A person throw an arrow on a target of radius $r$. The position of the arrow on the target is uniformly distributed. Let $X$ the distance between the arrow and the center of the circle. The score obtained by a person is $r-X$. What is the average score ? The answer is $\frac{r}{3}$, whereas I found $\frac{r}{2}$ as follow
We have that $X$ is uniform on $[0,r]$. If $Y=r-X$, then $$\mathbb E[Y]=\int_0^r (r-x)f_X(x)dx=\frac{1}{r}\int_0^r (r-x)dx=\frac{r}{2}.$$
Maybe there is a subtlety than I don't see ?
Best Answer
$$F_X(x) = \frac{\pi x^2}{\pi r^2}$$
$$f_X(x)=\frac{2x}{r^2}$$
\begin{align} E[r-X]&=r-E[X] \\ &=r - \frac1{r^2}\int_0^r 2x^2\, dx\\ &= r - \frac1{r^2}\frac{2r^3}3\\ &= \frac{r}{3} \end{align}