I want to solve the following:

Let $Q=(x, y)$ be a point chosen at random in a unit disc centered in $(0,0)$ and with radius $1$. Calculate the probabilities that $Q$ is within $0.5$ of the center; that $y > \frac{1}{\sqrt{2}}$; that both $||x-y||<1$ and $||x+y||<1$.

It seems that this is a problem of geometric probability, remember that $$P(Q)=\frac{m(Q)}{m(\Omega)}$$ with $m$ a geometric measure.

We can represent $\Omega :=$ set of all points in the interior of the circle of radius $1$.

The event to which we must find the probability $\Omega$ should be the set of points that belong to a ring that satisfies the conditions of the statement, but here is my problem, I can not see the area of that event.

Any suggestion to approach the problem? I appreciate it!

  • 5
    $\begingroup$ start with drawing all geometry, i.e. circles and linear constraints. stare at the drawing then the solution will come to you itself $\endgroup$
    – Aksakal
    Nov 19, 2023 at 13:47

1 Answer 1


Let $\Omega$ represent the set of all points inside that disc i.e. $(x,y)$ with the condition that $x^2 + y^2 < 1$. An ``event'' $E$ is then a subset of $\Omega$. To calculate the probability of this event $E$ you would compute, $$ P(E) = \frac{ \text{area}(E) }{\text{area}(\Omega) } $$

Here $\text{area}(\Omega) = \pi$, that part is easy to find.

In your problem you have three different events:

(i) $E_1$ where $x^2 + y^2 < (0.5)^2$

(ii) $E_2$ where $y > \frac{1}{\sqrt{2}}$

(iii) $E_3$ where $|x+y| < 1$ and $|x-y| < 1$

You can visualize these events by using WolframAlpha, for instance $E_3$ visualizes as:

enter image description here

So to find $P(E_3)$ simply calculate the area of that region and divide by $\pi$.


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