Correct Answer - C
5 persons A, B, C, D and E can stand in a queue in 5! Ways.
Persons A and E can occur together in 2! Ways.
Considering A and E as one person, there are 4 persons who can be queued in 4! Ways.
So, A and E can occur together in 4!×2!
ways.
So, required probability =
4!×2!
5!
=
2
5