If x belongs to [1,3] and y belongs to [1,4], find the probability that x<y.
Can anyone solve it…
If x belongs to [1,3] and y belongs to [1,4], find the probability that x<y.
Can anyone solve it…
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.
acd!
thanks for an ideal. I will work on this idea.
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Originally Posted by Naseer Asif
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.
probability is 3/4
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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