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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 squarenumber 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  
Senior Member Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 132 Thanks: 49 Math Focus: Area of Circle  Quote:
$\sqrt{111} = 10.53565375285273884840140466189966747674770065972 434833044...$, 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, $P(n=N^{2} \; \& \; n \equiv 1(mod 2) ) = 5/111$  

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involving, numbers, probability, square 
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