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#2 (permalink) |
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I JUST got here.
Join Date: Sep 2004
Posts: 3
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Re: Probability of Continuous Variables
Well, you need the joint density of the 2 variables. Then, draw the domain D={X<Y; 0<X<3; 0<Y<4} in the XY plane, and your probability is just the double integral of the joint density over D. For instance, if X and Y are uniformly distributed and independent, the probability is the area of D.
Hope this helps. |
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#4 (permalink) | |
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I JUST got here.
Join Date: Mar 2005
Posts: 2
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Re: Probability of Continuous Variables
Quote:
probability is 3/4. as the probability for interval [1,3] is same for both so probability for [1,4] is increase the 4th part . So probability is 3/4. |
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#6 (permalink) |
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I JUST got here.
Join Date: Mar 2005
Posts: 13
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Re: Probability of Continuous Variables
Ahsan, please don't mind i think your answer is probably wrong.
By using acd's idea i have solved the question, which is as follows. Area of the rectangle above the line y=x P(x<y) = -------------------------------------------------- Total area of the rectangle 2.5 = ---------- 2*3 5/2 = ---------- 6 5 = ------- 12 A better solution with figure is in attached file. I think it is right answer. Thanks acd, for the idea. |
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