6. (20 points) Each of the two baskets contains white and black balls, and the total number of balls in both baskets is 25. One ball is randomly drawn from each basket. It is known that the probability that both drawn balls will be white is 0.54. Find the probability that both drawn balls will be black.
Answer: 0.04.

Solution. Let in the $i$-th basket there be $n_{i}$ balls, among which $k_{i}$ are white, $i=1,2$. Then $\frac{k_{1}}{n_{1}} \cdot \frac{k_{2}}{n_{2}}=0.54=\frac{27}{50}$. Therefore, for some natural number $m$, the equalities $k_{1} \cdot k_{2}=27 m, n_{1} \cdot n_{2}=50 m$ hold. One of the numbers $n_{i}$ is divisible by 5, then the other one is also divisible by 5 (since $n_{1}+n_{2}=25$).

Let $n_{1} \leq n_{2}$. There are two possible cases:

1) $n_{1}=5, n_{2}=20$. Then $m=2, k_{1} \cdot k_{2}=54$, and $k_{1} \leq 5, k_{2} \leq 20$, so $k_{1}=3, k_{2}=18$.
2) $n_{1}=10, n_{2}=15$. Then $m=3, k_{1} \cdot k_{2}=81$, and $k_{1} \leq 10, k_{2} \leq 15$, so $k_{1}=9, k_{2}=9$. In both cases, the probability of drawing two black balls is $\left(1-\frac{k_{1}}{n_{1}}\right) \cdot\left(1-\frac{k_{2}}{n_{2}}\right)=0.04$.

| Points | Grading Criteria |
| :---: | :--- |
| 20 | The correct answer is obtained with justification. |
| 10 | Not all possible cases of ball distribution in the baskets are considered in the solution, or there is no justification that all possible cases have been considered. |
| 0 | The solution does not meet any of the above conditions. |