例 3.1.6medium一级题目发布者: ai-batch题干设 DDD 为平面上以原点为圆心、以 rrr 为半径的圆内区域，如今向该圆内随机投点，其坐标 (X,Y)(X,Y)(X,Y) 服从 DDD 上的二维均匀分布，其密度函数为 p(x,y)={1πr2,x2+y2⩽r2,0,x2+y2>r2.p(x,y) = \begin{cases} \dfrac{1}{\pi r^2}, & x^2+y^2 \leqslant r^2, \\ 0, & x^2+y^2 > r^2. \end{cases}p(x,y)=⎩⎨⎧πr21,0,x2+y2⩽r2,x2+y2>r2. 试求概率 P(∣X∣⩽r/2)P(|X| \leqslant r/2)P(∣X∣⩽r/2)。答案点击展开后可查看解析解析p(x,y)p(x,y)p(x,y) 的非零区域与 {∣X∣⩽r/2}\{|X| \leqslant r/2\}{∣X∣⩽r/2} 的交集部分见图 3.1.4，因此所求概率为 P (∣X∣⩽r2)=∫−r/2r/2 ∫−r2−x2r2−x21πr2 dydx=1πr2∫−r/2r/22r2−x2 dxP\!\left(|X| \leqslant \frac{r}{2}\right) = \int_{-r/2}^{r/2}\!\int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}} \frac{1}{\pi r^2}\,\mathrm{d}y\mathrm{d}x = \frac{1}{\pi r^2}\int_{-r/2}^{r/2} 2\sqrt{r^2-x^2}\,\mathrm{d}xP(∣X∣⩽2r)=∫−r/2r/2∫−r2−x2r2−x2πr21dydx=πr21∫−r/2r/22r2−x2dx =1πr2[xr2−x2+r2arcsinxr]−r/2r/2= \frac{1}{\pi r^2}\left[x\sqrt{r^2-x^2}+r^2\arcsin\frac{x}{r}\right]_{-r/2}^{r/2}=πr21[xr2−x2+r2arcsinrx]−r/2r/2 =1πr2(rr2−r24+2r2arcsin12)= \frac{1}{\pi r^2}\left(r\sqrt{r^2-\frac{r^2}{4}}+2r^2\arcsin\frac{1}{2}\right)=πr21(rr2−4r2+2r2arcsin21) =1π (32+π3)≈0.609.= \frac{1}{\pi}\!\left(\frac{\sqrt{3}}{2}+\frac{\pi}{3}\right) \approx 0.609.=π1(23+3π)≈0.609.
