First, consider the students from the first two schools as single units. This gives us three units to arrange (a group of 3 students, a group of 2 students, and 1 individual student). The number of ways to arrange these units is given by the permutation formula $A^n_n = n!$, where $n$ is the number of units. In this case, $n = 3$, so there are $3! = 6$ ways to arrange the units.

Within each unit, the students can also be arranged. For the first school (3 students), there are $A^3_3 = 3! = 6$ arrangements. For the second school (2 students), there are $A^2_2 = 2! = 2$ arrangements. The individual student from the third school has only 1 arrangement.

So, the total number of ways to arrange the students with same-school students together is the product of these arrangements, which is $6 \times 6 \times 2 \times 1 = 72$.

The total number of ways to arrange the 6 students regardless of school is $A^6_6 = 6! = 720$.

Therefore, the probability of same-school students standing together is given by the classical probability formula, $P = \dfrac{\text{number of desired outcomes}}{\text{total number of outcomes}} = \dfrac{72}{720} = \boxed{\dfrac{1}{10}}$.