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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.

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The notation P [r ≤ p| s ≤ p] means that
P (r ≤p) *given that s ≤ p
Since we know s ≤ p , then it means that s is drawn first.
Let us 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, there is one element missing from the entire set, so there are (n-1) altogether.

$p( r is among p-1 elements $)=\frac{p-1}{n-1}$

Among (1 . . s …. . p) the probability of drawing s is \frac{p}{p} .


P[r \leq p l s \leq p]$ is the probability that r \leq p$ when $s \leq p$.
Therefore, P[r \leq p \mid s \leq p]=\frac{(p-1)}{(n-1)} \times \frac{p}{p}$
=\frac{(p-1)}{(n-1)}$

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