Probability that two random variables have a product greater than 1

Question

I have two independent random variables $X$ and $Y,$ defined using the following formulas, respectively: $F_X(x)=(1/2)x,$ where $0<x<2;$ and $f_Y(y) = (2/9)y,$ where $0<y<3.$ What is the probability that $XY \ge 1$? `

Answer 1

For $XY\ge 1$, since $Y\le 3$, $X\ge \frac{1}{3}$. Similarly$Y\ge\frac{1}{2}$.

$P(XY\ge 1)$

$=\int_{\frac{1}{3}}^{2}\int_{\frac{1}{x}}^{3}P(X=x)P(Y=y)dydx$

$=\frac{1}{9}\int_{\frac{1}{3}}^{2}\int_{\frac{1}{x}}^{3}xy\cdot dydx$

$=\frac{1}{18}\int_{\frac{1}{3}}^{2}x(9-\frac{1}{x^2})dx$

$=\frac{1}{4}(4-\frac{1}{9})-\frac{1}{18}\ln 6$

$=\frac{35}{36}-\frac{1}{18}\ln 6$