Solutions - 5250-04

(1) Compute the marginal $f_Y(y)$:

$$\begin{array}{c}{f}_Y(y)={\int }_0^{1-y}24xydx\\ \\ \gg \gg {12x^{2}y|}_0^{1-y}\gg \gg 12{(1-y)}^{2}y\end{array}$$

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(2) Compute $f_{X|Y}(x|0.5)$:

$$\begin{array}{c}{f}_{X|Y}(x|0.5)=\frac{f_{X,Y}(x,0.5)}{f_Y(0.5)}\\ \\ \gg \gg \frac{24x\cdot \frac{1}{2}}{12{\left(\frac{1}{2}\right)}^2\cdot \frac{1}{2}}\gg \gg \frac{12x}{3/2}\gg \gg 8x\end{array}$$

The support is $x+0.5\le 1$, so $0\le x\le 0.5$.

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(3) Compute $P[X>0.3|Y=0.5]$:

$$\begin{array}{c}P[X>0.3|Y=0.5]={\int }_{0.3}^{0.5}8xdx\\ \\ \gg \gg {4x^2|}_{0.3}^{0.5}\\ \\ \gg \gg 4{(0.5)}^2-4{(0.3)}^2\gg \gg 1-0.36\gg \gg 0.64\end{array}$$