Bivariate distributions -- Example 3

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Bivariate distributions -- Example 3
  1. Lawrence Leemis
  2. Probability
  3. Bivariate random variables
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I paused and tried to solve on my own first. Didn't think of your idea so I evaluated the double integral from 0-60 and J-15 to J of the joint distribution which gave 1/4. Multiplied by 2 since the reciprocal is just as likely (i.e. intervals from 0-60 and J to J+15). Since this is a calculation of P(A or B) I then subtracted by P(A and B) which is 1/4 * 1/4. 1/4 + 1/4 - (1/4)^2 = .4375.

jwck