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A candy company distributes boxes of chocolates with a mixture of creams, toffees, and cordials. Suppose that the weight of each box is 1 kilogram, but the individual weights of the creams, toffees, and cordials vary from box to box. For a randomly selected box, let X and Y represent the weights of the creams and the toffees, respectively, and suppose that the joint density function of these variables is f(x, y) =  24xy, 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, x+ y ≤ 1, 0, elsewhere. (a) Find the probability that in a given box the cordials account for more than 1/2 of the weight. 

In letter a, it means x+y < 1/2, now how can we find the limits of the double integral? proper approach to solve this problem? thanks

PS: there would be no problem if the limits can be easily deciphered. but for this one its asking P(X + Y < 1/2)

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    Sorry, I misread the pdf as being nonzero for x+y<1/2. While the first part of my suggestion still holds (see David Mitra's answer to get details of how to do what I merely suggested), you _do_ need to set up an integral to get the probability, and again, David has explained in detail how to do so.2012-02-12

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The first thing you have to do is sketch the region of interest $ X+Y<\textstyle{1\over 2}, \quad X,Y\ge 0 $

This will be be the region of all points in in quadrant 1 whose coordinates $(x,y)$ satisfy: $y<-x+\textstyle{1\over 2}:$

enter image description here

The region is shown in pink above (note the density is 0 above the dashed line). If we call this region $A$, the integral is $ \int\kern-5pt\int_A f(x,y) dA. $

To set up the double integral as an iterated integral, you may think of the region as being generated by the vertical lines $\color{darkgreen}{\ell_x}$ as $x$ ranges from $x=0$ to $x=1/2$.

You first fix $x$ and "integrate along $\ell_x$" in the vertical direction. Then you integrate the $\ell_x$ integrals from $x=0$ to $x=1/2$.

So, fix $x$ and consider $\ell_x$. The limits of the inner integral are from the bottom of $\ell_x$ to the top. The bottom of $\ell_x$ is $y=0$ and the top is $y=-x+{1\over 2}$. Note the inner integral will be with respect to $y$.

So the inner integral is $ \int_0^{-x+{1\over 2}} xy \,dy $

Now set up the outer integral. As mentioned, we integrate the above expression from $x=0$ to $x=1/2$: $ \int_0^{1/2}\int_0^{-x+{1\over 2}} xy\, \,dy\,dx. $


Alternatively, you can think of the region as being generated by horizontal lines $\color{maroon}{\ell_y}$ that range from $y=0$ to $y=1/2$.

Here, you'd integrate along a horizontal line first, from its left endpoint, 0, to its right endpoint $-y+{1\over2}$. Then, integrate with respect to $y$ from $y=0$ to $y=1/2$:

$ \int_0^{1/2}\int_0^{-y+{1\over 2}} xy\, \,dx\,dy. $

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    @blackandyello The second double integral is just another way of setting it up. Note the $dx$ and $dy$ are interchanged; in the second double integral, we're integrating with respect to $x$ first (instead of wrt $y$ first as in the first double integral). For some problems, one way may be easier than the other. Here, it doesn't really matter.2012-02-17