Consider a group of three students $A, B$ and an other student in between $A$ and $B$. Choice for a student between $A$ and $B$ is 4. $A$ and $B$ can interchange their places in the group in 2 ways.
Now the group of three students (student $A$, student $B$ and a student in between $A$ and $B$ ) and the remaining 3 students can be stand in a row in 4 ! ways.
Hence total number of ways to stand in a row such that $A$ and $B$ are separated with one student in between them $=4 \times 2 \times 4$ !
Now total number of ways to stand 6 student stand in a row without any restriction $=6 !$
Hence required probability
$$
=\frac{4 \times 2 \times 4 !}{6 !}=\frac{4 \times 2}{6 \times 5}=\frac{4}{15}
$$