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 October 5th, 2019, 03:46 AM #1 Senior Member   Joined: Dec 2015 From: somewhere Posts: 734 Thanks: 98 Probability involving square numbers Let $\displaystyle n\leq 111 \:$ , $\displaystyle n\in \mathbb{N}$. Find the probability such that n is a square-number and odd . $\displaystyle P(n=N^{2} )$=? Last edited by idontknow; October 5th, 2019 at 04:09 AM. October 5th, 2019, 04:26 AM   #2
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Quote:
 Originally Posted by idontknow Let $\displaystyle n\leq 111 \:$ , $\displaystyle n\in \mathbb{N}$. Find the probability such that n is a square-number and odd . $\displaystyle P(n=N^{2} \& n=1(mod 2) )$=?
All square numbers below 111:
$\sqrt{111} = 10.53565375285273884840140466189966747674770065972 434833044...$, therefore biggest square number below 111 is $100=10^2$. Therefore, there are 10 square number below 111.

1,4,9,16,25,36,49,64,81,100. There 10 square numbers below 111. And five of them is odd.

Therefore, $P(n=N^{2} \; \& \; n \equiv 1(mod 2) ) = 5/111$ Tags involving, numbers, probability, square Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Carl James Mesaros Calculus 2 November 17th, 2017 10:45 AM EvanJ Advanced Statistics 0 December 5th, 2013 05:08 PM roberthun Algebra 12 March 10th, 2013 02:14 PM everk Real Analysis 2 September 5th, 2011 01:06 PM fantom.1040 Algebra 2 July 2nd, 2011 06:04 AM

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