Question 195, Let $\mathrm{E}$ be the midpoint of side $\mathrm{AB}$ of square $\mathrm{ABCD}$, and let points $P$ and $Q$ be chosen arbitrarily on sides $\mathrm{AD}$ and $\mathrm{BC}$, respectively. What is the probability that $\angle \mathrm{PEQ}$ is an acute angle?
Question 195, Answer: Let the side length of the square $\mathrm{ABCD}$ be 1. Establish a Cartesian coordinate system with $\mathrm{E}$ as the origin, $\overrightarrow{\mathrm{EB}}$ as the positive direction of the $\mathrm{x}$-axis, and $\overrightarrow{\mathrm{BC}}$ as the positive direction of the $\mathrm{y}$-axis. Let point $P\left(-\frac{1}{2}, t\right), Q\left(\frac{1}{2}, s\right)$. Then $\angle P E Q$ being an acute angle is equivalent to: $\overrightarrow{\mathrm{EP}} \cdot \overrightarrow{\mathrm{EQ}}>0 \Leftrightarrow\left(-\frac{1}{2}, \mathrm{t}\right) \cdot\left(\frac{1}{2}, \mathrm{~s}\right)>0 \Leftrightarrow \mathrm{ts}>\frac{1}{4}$. Also, $0 \leq \mathrm{t} \leq 1,0 \leq s \leq 1$, so $\mathrm{t} \geq \frac{1}{4}$. Therefore, the probability that $\angle \mathrm{PEQ}$ is an acute angle is: $\int_{\frac{1}{4}}^{1}\left(1-\frac{1}{4 \mathrm{t}}\right) \mathrm{dt}=\mathrm{t}-\frac{\ln t}{4} \int_{\frac{1}{4}}^{1}=\frac{3-\ln 4}{4}$.