# A company is reviewing tornado damage claims under a farm insurance policy. Let X be the portion of a claim representing damage to the house and let Y be the portion of the same claim representing damage to the rest of the property. The joint density function of X and Y is f(x,y)=6[1-(x+y)], x>0, y>0, x+y<1. Determine the probability that the portion of a claim representing damage to the house is less than 0.2.

A company is reviewing tornado damage claims under a farm insurance policy. Let X be the portion of a claim representing damage to the house and let Y be the portion of the same claim representing damage to the rest of the property. The joint density function of X and Y is

Determine the probability that the portion of a claim representing damage to the house is less than 0.2.
What I did was establish ${f}_{x}\left(X\right)=3-6x$ since X represents damage to the house. Then I integrated the function from 0 to 0.2 and I got an answer of 0.48, however, the actual answer is 0.488.
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Step 1
Let us do it like you did, first finding the density of X. To do this, we "integrate out" y. So we want
${\int }_{0}^{1-x}6\left(1-x-y\right)\phantom{\rule{thinmathspace}{0ex}}dy.$
The integration goes from $y=0$ to $y=1-x$ because of the condition $x+y<1$. The integral is
${6\left(y-xy-\frac{{y}^{2}}{2}\right)|}_{0}^{1-x}.$
Plug in. We get $6\left(1-x-x\left(1-x\right)-\frac{1}{2}\left(1-x{\right)}^{2}\right)$. This simplifies to $3\left(1-x{\right)}^{2}$.
Now integrate from $x=0$ to $x=0.2$. Equivalently, find the double integral of the joint density, $y=0$ to $1-x$, $x=0$ to $x=0.2$.