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(\(E_1/A\))=\(\Large\frac{\frac{p}{n}\times\frac{p-1}{n-1}}{\frac{p}{n}\times\frac{p-1}{n-1}+\frac{n-p}{n}\times\frac{p}{n-1}}\)

=\(\large\frac{p(p-1)}{p(p-1)+(n-p)p}\)

=\(\large\frac{(p-1)}{p-1+n-1)}\)

=\(\large\frac{p-1}{n-1}\)

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.