Chance a product of two uniforms exceeds one half

Let $X$ and $Y$ be independent, each uniform on $[0,1]$. Compute $\mathbb{P}(XY>\tfrac12)$.

$\tfrac12-\tfrac12\ln 2\approx0.153$. Condition on $X=x$: you need $Y>\tfrac{1}{2x}$, possible only when $x>\tfrac12$. So the probability is $\int_{1/2}^{1}\left(1-\tfrac{1}{2x}\right)dx=\tfrac12-\tfrac12\ln2$.