# Two natural numbers r,s are drawn one at a time,without replacement from the set S=1,2,3,......,n.Find $P[r\leq p|s\leq p],where \;p\in S$

Two number 'r' and 's' are taken successively without replacement from S=$$(1\;2\;3\dots\;n$$)
Let $$E_1$$$$;$$r$$\leq$$p$$;$$
$$E_2$$$$;r$$$$>$$p$$;$$
A$$;$$event 's' selected$$\leq$$p
P($$E_1/A$$)=P($$r\leq\;p/s\leq\;p)$$
=P(E$$_1$$/A)$$\large\frac{{P(E_1)}{P(A/E_1)}}{{P(E_1)}{P(A/E_1)}+{P(E_2)}{P(A/E_2)}}$$
P$$E_1$$=P($$'r'aney number\leq\;p$$)
=$$\large\frac{p}{n}$$
=P($$E_2$$)=P($$r>p$$)
=$$\large\frac{n-p}{n}$$
P($$A/E_1$$)=P($$s\leq\;p/r\leq\;p)$$
=$$\large\frac{p-1}{n-1}$$
P($$A/E_2$$)=$$\large\frac{p}{n-1}$$
P($$E_1/A$$)=$$\Large\frac{\frac{p}{n}\times\frac{p-1}{n-1}}{\frac{p}{n}\times\frac{p-1}{n-1}+\frac{n-p}{n}\times\frac{p}{n-1}}$$
=$$\large\frac{p(p-1)}{p(p-1)+(n-p)p}$$
=$$\large\frac{(p-1)}{p-1+n-1)}$$
=$$\large\frac{p-1}{n-1}$$

edited Jun 4, 2013
Could you  please explain what exactly is p ? Is it any other number randomly drawn from s ?
And also how did you get P(E1) = n/p. As in n an p are numbers drawn from n.
Thank you for this answer. It helped me a lot.