Random point $(X, Y, Z)$ is uniformly distributed in the unit cub. What is the probability that $Y^2-4XZ \geq 0$?
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1What have you tried? – Kavi Rama Murthy Jan 25 '20 at 11:59
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I tried to find the space in the cub where $Y^2−4XZ=0$. That is when $Y=1$ I have two points for $(X,Z)$ accordingly $(1;\frac{1}{4})$ and $(\frac{1}{4};1)$. When $Z=1$ then $Y^2=4X$ and when $X=1$ than $Y^2=4Z$. But I don't know how to continue? – GeorgiD Jan 25 '20 at 12:08
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Can you deal with double integrals? – almagest Jan 25 '20 at 12:08
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@almagest no, for now – GeorgiD Jan 25 '20 at 12:11
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@GeorgiD Can you do $\int_a^1\frac{1}{x}\ dx$? – almagest Jan 25 '20 at 12:20
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@almagest That should be $−ln(a)$ – GeorgiD Jan 25 '20 at 12:37
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I replaced the [tag:probability-theory] tag by [tag:probability]. Please avail yourself of the tag summaries when choosing tags. – joriki Jan 25 '20 at 12:57
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@ joriki Yes, thank you! – GeorgiD Jan 25 '20 at 12:58
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Since the unit cube has volume 1, the probability that $y^2- 4xz\ge 0$ is the volume of the region lying above the surface $y= \sqrt{xz}$ and inside the unit cube. That is equal to $\int_{x= 0}^1\int_0^1 x^{1/2}z^{1/2}dxdz= \left(\int_0^1 x^{1/2}dx\right)\left(\int_0^1 z^{1/2}dz\right)= \left(\int_0^1 x^{1/2}dx\right)^2= \frac{2}{3}$.
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Please note that this is a duplicate question; joining in the close vote to link it to the original is better for the site than adding a new answer in a different place. If your answer isn't covered by the answers to the original question, it would be preferable to add it there. – joriki Jan 25 '20 at 13:07
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