Someone gave me this problem.

If \(x\) and \(y\) are positive reals such that

\(x+y=2n\) [\(n\) is another positive real number],

what is the probability that \(xy>\frac{3}{4}n^2\)?

These kinds of problems are kind of my weak point. So I would appreciate it if you shared your thought process along with your solution.

Thanks in advance!

## Comments

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TopNewestClearly \(y\) is a dependant variable; you just want to identify the range of \(x\) values that do what you want. Thus, with \(y=2n-x\) you are interested in \(x\) such that \[ \begin{array}{rcl} x(2n-x) & > & \tfrac34n^2 \\ x^2 - 2nx + \tfrac34n^2 & < & 0 \\ (x - n)^2 & < & \tfrac14n^2 \\ |x-n| & < & \tfrac12n \end{array} \] and so you want \(\tfrac12n < x < \tfrac32n\). What you haven't told us is the probability distribution that \(x\) comes from. If \(x\) is uniformly distributed between \(0\) and \(2n\), the probability is \(\tfrac12\). – Mark Hennings · 3 years, 9 months ago

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– Mursalin Habib · 3 years, 9 months ago

Thanks! That was really helpful!Log in to reply