4. Solution. The first one can be right. For example, let there be 4 white chocolate candies out of 7 in the first box, and 3 white candies out of 5 in the second. Then, in the bag, there are 7 white candies out of 12 in total. The second one can also be right: if there were 8 white candies out of 14 in the first box, and 3 white candies out of 5 in the second, then in total, there are 11 white candies out of 19 in the bag. The third mathematician cannot be right, since the probability of drawing a white candy from the bag must be between $\frac{4}{7}$ and $\frac{3}{5}$, but $\frac{19}{35}<\frac{4}{7}<\frac{3}{5}$. 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

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