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I am a probability noob and I was solving the following problem, but my answer doesn't match the book's.

The length and width of panels used for interior doors(in inches) are denoted as Xand Y, respectively. Suppose that Xand Yare independent, continuous uniform random variables for 17.75<x<18.25 and 4.75<y<5.25, respectively.determine the probability that thearea of a panel exceeds 90 squared inches.

I thought the area random variable will have a minimum value of 17.75*4.75=84.3125 Maximum value of area will be 18.25*5.25=95.8125

Also if X and Y are uniformly distributed and independent then area should be uniformly distributed too. Every single value that area can take has equal chance. Please correct me here.

Now with this when I integrate area(A) from 90 to 95.8125 for a constant PDF f(A)=1/(95.8125-84.3125), my answer comes out to be 0.5054. But the answer according to the solutions given at the back of book is 0.499. I am not able to understand where I am doing a mistake. Can anyone please help me understand where I am going wrong?

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marked as duplicate by studiosus, drhab, userNaN, Dan, Alexander Gruber Jun 2 at 21:23

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A Start: Let random variable $W$ represent the area. The distribution of $W$ is not uniform, though it is not terribly far from being uniform.

If we assume independence (which is not entirely plausible), then the joint density of $X$ and $Y$ is $4$ on the rectangle given by $17.75\le x\le 18.25$ and $4.75\le y\le 5.25$.

Draw the rectangle, and draw carefully the hyperbola $xy=90$. The probability that the area is $\gt 90$ is the integral of $4$ over the part of the rectangle which is "above" the hyperbola. So it is $4$ times the area of a certain region.

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That's exactly the explanation given in my solution book. I couldn't get a feel for it though. Doesnt the rectangle just shows the range of values, x and y can take? So why double integrate over x and y? I am still not able to visualise how PDF of area will look like. And i am not even able to convince myself that my range of area variable that i thought are incorrect. So i am not able to have an understanding of it from the gut. –  Durin Jun 2 at 16:43
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In general, the probability that $(X,Y)$ lands in a certain region $D$ that is a subset of the region where our joint density "lives" is $4$ times the area of $D$. This is because the pair $(X,Y)$ has uniform distribution on the rectangle. To see the non-uniformity of $W$, imagine small increases $h$ in length and/or width towards the lower end of the range. This will have a smaller effect on the area of the panel than increases of $h$ in length and/or width towards the upper end of the range. –  André Nicolas Jun 2 at 16:58
    
I googled more and found this link cnx.org/content/m23332/latest. I am able to understand the solution analytically atleast for now. That gives a bit of moral boost(fake one probably). I think I will try to correlate what you said with my current understanding. But thank you very much for giving a reply. –  Durin Jun 2 at 17:38
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@Tom: You have discovered something important, that getting to grips with the geometry is very useful. –  André Nicolas Jun 6 at 13:47
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My comment (the first one) was meant to supply some geometric intuition. Roughly speaking imagine a small square, expand its side by $h$, and a bigger square, expand the side by $h$. There is a larger increase of area in the big square. –  André Nicolas Jun 6 at 15:00

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