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Let $X=(X_1,X_2,X_3)$ be a continuous random vector of the joint pdf $$f(x_1,x_2,x_3)= 12x_2 \;\mathrm f \mathrm o \mathrm r \; 0<x_3<x_2<x_1<1$$ and $0$ elsewhere.

I need to find the probability of event $B$ where $B=\{x_3\leq1/3\}$. I've been able to sketch the support and see that it's a sort of pyramidal shape, and know that I should be able to find the probability via a triple integral.

My instinct says to simply set it up like so: $$\int_0^{1/3}\int_{x_3}^{x_1}\int_{x_2}^{1}12x_2\;dx_1dx_2dx_3$$

but this is clearly wrong since the answer will still be in terms of a variable. If anyone could help me understand where I've been going wrong it would be greatly appreciated!

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It is $\int_0^{1/3} \int_{x_3}^{1} \int_{x_2}^{1} 12x_2 dx_1dx_2dx_3$. (The middle integral cannot involve $x_1$. The condition $x_2 <x_1$ is already taken care of in the limits for $x_1$).

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Change the last integral from ($x_2$ to 1) to ($x_1$ to 1). Does it make any sense?

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  • $\begingroup$ Sorry, I'm not sure I understand. Why would the bounds of $x_1$ be ($x_1$,1)? $\endgroup$ – sk13 Jan 20 at 22:49

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