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.

Thanks in advance!

#HelpMe! #MathProblem #Math

Easy Math Editor

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`2^{34}`	$2^{34}$
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Comments

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 $\tfrac12$ .

Mark Hennings - 7 years, 7 months ago

Thanks! That was really helpful!

Mursalin Habib - 7 years, 7 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$