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

Concept: A product p = a·b is odd only when every factor in it is odd, so P(p odd) = P(a odd) × P(b odd), and P(p even) = 1 − P(p odd). A sum X + Y is even exactly when X and Y share the same parity — both even, or both odd — so P(X + Y even) = P(X even)P(Y even) + P(X odd)P(Y odd).

1. In {1, 2, 3, 4, 5}, the odd values are 1, 3, 5 (3 of them) and the even values are 2, 4 (2 of them). Since the draws are with replacement, each draw is independent with P(odd) = 3/5 and P(even) = 2/5.

2. a·b is odd only when both a and b are odd: P(a·b odd) = (3/5) × (3/5) = 9/25. Hence P(a·b even) = 1 − 9/25 = 16/25.

3. a·b + c is even exactly when a·b and c share the same parity — both odd, or both even.

4. Both-odd branch: P(a·b odd) × P(c odd) = (9/25) × (3/5) = 27/125.

5. Both-even branch: P(a·b even) × P(c even) = (16/25) × (2/5) = 32/125.

6. Total probability = 27/125 + 32/125 = 59/125.

Cross-check: the complementary event (a·b + c odd) needs a·b and c to have different parity: P(a·b odd)P(c even) + P(a·b even)P(c odd) = (9/25)(2/5) + (16/25)(3/5) = 18/125 + 48/125 = 66/125. Since 59/125 + 66/125 = 125/125 = 1, the two results correctly account for the whole sample space.