# Probability, Random points in rectangle

There is a rectangle, the lower left is always fixed at co-ordinate $(0, 0)$. Let the width and height of the rectangle be $w$ and $h$. Let $P$ be a randomly chosen point from the rectangle with co-ordinates $(x,y)$, $x\geq 0$ and $x\leq w$, $y\geq 0$ and $y\leq h$. $x$ and $y$ can be any real number satisfying the constraint above.

What is the probability that the area of the rectangle whose lower left is $(0,0)$ and upper right is $P(x, y)$ (Hence, the area is $x\cdot y$) is greater than $A$? $w$, $h$ and $A$ will be provided.

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By "randomly chosen", do you mean "uniformly randomly chosen"? Also, I'd delete the sentence beginning "$x$ and $y$ can be ...", since at best it doesn't add to what was already said and at worst it sounds as if $x$ and $y$ are given or can be arbitrarily chosen, in contrast to the preceding sentence which states that they're randomly chosen. Also, the notation $P(x,y)$ doesn't make much sense, since $P$ is itself $(x,y)$ and not defined as a function of $x$ and $y$. –  joriki Feb 16 '13 at 10:09

The points that satisfy that are those where $xy>A$, that is, $y>\frac{A}{x}$ (you can do that because $x$ is positive and $x=0$ is not interesting because the area of the rectangle would be $0$, that is, if $A=0$ the probability is $1$).

So the probability is the are inside the rectangle that is above the curve $y=\frac{A}{x}$, divided by the total area of the rectangle, $wh$.

The bound points would be the intersection of $y=\frac{A}{x}$ with $y=h$ and with $x=w$, which are $(\frac{A}{h},h)$ and $(w,\frac{A}{w})$, so you're in fact interested in $$\int_{A/h}^w \left(h-\frac{A}{x}\right)dx = [hx-A\ln x]^w_{A/h} = wh-A - A\ln(wh) + A\ln A$$ And this has to be divided by $wh$ yielding the (arguably) beautiful $$1 + \frac{A}{wh} \left( \ln\left(\frac{A}{wh}\right) -1 \right)$$

(Note that if $A>wh$, this calculation breaks, but then the probability is clearly $0$)

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This is $\varphi\left(\frac{A}{wh}\right)$ where $\varphi(x)=1-x+x\log x$ if $0\leqslant x\leqslant 1$ and $0$ otherwise. Now you could explain what you tried to solve this problem...

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