# Probability of an event for a continuous random vector of three coordinates

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

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 $$x_2 is already taken care of in the limits for $$x_1$$).
Change the last integral from ($$x_2$$ to 1) to ($$x_1$$ to 1). Does it make any sense?
• Sorry, I'm not sure I understand. Why would the bounds of $x_1$ be ($x_1$,1)? – sk13 Jan 20 at 22:49