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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?

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Your calculation for the first problem is correct.

For the second, we condition on the result of the transfer.

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 $(4/10)(1)+(6/10)(1/2)$.

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