Geometric probability exercise

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This is a problem in 49p. of the book differential geometry of curves and surfaces - doCarmo

When a line $l$ in $\mathbb{E}^2$ meets a fixed circle $S$ of radius $1$ in $\mathbb{E}^2$, then what is the probability $P$ that the length of the chord is greater than $\sqrt{3}$

Proof : We assume that $S$ has the center $o=(0,0)$. Here we define $x_l\in l$ where (the distance $|x_l-o|$ between $x_l$ and $o$) is (the distance between $l$ and $o$).

Here when $l$ is divided into three pieces by the circle $S$, then the line segment of a finite length is called a chord.

So $|x_l-o| \leq \frac{1}{2}$ iff the length of the chord has a length $\geq \sqrt{3}$.

Hence the answer is the ratio between the probability that $ x_l$ is in $\frac{1}{2}$-ball $B_\frac{1}{2}$ and the probability that $x_l$ is in $1$-ball $B_1$, $P=\frac{1}{4}$

But solution in the book is as follows (to me) : $$ P = \frac{ {\rm length}\ \partial B_\frac{1}{2} }{{\rm length}\ \partial B_1} = \frac{1}{2} $$

Then which one is right ?