27.1. (Belgium, 80). Each of the two urns contains white and black balls, and the total number of balls in both urns is 25. One ball is randomly drawn from each urn. Knowing that the probability that both drawn balls will be white is 0.54, find the probability that both drawn balls will be black.
27.1. Let the total number of balls in the first and second urns be \( m_{\mathrm{i}} \) and \( m_{2} \) respectively (for definiteness, assume \( m_{1} \leqslant m_{2} \)), and the number of white balls in these urns be \( k_{1} \) and \( k_{2} \) respectively. Then the probability that both drawn balls are white is \( \left(k_{1} / m_{1}\right) \cdot \left(k_{2} / m_{2}\right) \). We obtain the following relationships:

\[
\frac{k_{1}}{m_{1}} \cdot \frac{k_{2}}{m_{2}} = 0.54 = \frac{27}{50}, \quad m_{1} + m_{2} = 25
\]

Since \( 27 m_{1} m_{2} = 50 k_{1} k_{8} \), at least one of the numbers \( m_{1}, m_{2} \) is divisible by 5. But the sum \( m_{1} + m_{2} \) is also divisible by 5, so each of the numbers \( m_{\mathrm{i}}, m_{2} \) is divisible by 5. Thus, we have only two possibilities: either \( m_{\mathrm{I}} = 5, m_{2} = 20 \), or \( m_{1} = 10, m_{2} = 15 \). In the case \( m_{1} = 5, m_{2} = 20 \), we get

\[
k_{1} k_{2} = 54, \text{ where } 0 \leq k_{1} \leqslant 5, \quad 0 \leqslant k_{2} \leqslant 20.
\]

By trying all possible values of \( k_{1} \), we find \( k_{1} = 3, k_{2} = 18 \). Then in the first urn, there are 2 black balls, and in the second urn, there are also 2 black balls, and the probability of drawing two black balls is \( (2 / 5) \cdot (2 / 20) = 0.04 \). Similarly, in the case \( m_{1} = 10, m_{2} = 15 \), we find \( k_{1} = 9, k_{2} = 9 \). Then in the first urn, there is 1 black ball, and in the second urn, there are 6 black balls, and the probability of drawing two black balls is again \( (1 / 10) \cdot (6 / 15) = 0.04 \) (in both cases, the answers are the same).