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This is what I did:

Total number of possibilities nc3 .

Now suppose that the stations train stops at is such that exactly two of these are consecutive.

(1,2) then it may stop at (4,5,6…..n) and no. of possibilities are n-3

(2,3) then it may stop at (5,6,7…..n) and no. of possibilities are n-4

…………………………………….

(n-3,n-2) then it may stop only at n and no. of possibilities are 1

So the number of ways in which the train may stop at exactly two consecutive stations is

(n-3) + (n-4) + ……… + 1 n-3 terms

using formula of A.P. [n/2 (a+l)] n is no. of terms, a is first term, l last ]

possibilities are (n-3)(n-2)/2

Now if exactly 3 stations are consecutive then (1,2,3) (2,3,4)….. (n-2,n-2,n)

Hence here the no. of possibilities are n-2

So the things I have to exclude are (n-3)(n-2)/2 + (n-2)

Which is equal to (n-1)(n-2)/2

So we have Pr{event} = 1 – (n-1)(n-2)/2*nC3 = (n-3)/n

But I am not getting the right answer. Any help where my analysis is wrong?