(a) Two reds: only RR counts, so P(two reds) = 16/49. Adding 4/7 + 4/7 would give more than 1; those are successive branches, not alternative complete outcomes.
(b) Exactly one red: RB or BR counts. P(exactly one red) = 12/49 + 12/49 = 24/49. Counting only RB misses getting the red second.
(c) At least one red: RR, RB or BR counts, giving (16 + 12 + 12)/49 = 40/49. A shorter route uses the complement: the only excluded outcome is BB, so 1 − 9/49 = 40/49.
As a check, all four path probabilities total (16 + 12 + 12 + 9)/49 = 1. Exactly one and at least one are not interchangeable: at least one includes two reds.