6. Evaluate the three solutions and answers provided below. Identify all errors and shortcomings, and provide the correct solution to this problem.

From a deck of cards, six cards were selected - three of black suits and three of red suits. The selected cards were shuffled and stacked. Find the probability that the three bottom cards will be of the same suit.

Let the red suit cards be denoted by k, and the black suit cards by ч.

Solution 1: List all possible cases. The bottom cards can be: a) $3 \mathbf{k}$; b) $2 \mathbf{k}$ and $1 \mathbf{ч}$; c) $1 \mathbf{k}$ and $2 \mathbf{ч}$; d) $3 \mathbf{ч}$. Out of these 4 cases, two are suitable. Therefore, the required probability is $\frac{1}{2}$.

Solution 2: List all possible cases. Among the top three cards, there can be: a) 1 more k than among the bottom cards; b) 1 more ч than among the bottom cards; c) all cards of the same suit. Out of these three cases, only case c) satisfies the condition. Therefore, the required probability is $\frac{1}{3}$.

Solution 3: There are only eight possible cases for the arrangement of the three bottom cards: a) ччч; b) ччk; c) чkч; d) kчч; e) чkк; f) kчk; g) ккч; h) ккк. Out of these 8 cases, only 2 cases satisfy the condition - cases a) and h). Therefore, the required probability is $\frac{1}{4}$.
6.II.6. Comment. In all the given solutions, the same mistake has been made. Namely, the systems of events considered in them do not consist of equally probable events.

Example of a correct solution. The number of all possible arrangements of the cards is $6!=720$. In $36=3!\cdot 3!$ of these arrangements, the red cards lie on top and the black cards lie at the bottom, and there are just as many arrangements where the top three cards are black. Therefore, there are $36+36=72$ suitable arrangements, so the desired probability is $\frac{1}{10}$.