suppose X and Y have the joint pdf $\displaystyle f_{X, Y} = x + y, 0 < x, y < 1 $ Find P(X < 2Y)
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Originally Posted by wik_chick88 suppose X and Y have the joint pdf $\displaystyle f_{X, Y} = x + y, 0 < x, y < 1 $ Find P(X < 2Y) $\displaystyle \displaystyle \Pr(X < 2Y) = \int_{y=0}^{y = 1} \int_{x = 2y}^{x = 2} f(x, y) \, dx \, dy$.
Right, but X cannot exceed 1, with any probability, so the imposter must draw the line in the unit square. I'd switch the order of integration... $\displaystyle 1-\int_0^1\int_0^{x/2} (x+y)dydx$
Originally Posted by matheagle Right, but X cannot exceed 1, with any probability, so the imposter must draw the line in the unit square. I'd switch the order of integration... $\displaystyle 1-\int_0^1\int_0^{x/2} (x+y)dydx$ Right you are. I misread the support.
thats why I'm the math beagle
the final answer i got is $\displaystyle \frac{19}{24}$ is this correct? thanks so much!
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