Question:

**Step 1: Determine the total number of ways to select three distinct numbers.** The set of available numbers is $S = \{-1, 0, 1, 2, 3\}$. We are selecting three distinct numbers from this set without repetition. The order of selection does not matter, so this is a combination problem. The total number of ways to select 3 numbers from 5 is given by the combination formula $\binom{n}{k}$: $$ \text{Total combinations} = \binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4 \times 3!}{3! \times 2!} = \frac{5 \times 4}{2 \times 1} = 10 $$ There are 10 possible unique combinations of three distinct numbers from the given set. **Step 2: Determine the number of combinations where the sum of the three numbers is 0.** We need to find all subsets $\{a, b, c\}$ from $S$ such that $a+b+c=0$. Let's list all possible combinations and calculate their sums: * $\{-1, 0, 1\} \implies -1 + 0 + 1 = 0$ * $\{-1, 0, 2\} \implies -1 + 0 + 2 = 1$ * $\{-1, 0, 3\} \implies -1 + 0 + 3 = 2$ * $\{-1, 1, 2\} \implies -1 + 1 + 2 = 2$ * $\{-1, 1, 3\} \implies -1 + 1 + 3 = 3$ * $\{-1, 2, 3\} \implies -1 + 2 + 3 = 4$ * $\{0, 1, 2\} \implies 0 + 1 + 2 = 3$ * $\{0, 1, 3\} \implies 0 + 1 + 3 = 4$ * $\{0, 2, 3\} \implies 0 + 2 + 3 = 5$ * $\{1, 2, 3\} \implies 1 + 2 + 3 = 6$ From this exhaustive list, only one combination, $\{-1, 0, 1\}$, results in a sum of 0. Thus, there is 1 favorable outcome. **Step 3: Calculate the probability.** The probability that $a+b+c=0$ is the ratio of the number of favorable outcomes to the total number of possible outcomes: $$ P(a+b+c=0) = \frac{\text{Number of combinations with sum 0}}{\text{Total number of combinations}} = \frac{1}{10} $$