Hello, oswaldo!
A random process starts from point $\displaystyle O,$
Point $\displaystyle A$ is obtained in a uniformly selected direction, 1 unit from $\displaystyle O.$
The same process is repeated at point $\displaystyle A$ to obtain point $\displaystyle B$,
and at point $\displaystyle B$ to obtain point $\displaystyle C.$
What is the probability that a triangle is obtained when point $\displaystyle C$ is generated?
. . $\displaystyle (A)\;\frac{1}{8} \qquad (B)\;\frac{1}{9} \qquad (C)\;\frac{1}{12} \qquad (D)\;\frac{1}{16} \qquad (E)\;\frac{1}{18} $ Code:
B
o
* *
* *
* *
O o * * * * * * * * o A
*
o
C
I got: .$\displaystyle {\color{blue}(E)\;\frac{1}{18}}$
Segment $\displaystyle OA$ can be any direction.
Code:
B
o
\
\
\
\
60° \
O o * * * * * o A
60° /
/
/
/
/
o
B
Segment $\displaystyle AB$ must be within 60° on either side of $\displaystyle OA$.
Otherwise, $\displaystyle OB > 1$ and we will have no triangle.
$\displaystyle P(\angle OAB \leq 60^o) \:=\:\frac{120^o}{360^o} \:=\:\frac{1}{3}$
Code:
B
o
/ *
/60°*
/ *
/ *
*
O o * * * * * o A
Segment $\displaystyle BC$ must be within 60° of $\displaystyle AB.$
$\displaystyle P(\angle ABC \leq 60^o) \:=\:\frac{60^o}{360^o} \:=\:\frac{1}{6}$
Therefore: .$\displaystyle P(OABC\text{ forms a triangle}) \;=\;\frac{1}{3}\cdot\frac{1}{6} \;=\;\frac{1}{18}$