A box contains $2n$ balls of $n$ different colors, with 2 of each color. Balls are picked at random from the box with replacement until two balls of the same color have appeared. Let $X$ be the number of draws made.
a) Find a formula for $P(X > k)$, $k = 2, 3, ...$
$P(X>k) = 1 - P(X\le k) = 1- (P(X=2)+P(X=3)+P(X=4)+...P(X=k))$
So I manaully find $P(X=2)$, $P(X=3)$, $P(X=4)$, ... and try to find a pattern to make an equation?
Edit 1:
For each $n$ color, trials = 2, success rate = $\frac{2}{2n}$, # of needed success = 2, ordering doesn't matter, with replacement
$P(X = 2) = n \cdot Binomial(2, \frac{2}{2n}) = \binom{2}{2} (\frac{2}{2n})^2(1-\frac{2}{2n})^0$
For each $n$ color, trials = 3, success rate = $\frac{2}{2n}$, # of needed success = 2, ordering doesn't matter, with replacement
$P(X = 3) = n \cdot Binomial(3, \frac{2}{2n}) = \binom{3}{2} (\frac{2}{2n})^2(1-\frac{2}{2n})^1$
For each $n$ color, trials = 4, success rate = $\frac{2}{2n}$, # of needed success = 2, ordering doesn't matter, with replacement
$P(X = 4) = n \cdot Binomial(4, \frac{2}{2n}) = \binom{4}{2} (\frac{2}{2n})^2(1-\frac{2}{2n})^2$
For each $n$ color, trials = 5, success rate = $\frac{2}{2n}$, # of needed success = 2, ordering doesn't matter, with replacement
$P(X = 5) = n \cdot Binomial(5, \frac{2}{2n}) = \binom{5}{2} (\frac{2}{2n})^2(1-\frac{2}{2n})^3$
$$P(X>k)=1-(n \cdot \sum_{i=2}^k \binom{i}{2} (\frac{2}{2n})^2(1-\frac{2}{2n})^{i-2})$$
Edit 2:
Last ball has to be a color already chosen.
For X = 2, no restrictions on the 1st ball, 2nd ball must be a color already chosen.
$P(X = 2) = \frac{2n}{2n} \cdot \frac{2}{2n}$
For X = 3, no restrictions on the 1st ball, no restrictions on the 2nd ball, 3rd ball must be a color already chosen.
$P(X = 3) = \frac{2n}{2n} \cdot \frac{2n}{2n} \cdot \frac{4}{2n}$
For X = 4, no restrictions on the 1st ball, no restrictions on the 2nd ball, no restrictions on the 3rd ball, 4th ball must be a color already chosen.
$P(X = 4) = \frac{2n}{2n} \cdot \frac{2n}{2n} \cdot \frac{2n}{2n} \cdot \frac{6}{2n}$
$$P(X > k) = 1 - (\sum_{i=2}^k \frac{2^{i-1}}{2n})$$
Edit 3:
Last ball has to be a color already chosen AND other balls must be different!
For X = 2, no restrictions on the 1st ball, 2nd ball must be a color already chosen.
$P(X = 2) = \frac{2n}{2n} \cdot \frac{2}{2n}$
For X = 3, no restrictions on the 1st ball, 2nd ball must be different from the 1st ball, 3rd ball must be a color already chosen.
$P(X = 3) = \frac{2n}{2n} \cdot \frac{2n-2}{2n} \cdot \frac{4}{2n}$
For X = 4, no restrictions on the 1st ball, 2nd ball must be different from the 1st ball, 3rd ball must be different from the 1st and 2nd ball, 4th ball must be a color already chosen.
$P(X = 4) = \frac{2n}{2n} \cdot \frac{2n-2}{2n} \cdot \frac{2n-4}{2n} \cdot \frac{6}{2n}$
$$P(X>k)=1-(\frac{1}{n} + \frac{2(n-1)}{n^2} + \frac{3\left(n-1\right)\left(n-2\right)}{n^3} + ... + \frac{(k-1)(n-1)(n-2)\cdot\cdot\cdot(n-(k-2))}{n^{k-1}})$$
Answer from Textbook:
$\frac{2n-2}{2n} \cdot \frac{2n-4}{2n} \cdot ... \cdot \frac{2n-2(k-1)}{2n}$
You interpreted the problem as if you pre-selected one of the colors before drawing any balls and require to draw two balls of that color, for example if there are red balls you might choose red and then you will stop only when you have drawn red twice. I would interpret it differently--if any color is drawn that has been drawn before, you stop. By the pigeonhole principle the maximum possible value of $X$ is $n +1.$
The textbook apparently interpreted the question the way I did.
For example, you draw one ball and observe its color. Put it back and draw again. In order for $X > 2,$ the second ball you draw must be different from the first. There are $2n - 2$ such balls, so $P(X> 2) = \frac{2n - 2}{2n}.$
If the second ball is a different color from the first, then on the third draw there are now only $2n - 4$ balls that are different from any color already drawn. (Either of the two previously seen colors is a match.) To have $X > 3$ you must avoid a match on draw number $2$ and again on draw number $3,$ so $P(X > 3) = \frac{2n - 2}{2n} \cdot \frac{2n - 4}{2n}.$ And so forth.