Let the number of blue marbles be b.
Let the total number of marbles be n.
P(3 blue marbles in a row) =b/n*(b-1)/(n-1) * (b-2)/(n-2)=1/3417
P(2 blue marbles in a row) =b/n * (b-1)/(n-1) = 1/204
If we replace the second relation into the first one, we get:
1/204 * (b-2)/(n-2) = 1/3417
Hence: 3417 * (b-2) = 204 * (n-2)
Finding the least number of marbles in the bag boils down to finding the LCM(204, 3417) = 2 * 2 * 3 * 17 * 67 = 13668
So: 3417 * 4 = 204 * 67
One possible solution: b-2 = 4 and n-2 = 67
b=6 and n= 69. When replaced back in the initial condition b/n*(b-1)/(n-1) * (b-2)/(n-2)=1/3417 it does not work
Therefore we have to go for multiples of 4 and 67
The next possible solution: b-2 = 2*4 and n-2 = 2*67
b=10 and n=136 - which is the correct one.
For your third question, just try to play woth multiples of 4 and 67.