6. Suppose there are 5 cards, each with a different number written on it, and the numbers are $2,3,4,5,6$. When these cards are randomly placed in a row, the probability that for each $i(1 \leq i \leq 5)$, the number at the $\mathrm{i}$-th position from the left is greater than or equal to $\mathrm{i}$ is
6. $\frac{2}{15}$.

Analysis: First consider the number in the 5th position. If it is 6, then consider the position of 2; 2 can only be in the 1st or 2nd position, so there are two ways to place it; then consider 3, which can only be in the 1st, 2nd, or 3rd position, so after placing 2, there are 2 ways to place 3; the remaining 4 and 5 can be placed arbitrarily, and there are 2 positions left, so there are a total of 8 ways to place them. Similarly, if the 5th position is 5, there are also 8 ways to place the numbers, so there are a total of 16 ways, and the probability is $p=\frac{16}{5!}=\frac{2}{15}$.