The contents of urns I, II and III are as follows: and red. 2 white, 2 black and 4 red balls, 4 white, 2 black and 2 red balls, 3 white, 6 black and 2 red balls. One urn is chosen at random and two balls drawn from it. They happen to be white Use Bayes' theorem to find the probability that they come from urns II.

Sure, let's use Bayes' theorem to solve this problem step by step.

First, let's define some probabilities:

P(I) = P(II) = P(III) = 1/3, because one urn is chosen at random.

P(WW|I) = Probability of drawing two white balls from urn I = 0, because urn I has no white balls.

P(WW|II) = Probability of drawing two white balls from urn II = (4/8) * (3/7) = 6/28 = 3/14, because urn II has 4 white balls out of 8 total balls, and after drawing one white ball, it has 3 white balls out of 7 total balls.

P(WW|III) = Probability of drawing two white balls from urn III = (3/11) * (2/10) = 6/110 = 3/55, because urn III has 3 white balls out of 11 total balls, and after drawing one white ball, it has 2 white balls out of 10 total balls.

Now, we can use Bayes' theorem to find the probability that the two white balls come from urn II. Bayes' theorem is given by:

P(II|WW) = P(WW|II) * P(II) / P(WW)

We already know P(WW|II) and P(II), but we need to find P(WW), the total probability of drawing two white balls. We can find this by adding up the probabilities of drawing two white balls from each urn, weighted by the probability of choosing each urn:

P(WW) = P(WW|I) * P(I) + P(WW|II) * P(II) + P(WW|III) * P(III) = 0 * (1/3) + (3/14) * (1/3) + (3/55) * (1/3) = 1/14 + 1/55 = 4/165 + 3/165 = 7/165

Now we can substitute these values into Bayes' theorem:

P(II|WW) = P(WW|II) * P(II) / P(WW) = (3/14) * (1/3) / (7/165) = 1/14 / (7/165) = 165/14 * 1/7 = 165/98 = 15/9 = 5/3

So, the probability that the two white balls come from urn II is 5/3. However, this is greater than 1, which is not possible for a probability. There seems to be a mistake in the calculation. Let's go back and check our work.

The mistake is in the calculation of P(WW). The correct calculation is:

P(WW) = P(WW|I) * P(I) + P(WW|II) * P(II) + P(WW|III) * P(III) = 0 * (1/3) + (3/14) * (1/3) + (3/55) * (1/3) = 0 + 1/14 + 1/55 = 4/165 + 9/165 = 13/165

Now let's substitute the correct values into Bayes' theorem:

P(II|WW) = P(WW|II) * P(II) / P(WW) = (3/14) * (1/3) / (13/165) = 1/14 / (13/165) = 165/14 * 1/13 = 165/182 = 15/14 = 1.0714

This is still greater than 1, which is not possible for a probability. There seems to be another mistake in the calculation. The mistake is in the calculation of P(WW|II). The correct calculation is:

P(WW|II) = (4/8) * (3/7) = 1/2 * 3/7 = 3/14

Now let's substitute the correct values into Bayes' theorem:

P(II|WW) = P(WW|II) * P(II) / P(WW) = (3/14) * (1/3) / (13/165) = 1/14 / (13/165) = 165/14 * 1/13 = 165/182 = 0.9066

So, the correct probability that the two white balls come from urn II is approximately 0.9066.