Evaluating probabilities with joint density. I have a density function $f(x,y)=\alpha x^2y^2$ for $y\in (0,1)$ and $x\in (-y,y)$. To find $\alpha$ I evaluated $$\alpha \int_{-y}^y\int_0^1 x^2y^2 \ dydx=1$$ and ascertained $\alpha=9/2y^3$. How would I then calculate the probability $P(X^2\ge 1/8 | Y\le 1/2)$? I know that I find the conditional density $f_{X|Y}(x,y)$ over the marginal $f_Y(y)$, but this is not a case of $P(X\in A|Y=y)$ so I am unsure how to proceed.

$$\alpha \int_0^1\int_{-y}^y x^2y^2 \ dxdy=1$$

Let's carry out the integration.

$$\begin{align} \int_0^1\int_{-y}^y x^2y^2 \ dxdy&=\int_0^1y^2\left(\frac23 y^3\right)dy\\\\\\\ &=\frac19 \end{align}$$

Thus, setting $\alpha\left(\frac19\right)=1$ reveals that $\alpha = 9$.

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To calculate the probability $P(X^2\ge 1/8 | Y\le 1/2)$, note that this is equivalent to $2P(X\ge \sqrt{1/8} | Y\le 1/2)$. This is given by

$$2\int_{\sqrt{1/8}}^{1/2} \int_x^{1/2} 9x^2y^2$$

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