# Geometric Probability Problem

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!

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

## 1 Answer

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:

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