Urn $U_i \;\;(i = 1, 2)$ contains $N_i$ balls out of which $r_i$ are red and $N_i − r_i$ are black. A sample of $n \;(1 ≤ n ≤ N_1)$ balls is chosen at random (without replacement) from urn $U_1$ and all the balls in the selected sample are transferred to urn $U_2$. After the transfer two balls are drawn at random from the urn $U_2$. Find the probability that both the balls drawn from urn $U_2$ are red.
My attempt:
Let $Y $ be the number of red balls transferred from $U_1$ to $U_2$
Support of $Y $ is $S_Y =\{0,1,2\ldots r_1\}$ with pmf
$$f_Y (k) = \frac{{r_1\choose k}{N_1-r_1\choose n-k}}{{N_1\choose n}} \; \text{for } k\in S_Y \text{ and } 0 \text{ otherwise}$$
Let $Z$ be the number of balls drawn out of $U_2$
$S_Z = \{0,1,2\} $ with the pmf
$$f_Z(k) = \frac{{r_2 +Y\choose k}{N_2 + n -r_2 -Y\choose 2-k}}{{N_2+n\choose 2}} \; \text{for } k\in S_Z \text{ and } 0 \text{ otherwise} $$ Then, the probability of drawing both red balls from $U_2 $ is
$$Pr(Z=2)$$
$$ =\sum_{i\;=\;0}^{r_1} Pr(Z=2 |Y = i)P(Y=i) $$
$$ =\sum_{i\;=\;0}^{r_1}\frac{{r_2 + i\choose 2}}{{N_2+n\choose 2}} \frac{{r_1\choose i}{N_1-r_1\choose n-i}}{{N_1\choose n}} $$
Is this correct?
However, my instructor has given the following answer:
$$\frac1{(N_2+1)(N_2 +n-1)} \big[r_2(r_2-1) + 2r_2\frac{n_1}{N_1} + n(n-1)\frac{r_1(r_1-1)}{N_1(N_1-1)}\big]$$
I can't see how to reduce my answer to this above form.