Lets start with a more general question: "An urn contains B black, W white, and G gray marbles. What is the probability that a given color is the first/second/third color to be exhausted?"
The easy part is the third color. Draw the marbles out and put them in a row, an keep on drawing until the entire urn is empty. The third color exhausted will be just the color of the last marble. Then we have:
P_B_3rd(B,G,W) = B/(B+G+W)
P_G_3rd(B,G,W) = G/(B+G+W)
P_W_3rd(B,G,W) = W/(B+G+W)
Now lets look at a specific case: "What is the probability that Black is the third color exhausted and Gray is the second color exhausted?" Alternately "What is the probability that the colors are exhausted in the order White, Gray, Black?"
We know the first part. If we assume Black will be the third color exhausted then the probability that Gray is the second color exhausted is simply G/(G+W).
Then we can multiply by P_B_3rd to get P_WGB=G*B/((G+W)*(B+G+W))
Then by permutation of variables we can say:
P_WGB(B,G,W) = G/(G+W) * B/(B+G+W) = G*B/((G+W)*(B+G+W))
P_GWB(B,G,W) = W/(G+W) * B/(B+G+W) = W*B/((G+W)*(B+G+W))
P_WBG(B,G,W) = B/(W+B) * G/(B+G+W) = G*B/((W+B)*(B+G+W))
P_BWG(B,G,W) = W/(W+B) * G/(B+G+W) = W*G/((W+B)*(B+G+W))
P_GBW(B,G,W) = B/(G+B) * W/(B+G+W) = W*B/((G+B)*(B+G+W))
P_BGW(B,G,W) = G/(G+B) * W/(B+G+W) = W*G/((G+B)*(B+G+W))
Now to our desired questions: "What is the probability an given color is the second color exhausted?" and "What is the probability an given color is the first color exhausted?"
These are now simple to calculate, just add the two probabilities where the chosen color is in the desired order, such as Black being exhausted first will be just P_BWG + P_BGW. Thus:
P_W_1st(B,G,W) = G*B/((G+W)*(B+G+W)) + G*B/((W+B)*(B+G+W)) = G*B*(2W+G+B)/((W+B)*(G+W)*(B+G+W))
P_W_2nd(B,G,W) = W*B/((G+W)*(B+G+W)) + W*G/((W+B)*(B+G+W)) = W*(W*B+B^2+W*G+G^2)/((W+B)*(G+W)*(B+G+W))
P_G_1st(B,G,W) = W*B/((G+W)*(B+G+W)) + W*B/((G+B)*(B+G+W)) = W*B*(W+2G+B)/((G+B)*(G+W)*(B+G+W))
P_G_2nd(B,G,W) = G*B/((G+W)*(B+G+W)) + W*G/((G+B)*(B+G+W)) = G*(G*B+B^2+W*G+W^2)/((G+B)*(G+W)*(B+G+W))
P_B_1st(B,G,W) = W*G/((W+B)*(B+G+W)) + W*G/((G+B)*(B+G+W)) = G*W*(W+G+2B)/((W+B)*(G+B)*(B+G+W))
P_B_2nd(B,G,W) = G*B/((W+B)*(B+G+W)) + W*B/((G+B)*(B+G+W)) = B*(W*B+W^2+G*B+G^2)/((W+B)*(G+B)*(B+G+W))
Now with that done we can turn to the problem as posed by KS.
Part 1A can be split into (first draw is Black and first exhaust is Black) + (first draw is Gray and first exhaust is Gray) + (first draw is White and first exhaust is White)
Then the probability can be calculated as
3/63 * P_B_1st(2,48,12) + 48/63 * P_G_1st(3,47,12) + 12/63 * P_W_1st(3,48,11)
= 214848/2240525 = 0.0958918110710659331
Similarly, Part 1B can be split into (first draw is Black and second exhaust is Black) + (first draw is Gray and second exhaust is Gray) + (first draw is White and second exhaust is White)
Then the probability can be calculated as
3/63 * P_B_2nd(2,48,12) + 48/63 * P_G_2nd(3,48,11) + 12/63 * P_W_2nd(3,47,12)
= 652452/2240525 = 0.2912049631224824538
Part 2A can be split into nine cases covering the nine different first two draws (draw Black, Black, predict Black) + (draw Black, Gray, predict Gray) + (draw Black, White, predict White) + (draw Gray, Black, predict Black) + (draw Gray, Gray, predict Gray) + (draw Gray, White, predict White) + (draw White, Black, predict Black) + (draw White, Gray, predict Gray) + (draw White, White, predict White).
Then the probability can be calculated as
3/63 * 2/62 * P_B_1st(1,48,12) + 3/63 * 48/62 * P_G_1st(2,47,12) + 3/63 * 12/62 * P_W_1st(2,48,11)
+ 48/63 * 2/62 * P_B_1st(2,47,12) + 48/63 * 47/62 * P_G_1st(3,46,12) + 48/63 * 12/62 * P_W_1st(3,47,11)
+ 12/63 * 2/62 * P_B_1st(2,48,11) + 12/63 * 48/62 * P_G_1st(3,47,11) + 12/63 * 11/62 * P_W_1st(3,48,10)
= 214848/2240525 = 0.0958918110710659331
Similarly, Part 2B can be split into nine cases covering the nine different first two draws (draw Black, Black, predict Black) + (draw Black, Gray, predict Gray) + (draw Black, White, predict White) + (draw Gray, Black, predict Black) + (draw Gray, Gray, predict Gray) + (draw Gray, White, predict White) + (draw White, Black, predict Black) + (draw White, Gray, predict Gray) + (draw White, White, predict White).
Then the probability can be calculated as
3/63 * 2/62 * P_B_2nd(1,48,12) + 3/63 * 48/62 * P_G_2nd(2,47,12) + 3/63 * 12/62 * P_W_2nd(2,48,11)
+ 48/63 * 2/62 * P_B_2nd(2,47,12) + 48/63 * 47/62 * P_G_2nd(3,46,12) + 48/63 * 12/62 * P_W_2nd(3,47,11)
+ 12/63 * 2/62 * P_B_2nd(2,48,11) + 12/63 * 48/62 * P_G_2nd(3,47,11) + 12/63 * 11/62 * P_W_2nd(3,47,11)
= 652452/2240525 = 0.2912049631224824538
Solution
