3. Among 10 balls, 3 are red. If the balls are randomly distributed to 10 children, each getting one ball, then the probability that exactly one of the last three children gets a red ball is
A. $C_{4}^{1} \cdot \frac{3}{10}$
B. $\frac{C_{3}^{1} C_{7}^{2}}{C_{10}^{3}}$
C. $C_{3}^{1} \cdot \frac{3}{10} \cdot\left(\frac{7}{10}\right)^{2}$
D. $\frac{3}{10} \cdot\left(\frac{7}{10}\right)^{2}$
10. This problem belongs to geometric probability. Let the times they arrive at the agreed time be $x, y$, then $0 \leqslant x, y \leqslant$ 60, to meet each other, they must satisfy $|x-y| \leqslant 20$. As shown in the figure, the shaded area represents the times when they can meet, the area of which is $60 \times 60-40 \times 40$, so its probability is
$$
1-\frac{40 \times 40}{60 \times 60}=\frac{5}{9} \text {. }
$$
3. $\mathrm{C}$ Among them, selecting one child to take the red ball has 3 ways, the probability of taking the red ball is $\frac{3}{10}$, and the probability of not taking the red ball is $\frac{7}{10}$.