Conditional Probability Uniform Bivariate Transformation Distribution

Question

I'm reviewing probability theory from years ago and am a bit rusty. I'm not sure how to calculate the conditional probability for a uniform distribution after a bivariate transformation.

Suppose X and Y both follow a Uniform$(0,2)$ distribution. For the transformation $U=X+Y$ and $V=X$, I'll then have the following joint PDF: \begin{cases} \frac{1}{4}, & \text{if $0<V<U<4$}.\\ 0, & \text{otherwise}. \end{cases}

How would I calculate the conditional probability $P(V<.2|U=1.5)$? I'm likely missing something very intuitive. I've had someone say that U follows a $U(0,1.5)$ distribution and thus $P(V<.2) =\frac{1}{1.5} * .2$ but that doesn't seem right to me. I'd have to multiply by $\frac{1}{4}$ at some point, right?

Edit:

Conditional. on $f(V|U)$

First found the marginal of $U$ as follows:

\begin{cases} \frac{u}{4}, & \text{if $0<U<2$}\\ \frac{1}{2}-\frac{u-2}{4}, & \text{if $2<U<4$}\\ \ 0, & \text{otherwise} \end{cases}

Since $U=1.5$, the conditional $f(V|U=1.5)=\frac{f(U,V)}{f(U=1.5)}=\frac{\frac{1}{4}}{\frac{u}{4}}=\frac{1}{1.5}$

So the probability $(V<.2|U=1.5)=$

$$\int_{0}^{.2} \frac{1}{1.5} dv = .2*\frac{1}{1.5} - 0*\frac{1}{1.5}=.133$$

Answer 1

This follows my comment (using the approach shown in the Youtube video). Once you have the joint you can calculate the conditional.

Sorry for the typo. Of course, $g_2(X,Y)=X$, and not $Y$.