Probability question

(a)
chewwy's approach is right. I'll fill in the details.

i = 1 (probability 1/21)
One way of getting 4 (probability = 1/6).

i = 2 (probability 2/21)
Three ways of getting 4: 1 + 3, 2 + 2, 3 + 1 (probability = 3/36).

i = 3 (probability 3/21)
Three ways of getting 4: 2 + 1 + 1, 1 + 2 + 1, 1 + 1 + 2 (probability = 3/6^3).

i = 4 (probability 4/21)
One way of getting 4: 1 + 1 + 1 + 1 (probability = 1/6^4).

i >= 5
Impossible to get 4.

Answer
= (1/21)(1/6) + (2/21)(3/36) + (3/21)(3/6^3) + (4/21)(1/6^4)
= 35/1944

(b)
P(2 fair dice thrown, score is 4) / P(score is 4)
(2/21)(3/36) / (35/1944)
= 108/245

i = 3 (probability 3/21)
Three ways of getting 4: 2 + 1 + 1, 1 + 2 + 1, 1 + 1 + 2 (probability = 3/6^3).
Should there be 6 ways of getting 4?
2+1a+1b, 1a+2+1b,1a+1b+2
2+1b+1a, 1b+2 +1a,1b+1a+2