4. A bag contains 20 balls, of which 9 are white, 5 are red, and 6 are black. Now, 10 balls are randomly drawn from the bag, such that the number of white balls is no less than 3 and no more than 7, the number of red balls is no less than 2 and no more than 5, and the number of black balls is no less than 1 and no more than 3. The probability of this is $\qquad$ .
4. Solution: From $f(x)=\left(x^{3}+x^{4}+x^{5}+x^{6}+x^{7}\right)\left(x^{2}+\right.$
$$
\begin{aligned}
x^{3}+ & \left.x^{4}+x^{5}\right)\left(1+x+x^{2}+x^{3}\right) \\
& =\left(x^{3}+x^{4}+x^{5}+x^{6}+x^{7}\right)\left(x^{2}+2 x^{3}+3 x^{4}+4 x^{5}\right.
\end{aligned}
$$
$+3 x^{6}+2 x^{7}+x^{8}$ ), the coefficient of $x^{10}$ in the expanded form: $2+3+4+3$ $+2=14$, meaning there are 14 ways that satisfy the condition, so the required probability is $p$
$$
=\frac{14}{\mathrm{C}_{20}^{10}}=\frac{7}{92378} .
$$