If six students, including two particular students A and B, stand in a row, then the probability that A and B are separated with one student in between them is

$\frac{4}{15}$

Total number of ways in which 6 students can stand in a row is 6!.

One student can stand between A and B in ${^4C}_1 × 2!$ ways.

Considering these three as one individual we have 4 students who can stand in a row in 4! ways.

Therefore, number of ways in which one student is there between A and B and is ${^4C}_1 × 2!× 4!$

Hence, required probability $\frac{^4C_1 × 2!× 4!}{6!}=\frac{4}{15}$