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)?

  • $\begingroup$ How exactly are the two balls being drawn? One per urn or any two? What does "drawn wr" mean? $\endgroup$ – whuber Apr 25 '18 at 13:20
  • $\begingroup$ An Urn is selected first and then balls are drawn with replacement . $\endgroup$ – Harry Apr 25 '18 at 13:21
  • $\begingroup$ 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. $\endgroup$ – whuber Apr 25 '18 at 13:40
  • $\begingroup$ Sorry for omitting details. I have edited the question. $\endgroup$ – Harry Apr 25 '18 at 13:59
  • 2
    $\begingroup$ Can you see that the probability that the two balls have different colors must be the same for both urns? $\endgroup$ – 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.

| cite | improve this answer | |

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.