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Postby » Thu Mar 21, 2013 2:01 am

The joint probability distribution of random variables X and Y is given by f(x,y) = kxy for x = 1, 2, 3, y = 1, 2, 3, and k is a positive constant. What is the probability that X + Y will exceed 5?
Choose one answer.
a. 1/4
b. 1/9
c. 1/36
d. Cannot be determined

Finance Junkie
Posts: 356
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Postby content.pristine » Mon Mar 25, 2013 1:51 pm

Substituting the various values of x and y, we get:
f(1,1)=k, f(1,2)=2k, f(1,3)=3k, f(2,1)=2k, f(2,2)=4k, f(2,3)=6k, f(3,1)=3k, f(3,2)=6k, and f(3,3)=9k. Therefore,
k1 + 2k + 3k + 2k + 4k + 6k + 3k + 6k + 9k = 1
so that, 36k = 1 and k=1/36 = 2.78%
f(1,1)=2.78%, f(1,2)=5.56%, f(1,3)=8.33%, f(2,1)=5.56%, f(2,2)=11.11%, f(2,3)=16.67%, f(3,1)=8.33%, f(3,2)=16.67%, and f(3,3)=25.00%.
f(X+Y)>5 = (the only case that is possible is X=3, Y=3, no other combination)
= x*y*k = 3*3/36 = 1/4

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