# Find probability that balls drawn are of different colours

Consider two urns containing 2 red, 3 white and 3 red, 2 white respectively. If an urn is selected randomly and two balls are drawn from it with replacement, then what is the probability that balls drawn are of different colours?

I know to find the probability but my confusion is whether it is P(RW)+P(WR) or just P(RW)?

• How exactly are the two balls being drawn? One per urn or any two? What does "drawn wr" mean? – whuber Apr 25 '18 at 13:20
• An Urn is selected first and then balls are drawn with replacement . – Harry Apr 25 '18 at 13:21
• Are the urns selected randomly with equal probabilities? You need to include all these details in the question itself, along with the self-study tag. – whuber Apr 25 '18 at 13:40
• Sorry for omitting details. I have edited the question. – Harry Apr 25 '18 at 13:59
• Can you see that the probability that the two balls have different colors must be the same for both urns? – kjetil b halvorsen Apr 25 '18 at 19:10

Let $k_i$ be the number of red balls in urn $i$, and $n_i$ be the total number of balls in urn $i$. Thus, the number of white balls in urn $i$ is $n_i-k_i$.
The probability of picking different colored balls is $P(RW) + P(WR)$. For urn $i$, this is $$\frac{k_i}{n_i}·\frac{n_i-k_i}{n_i} + \frac{n_i-k_i}{n_i}·\frac{k_i}{n_i} = \frac{2k_i(n_i-k_i)}{n_i^2}$$
If the probability of picking the first urn is $p$, then the probability of picking the second urn is $1-p$. Thus, the probability of picking two different colored balls is $$p·\frac{2k_1(n_1-k_1)}{n_1^2} + (1-p)·\frac{2k_2(n_2-k_2)}{n_2^2}$$
The curious element of this problem is that if there is a symmetry to the coloring distributions (e.g., $k_1 = n_2-k_2$ and $n_1 = n_2$), then the value of $p$ is irrelevant.