a, b, c are chosen randomly and with replacement from the set {1, 2, 3, 4, 5}.…

Concept: For integers p and q, the sum p + q is even exactly when p and q have the same parity (both odd or both even). A product p × q is odd only when both factors are odd; if either factor is even, the product is even. When a value is drawn uniformly with replacement from {1, 2, 3, 4, 5}, it is odd with probability 3/5 (the odd values 1, 3, 5) and even with probability 2/5 (the even values 2, 4), and repeated draws are independent of each other.

Application: a×b + c is even exactly when a×b and c share the same parity, so there are two disjoint ways this can happen:

1. a×b is odd and c is odd. a×b is odd only when both a and b are odd, so P(a×b odd) = (3/5) × (3/5) = 9/25. Combined with c odd: (9/25) × (3/5) = 27/125.

2. a×b is even and c is even. P(a×b even) = 1 − 9/25 = 16/25. Combined with c even: (16/25) × (2/5) = 32/125.

Adding the two disjoint cases: 27/125 + 32/125 = 59/125.

Cross-check: listing all 8 parity patterns of (a, b, c), weighting odd draws by 3 and even draws by 2 (out of 5 values), the even-sum patterns are (odd, odd, odd), (odd, even, even), (even, odd, even) and (even, even, even), with weights 27, 12, 12 and 8. These add to 59 out of 5³ = 125 total outcomes, confirming the probability is 59/125.