"A dart is thrown at a circular target of radius $10$ inches, given that it falls in the upper half of the target."
I know that a conditional density function is given by the formula $$f(x|E) = \begin{cases} \dfrac{f(x)}{P(E)}, & \text{if $x$ is in E} \\\\ 0, & \text{if $x$ is not in E} \end{cases}$$
Now from what I understand, all you do is restrict the unconditional probability density function to the new domain (i.e. divide by $1/2$ because it is half the board). But other than that I'm not sure what to do...
I know the fact that $x^2+y^2 < 10^2$ and thought about integrating using polar coordinates: $$f(x)=\int_0^\pi \int_0^{10} r drd\theta = 50\pi$$ to get the density function for the top half of the board. But I know this answer is wrong (correct answer is $\frac\pi{50}$) and I'm not sure what to do...
Although not stated in the exercise, we can arrive to the correct answer if we assume that $(X,Y)$ is a uniformly distributed random vector in the circle with radius $r=10$. In that case $$f_{X,Y}(x,y)=\dfrac{1}{10^2\pi}$$ for all $(x,y)$ with $x^2+y^2\le 10$. Then $$f(x,y|E) = \begin{cases} \dfrac{f(x,y)}{P(E)}, & \text{if } (x,y) \in E \\\\ 0, & \text{if }(x,y)\notin E \end{cases}$$ which implies that $$f(x,y|E) = \begin{cases} \dfrac{\frac{1}{100π}}{\frac{1}{2}}, & \text{if }(x,y)\in E \\\\ 0, & \text{if }(x,y)\notin E \end{cases}=\begin{cases} \dfrac{1}{50π}, & \text{if }(x,y) \in E \\\\ 0, & \text{if }(x,y)\notin E \end{cases} $$