# We sample two balls from the second urn with replacement.

We have an urn with three green balls and two yellow balls. We pick a sample of two without replacement and put these two balls in a second urn that was previously empty. Next we sample two balls from the second urn with replacement.

$(a).$ What is the probability that the first sample had two balls of the same color?

My answer for this is... $$\frac{\binom{3}{2}}{\binom{5}{2}}+\frac{\binom{2}{2}}{\binom{5}{2}}=\frac{4}{10}$$ which I believe to be correct.

$(b).$ What is the probability that the second sample had two balls of the same color?

For this part, I have tried lots of things, none of which seem to give me an appropriate answer. Is this conditional, or do I use inclusion exclusion?

Given that two balls of the same colour were transferred, the probability the balls sampled from the second urn are of the same colour is $1$.
Given that the transferred balls were of different colours, the probability the balls sampled from the second urn are of the same colour is $1/2$.
Thus by the Law of Total Probability, the probability the balls sampled from the second urn are of the same colour is $(4/10)(1)+(6/10)(1/2)$.