4. Two boxes are each filled with black and white balls, and the total number of balls in the two boxes is 25. Each time, a ball is randomly drawn from each box. If the probability of drawing two black balls is $\frac{27}{50}$, then, the probability of drawing two white balls is
4. $\frac{1}{25}$.

Let the events of drawing a black ball from the two boxes be $A$ and $B$, respectively, then
$$
P(A \cdot B)=P(A) \cdot P(B)=\frac{27}{50} \text {. }
$$

On the other hand, suppose one of the boxes contains $n$ balls, of which $a$ are black, and the other box contains $b$ black balls. Then
$$
P(A \cdot B)=P(A) \cdot P(B)=\frac{a}{n} \cdot \frac{b}{25-n} .
$$

Comparing equations (1) and (2), we get
$$
n=10, a=b=9 \text { or } n=5, a=3, b=18 \text {. }
$$
(1) When one box contains 10 balls, and the other box contains 15 balls, with 9 black balls in each.
Thus, the probability that both balls drawn are white is
$$
\begin{array}{l}
P(\bar{A} \cdot \bar{B})=P(\bar{A}) \cdot P(\bar{B}) \\
=(1-P(A)) \cdot(1-P(B))=\frac{1}{10} \times \frac{6}{15}=\frac{1}{25} .
\end{array}
$$
(2) When one box contains 5 balls, 3 of which are black, and the other box contains 20 balls, 18 of which are black.
Thus, the probability that both balls drawn are white is
$$
\begin{array}{l}
P(\bar{A} \cdot \bar{B})=P(\bar{A}) \cdot P(\bar{B}) \\
=(1-P(A)) \cdot(1-P(B))=\frac{2}{5} \times \frac{2}{20}=\frac{1}{25} .
\end{array}
$$