1. Solution. The first one can be right. For example, let there be 2 blue candies out of 5 in the first packet, and 3 blue candies out of 8 in the second. Then, in the box, there are 5 blue candies out of 13. The second one can also be right: for example, if there are 4 blue candies out of 10 in the first packet, and 3 blue candies out of 8 in the second, then in total, there are 7 blue candies out of 18 in the box. The third mathematician cannot be right, since the probability of drawing a blue candy from the box must be between the numbers $\frac{3}{8}$ and $\frac{2}{5}$, but $\frac{3}{8}<\frac{2}{5}<\frac{17}{40}$. Contradiction.
Answer: the first and second mathematicians may be right; the third is wrong.

Comment. Other methods of solving are possible. For example, the problem can be solved using a system of equations.

## Evaluation Criteria

| The correctness/incorrectness of all three is justified | 3 points |
| :--- | :---: |
| The correctness/incorrectness of two is justified | 2 points |
| The correctness/incorrectness of only one is justified | 1 point |
| Only the answer (correct or incorrect) without justification | 0 points |