1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Did you mean "with replacement"? $\endgroup$ – Trevor Wilson Oct 17 '13 at 5:37
  • $\begingroup$ no, the same ball can be chosen twice $\endgroup$ – Trolol Oct 17 '13 at 5:38
  • 1
    $\begingroup$ How can you choose the same ball again if you don't replace it after the first time? $\endgroup$ – Trevor Wilson Oct 17 '13 at 5:40
  • $\begingroup$ sorry, I meant they are not taken out from the box, so according to your definition, with replacement. :P $\endgroup$ – Trolol Oct 17 '13 at 5:41
  • $\begingroup$ Is there another definition that means the opposite of what the words say? It must be like when people say they "could care less" :) $\endgroup$ – Trevor Wilson Oct 17 '13 at 5:43
1
$\begingroup$

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 $(Y/(X+Y))^Z$.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.