Imagine that the balls have secret identifying numbers. and imagine that we draw out all the balls, whether someone has already won or not.
There are $10!$ ways to draw out the balls, all equally likely.
Player A could win on the first draw. How many sequences correspond to this? The winning red ball could be any of the $3$, and then the rest of the balls could come in any one of $9!$ possible orders, for a total of $(3)(9!)$.
Or else Player A could win on the third draw. That happens if the first ball is black, the second is black, the third is red, and the rest are arranged in any order. The first ball can then be chosen in $7$ ways, and for each choice the second can be chosen in $6$ ways. Then we have $3$ ways for the third, since it has to be red, and then $7!$ for the rest, for a total of $(7)(6)(3)(7!)$.
This should be the second entry in the numerator. It isn't, presumably a typo. No wonder you are puzzled.
Or else Player $A$ could win on the fifth draw. The same reasoning gives $(7)(6)(5)(4)(3)(5!)$ ways. That's correctly written down.
Finally, there could be a win on the $7$-th draw. The number of sequences that give this is correctly written down.
Randomly select an urn, draw a ball without replacement, then again randomly select an urn and draw a second ball. Let $D_n$ be the colour of the $n^{th}$ draw, and $U_n$ be the urn from which it is drawn.
Using k for black, the urns are: $a = \{(w,4), (r,2)\}, b = \{(r,3), (k,3)\}$
Now, if we know a black ball is going to be removed in the second draw, then only one of the other two black balls, or one of the three red balls, could be removed from the second urn during the first draw.
So the probability of drawing a red ball in the first draw give that knowledge is:
$$P(D_1=r \mid D_2=k) \\ = P(D_1=r \cap U_1=a \mid D_2=k)+P(D_1=r \cap U_1=b \mid D_2=k) \\ = P(U_1=a)P(D_1=r \mid U_1=a \cap D_2=k)+P(U_1=b)P(D_1=r \mid U_1=b \cap D_2=k) \\ = \frac{1}{2}\frac{2}{6}+\frac{1}{2}\frac{3}{5} \\ = \frac{7}{15}$$
Best Answer
Let $p(n)$ be the probability that the first player to play, as $n$ white balls remain, wins.
We have $$ p(n) = \frac{7}{n+7}\times 1 + \frac{n}{n+7}\times(1-p(n-1))\\p(0) = 1 $$ (according to the result of the first turn).
From this it should be easy.