosing the balls with replacement (so you can choose the same ball multiple times) then your answer is correct.
Because the events "all $Z$ balls are white" and "all $Z$ balls are red" are mutually exclusive, the probability that one of them happens is the sum of the probabilities of each.  Because the colors of successive balls are independent from each other (this is where we use that the balls are being replaced) the probability that all $Z$ balls are white is the $Z^\text{th}$ power of the probability that the first ball we draw is white, namely $(X/(X+Y))^Z$.  Similarly the probability that all $Z$ balls are red is $(Y/(X+Y))^Z$.
A: I think you're answer ( and what I think is your reasoning) is correct; you can either choose Z Red balls, or you can choose Z White balls. The probability of choosing a White ball on any "trial" is $\frac {X}{X+Y}$ ; the number of white balls to the total; similarly, the probability of selecting a Red ball is $\frac {Y}{X+Y}$. The two events choosing a Red ball and choosing a White ball are clearly disjoint so that $P(Z)$ balls of same color= $P$(choose Z Red balls)+$P$ (choose Z White balls ). Assuming the trials are indepependent, the probability of choosing $Z$ Red is , as you said $(\frac{X}{X+Y})^Z$ and the probability of choosing a White ball is $(\frac{Y}{X+Y})^Z$ , and, by disjointness of the events, the probability is the sum of the respective probability, so equal to :
$(\frac{X}{X+Y})^Z$ + $(\frac{Y}{X+Y})^Z$ 
As you said.
