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}}$