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Home  >>  CBSE XII  >>  Math  >>  Probability
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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 $P[r\leq p|s\leq p],where \;p\in S$

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

 

answered Mar 1, 2013 by poojasapani_1
edited Jun 4, 2013 by poojasapani_1
 

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