Left side represents first in the queue, right side represents last in queue.

Number of ways to arrange them with no restriction = 7! = 5040


Required cases :

① Frank is first, Edward is not last.

Number of ways to choose the last person = 5 (from the remaining 5 people, we cannot choose Edward)

Number of ways to arrange the other positions = 5! = 120

Total number of ways here = 5 × 120 = 600


② Frank is not first, Edward is last.

The number of ways is also the same (600) as above since we are just swapping Frank with Edward and 'first' with 'last'


③ Frank is first and Edward is last.

Number of ways to arrange the middle 5 people = 5! = 120


Total number of ways for the three cases

= 600 + 600 + 120
= 1320


Required probability

= 1320 / 5040
= 132 / 504
= 11/42