Two natural numbers r, s are drawn one at a time, without replacement from the set S= {1, 2, 3, ...., n}. Find P [r ≤ p|s ≤ p], where p ∈ S.

The notation P [r ≤ p| s ≤ p]
means
P (r ≤p) *given that s ≤ p
Since we're told s ≤ p , then it means s is drawn first.
Let we have n numbers before s is drawn:
(1 . . s …. . p . . .. n)
After s is drawn,
[ 1 ... p] has one element missing, so there are (p-1) elements.
Also, the entire set has one element missing, so there are (n-1) altogether.
P(r is among p – 1 elements )

Among (1 . . s …. . p) the probability of drawing s is .
Now,

P [r ≤ p|s ≤ p] is the probability that r ≤ p when s ≤ p.

So,