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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.

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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Solution

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 | 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)}

Posted by

infoexpert22

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