# Find the probability distribution of the number of red balls chosen in the following case. Urn A contains four red balls, urn B contains three red balls and one green ball, urn C contains two red balls and two green balls. One of these urns is selected at random, and then two balls are chosen at random(without replacement) from the selected urn. There are three urns with urn A containing 4 red balls, urn B containing 3 red and 1 green ball and urn C containing 2 red and 2 green balls.

One of the urns is selected at random and then 2 balls are picked from it at random. The probability distribution of the number of red balls picked is required.

The number of red balls that are picked can have three values, 0, 1 and 2.

Let's see what is the probability of picking 2 red balls. If urn A is selected, the probability of which is 1/3, we are certain to get 2 red balls. If urn B is selected, the probability of which is again 1/3, the probability of getting 2 red balls is (1/3)*(3/4)*(2/3) = 1/6. For urn C, the probability of getting 2 red balls is (1/3)*(2/4)*(1/3) = 1/18. Adding 1/3 + 1/6 + 1/18 = 5/9. The probability of picking 2 red balls is 5/9.

Now, let's find the probability of picking only 1 red ball. If urn A is selected, the probability of picking 1 red ball is 0. If urn B is selected, the probability of picking 1 red ball is (1/3)*2*(3/4)*(1/3) = 1/6. If urn C is selected, the probability of picking 1 red ball is (1/3)*2*(2/4)*(2/3) = 2/9. 1/6 + 2/9 = 7/18

Finally, let's find the probability of picking 0. If the urn A is selected, the required probability is 0. Again if urn B is selected, the required probability is 0. When urn C is selected, the probability is (1/3)*(2/4)*(1/3) = 1/18

It may be seen that 5/9 + 7/18 + 1/18 = 1.

The probability distribution of picking red balls is:

0: 1/18, 1: 7/18, 2: 5/9