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Two random variables X and Y have the following joint pdf: $$f_{X,Y}(x,y)\begin{cases}10x^{2}y & 0<x<1,0<y<x\\0 & \text{otherwise}\end{cases}$$

I am asked to find the marginal pdf of $X$, followed by the conditional pdf of $Y$ given $X=x$, and finally to evaluate $\Pr(X+Y\geq1)$.

So far, I've managed the following (and am hoping it is correct):

For the marginal pdf of $X$, \begin{align*} f_{X}(x)&=\int_{-\infty}^{\infty}f_{X,Y}(x,y)\;dy\\ &=\int_{0}^{x}10x^{2}y\;dy\text{ [do I integrate from 0 to $x$ here?]}\\ &=10x^{2}\frac{y^{2}}{2}\bigg|^{y=x}_{y=0}\\ &=5x^{4} \end{align*}

Assuming that the above is correct, for the conditional pdf of $Y$ given $X$, I have: \begin{align*} f_{Y|X}(y|x)&=\frac{f_{X,Y}(x,y)}{f_{X}(x,y)}\\ &=\frac{10x^{2}y}{5x^{4}}\\ &=\frac{2y}{x^{2}} \end{align*}

I'm not sure whether the above is correct, and any help would be appreciated if I have any errors in my understanding. But I am not sure at all how to proceed with $\Pr(X+Y\geq1)$. I've noticed this question seems relevant but the contents within are pretty much new to me and the mathematics involved is a little bit too advanced for my level (I'm only first-year under-grad).

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Old mate
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You are correct in deriving the marginal pdf of $X$ and the conditional pdf of $Y$ given $X=x$. Now, for the last part, you want to integrate the joint pdf of $X$ and $Y$ over all points $(x,y)$ such that $x+y\geq 1$. We have \begin{align*} P(X+Y \geq 1) &= P(Y \geq 1-X) \\ &= \int\limits_{\frac12}^{1} \int\limits_{1-x}^{x} 10x^2 y \;\mathrm{d}y \;\mathrm{d}x. \end{align*}

The limits of integration for $y$ should make sense. I urge you to draw a picture of the support, and the region you wish to integrate to make it clearer to yourself. You will then see that $x$ must be greater than $\frac12$, but of course less than $1$, given the support of $f_{X,Y}(x,y)$.

kokocijo
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