What's wrong with this 'proof' that probability of being $2$ m away is $1/9$? Suppose there is a chicken (which we will assume to simply be a point), which is encolsed in a circular barn of radius $6$m. At the centre of the barn, there is a well, (which we will also assume to simply be a point). If the chicken was equally likely to be at any point within the barn, what is the probability that the chicken is exactly $2$m away from the well? I know it should be zero, since this is a continuous distribution.
However, I was wondering, what goes wrong in the following reasoning to obtain an answer which isn't zero. The only way for the chicken to be exactly $2$m away from the well, is if it were confined to a concentric circle of radius $2$m at the well. The length of the circumference can then be thought of as all the possible points the chicken could be at in order to be $2$m away from the well, so that's $2\pi r' = 2\pi(2) = 4\pi$. All the possible points in the barn is simply the area, so that's $\pi r^2 = \pi(6^2) = 36\pi$.
Hence, if we let $X$ be the random variable of how far the chicken is away from the well, then
$$ \mathbb{P}(X = 2) = \dfrac{4\pi}{36\pi} = \dfrac{1}{9}. $$
So to be extra clear:

Question: What goes wrong in the reasoning above?

Thank you for any help.
 A: Roughly speaking, the circumference as a subset of a plane has measure zero.
On the other hand, because an arc of the circumference as a subset of the circumference does not have measure zero, if we land on the circumference and every point on it is equally probable, then it is valid to say that the probability of landing on the circumference's top half equals $\dfrac{\pi r}{2\pi r}.$
A: What went wrong is that you compared a length with an area. It might help to put units in there. The length of the circle is 4$\pi m$ but the area of the circle is 36$\pi m^2$ so when you divide you get
$$
{1\over 9 m},
$$
and the $m^{-1}$ hopefully shows you it doesn't make sense.
A: You could use the continuity of the probability function here. For instance if you let X be a continuous r.v. denoting the chicken's radial distance from the well, then $\{ X=2\} \subset E_{n}:= \{|X-2| \leq \frac{1}{n}\}$ for $\forall n \geq 0$ (i.e. it is a subset of the event that the chicken is in a ring of inner radius $2-\frac{1}{n}$ and outer radius $2+\frac{1}{n}$). Note that $E_{n+1} \subset E_{n}$ so that we have a strictly decreasing set of events. Now,
\begin{align}P(X=2) = P(\lim_{n \to \infty} E_{n}) = \lim_{n \to \infty} P(E_{n}) = \lim_{n \to \infty} \frac{\pi(2/n)}{36\pi} = 0\end{align}
Here the second equality comes from the continuity of the probability function.
