4. There are 7 athletes who respectively won first, second, and third prizes in a competition. It is known that the number of people who won the first prize is no less than 1, the number of people who won the second prize is no less than 2, and the number of people who won the third prize is no less than 3. Then the probability that exactly 2 people won the first prize is $\qquad$
4. $\frac{6}{13}$.
In the order of first, second, and third prizes, there are three scenarios for the number of winners: $(1,2,4),(1,3,3),(2,2,3)$.
When $(1,2,4)$, the number of ways to award prizes is
$\mathrm{C}_{7}^{1} \mathrm{C}_{6}^{2} \mathrm{C}_{4}^{4}=7 \times \frac{6 \times 5}{2} \times 1=105$ (ways);
When $(1,3,3)$, the number of ways to award prizes is
$\mathrm{C}_{7}^{1} \mathrm{C}_{6}^{3} \mathrm{C}_{3}^{3}=7 \times \frac{6 \times 5 \times 4}{3 \times 2} \times 1=140$ (ways);
When $(2,2,3)$, the number of ways to award prizes is
$C_{7}^{3} C_{4}^{2} C_{2}^{2}=\frac{7 \times 6 \times 5}{3 \times 2} \times \frac{4 \times 3}{2} \times 1=210$ (ways).
Therefore, the probability that 2 people win the first prize is
$$
\frac{210}{210+140+105}=\frac{6}{13} \text {. }
$$