(7) A bag contains $n(n>1)$ red balls and 5 white balls. Each time a prize is drawn, two balls are taken out of the bag at once. If the two balls are of different colors, it is considered a win. After each draw, the balls are put back into the bag. Let the probability of winning exactly once in three draws be $p$. Then $n=$ $\qquad$ when $p$ reaches its maximum value $\qquad$ .
(7) $p_{\text {max }}=\frac{4}{9}$ Hint: Let the probability of winning in one draw be $p_{0}$. Then
$$
p_{0}=\frac{5 n}{\mathrm{C}_{n+5}^{2}}=\frac{10 n}{(n+5)(n+4)} .
$$

Thus, the probability of winning exactly once in three draws is
$$
p=\mathrm{C}_{3}^{1} \cdot p_{0}\left(1-p_{0}\right)^{2}=3 p_{0}\left(1-p_{0}\right)^{2} \leqslant \frac{3}{2}\left(\frac{2}{3}\right)^{3}=\frac{4}{9} \text {. }
$$

The equality holds if and only if $2 p_{0}=1-p_{0}$, i.e., $p_{0}=\frac{1}{3}$.
From
$$
\left.\frac{10 n}{(n+5)(n+4)}=\frac{1}{3} \Rightarrow n=20 \text { (since } n>1\right) \text {, }
$$

at this point $p_{\text {max }}=\frac{4}{9}$.