2
$\begingroup$
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)

  • 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?2012-02-12
  • 0
    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

1 Answers 1