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# A Probability Problem: A Little Help Would Be Good

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.

Note by Mursalin Habib
4 years, 2 months ago

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Clearly $$y$$ 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 $$\tfrac12$$.

- 4 years, 2 months ago