Urn Probability Combination Problem

Question

I am in my first year of Comp Sci and I am reviewing for my Math Final.

There are two Urns $U_1$ and $U_2$. $U_1$ has $10$ red balls and $8$ blue balls. $U_2$ has $16$ red and $4$ blue. Suppose you choose one of the two urns at random pick two balls without replacement.

1. What is the probability that two red balls are selected?

2. What is the probability the balls picked have different color

3. Suppose the balls were put into one urn would the probability of choosing two red balls (without replacement) be higher?

I have done simpler versions of this question but not with indistinct balls. Any help would be awesome.

Answer 1

1. Let's look at the following hypotheses: $H_1$: $U_1$ was picked, $H_2$: $U_2$ was picked. By the problem's definition, $\mathrm{P}(H_1) = \mathrm{P}(H_2) = \frac12$.
   Let $R$ denote the event that two red balls were chosen. Now, observe that $\displaystyle \mathrm{P}(R \mid H_1) = \frac{\binom{10}{2}}{\binom{18}{2}}$ and $\displaystyle \mathrm{P}(R \mid H_2) = \frac{\binom{16}{2}}{\binom{20}{2}}$. Now, by the formula for total probability, we have that $\mathrm{P}(R) = \mathrm{P}(R \mid H_1) \mathrm{P}(H_1) + \mathrm{P}(R \mid H_2) \mathrm{P}(H_2) = \ldots$;
2. As previously, let $R$ denote the event that two red balls were chosen, and $B$ the event that two black balls where chosen. Now, the probability that the two balls are of different color is $1 - \mathrm{P}(R) - \mathrm{P}(B)$;
3. I think you can handle this one yourself.