SOLUTION: Find k so that f (x, y) = kxy, 1 ≤ x ≤ y ≤ 2 will be a probability density function. Also find (i) P(X ≤ 3/2 , Y ≤ 3/2 ), and (ii) P(X + Y ≤ 3/2 ).
Thank you :)
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-> SOLUTION: Find k so that f (x, y) = kxy, 1 ≤ x ≤ y ≤ 2 will be a probability density function. Also find (i) P(X ≤ 3/2 , Y ≤ 3/2 ), and (ii) P(X + Y ≤ 3/2 ).
Thank you :)
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Question 1178058: Find k so that f (x, y) = kxy, 1 ≤ x ≤ y ≤ 2 will be a probability density function. Also find (i) P(X ≤ 3/2 , Y ≤ 3/2 ), and (ii) P(X + Y ≤ 3/2 ).
Thank you :) Answer by robertb(5830) (Show Source):
(ii) Quickly sketching the region defined by as well as the square region [1,2]x[1,2], you will see that these two regions do not intersect at all. Hence .
***In general, for this problem, ,
where and .
