# A stone is thrown into a circular pond of radius 1 meter. Suppose the stone falls uniformly at random on the area of the pond. What will be the expected distance of the stone from the center of the pond?

## Options are: A. 1/3 B. 1/2 C. 2/3 D. $\frac{1}{2} ^ \left(\frac{1}{2}\right)$

Jun 6, 2018

Option: C

#### Explanation:

$A \left(r\right) = \pi {r}^{2} , q \quad \mathrm{dA} \left(r\right) = 2 \pi r \setminus \mathrm{dr}$

By definition of the PDF $m a t h \boldsymbol{P} \left(r\right)$:

$m a t h \boldsymbol{P} \left(r\right) \cdot \mathrm{dr} = P \left(r < X < r + \mathrm{dr}\right)$

$= \frac{\mathrm{dA}}{\pi \cdot {1}^{2}} = 2 r \mathrm{dr} \implies m a t h \boldsymbol{P} \left(r\right) = 2 r$

Check:

• ${\int}_{0}^{1} m a t h \boldsymbol{P} \setminus \mathrm{dr} = 1$

Expected value of $r$:

$\left\langle r\right\rangle = {\int}_{0}^{1} \setminus r \setminus m a t h \boldsymbol{P} \setminus \mathrm{dr}$

$= 2 {\int}_{0}^{1} {r}^{2} \setminus \mathrm{dr} = \frac{2}{3}$