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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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1  
Draw a sketch of the plane with coordinate axes $x$ and $y$, and indicate on it the region where $f(x,y) > 0$. (Hint: it is a triangle). The point $(X,Y)$ always lies inside this region. Draw on the same sketch the region where $X+Y < 1/2$. (Hint: it is the region below a straight line). Now you have to find the probability that the point $(X,Y)$ lies in the region defined by $X+Y<1/2$ and you know that $(X,Y)$ must lie in the triangle sketched before. Do you necessarily have to integrate to find the answer or do you think a guess might suffice? –  Dilip Sarwate Feb 12 '12 at 13:52
    
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. –  Dilip Sarwate Feb 12 '12 at 15:40

1 Answer 1

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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why is that the outer integral is using constnts as limits unlike the inner integral which uses some variables. cant get the logic behind on why not follow the way the 1st integral was set up –  IvanMatala Feb 16 '12 at 11:50
    
@blackandyello The inner integral in the first double integral above integrates over the line $\ell_x$ from its bottom to its top. The bottom of the line $\ell_x$ always at $y=0$. The top depends where $\ell_x$ is exactly: the top is at $y=-x+1/2$ since it touches the line $x+y=1/2$. The outer integral will always have constants as its limits. They range from the leftmost of the $\ell_x$ lines, $x=0$, to the right most, $x=1/2$. –  David Mitra Feb 17 '12 at 1:28
    
@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. –  David Mitra Feb 17 '12 at 1:30

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