Question 1192023
If a ball is returned, this is like two 10-side dice being rolled.
There are 100 different outcomes from 1/1 to 10/10
The smallest sum is 2, the largest 20, each of those two values with 1/100 probability 
3 will have 2/100 probability (1/2 an 2/1)
4 will have 3/100...
from the other side
20 will have 1/100 probability
19 will have 2/100 (9/10. 10/9)
18 will have 3/100 (8/10, 10/8, 9/9)
17 will have 4/100
16 will have 5/100
15 will have 6/100 probability (7/8,8/7,6/9,9/6,5/10,10/5).
The answer is 21/100.
the most probable outcome is a sum of 11 with 6 and 5; 7 and 4; 8 and 3; 9 and 2; 10 and 1, each twice or 1/10 probability. Contrast this with two cubic die with a maximum sum of 12, the most probable one of 7 (1 more than the number of sides), with a probability of 1/6 (as if one die were rolled).
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if the ball has an even number, there are 5 possibilities, 2,4,6,8,10
Two of these possibilities can be excluded immediately, 2 and 4, since ball 2 won't make a difference.
if it is a 10, with probability 1/10, 6 of the 10 possibilities of ball 2 will work, so that probability is (1/10)(3/5)=3/50.
If it is an 8, 4 of 10 possibilities will work, so that probability is (1/10)(2/5)=2/50.
If it is a 6, 2 of 10 possibilities will work, so that probability is (1/10)(1/5)=1/50.
The answer is the sum of those 3 or 6/50 or 12%.