**Geometric probability** is the idea of looking at probability in terms of lengths, area, or volumes. This technique is useful when the set of possible outcomes is infinite and we are unable to use the formula

\[ p(X) = \frac{\mbox{desired outcomes}}{\mbox{total outcomes}} .\]

The following problem illustrates why this is useful and important.

## A point is chosen uniformly at random on the real line in the interval \( (0,3) \). What is the probability that the chosen point is closer to the point 0 than it is to the point 1?

Solution: It’s pretty easy to see that a point in this interval will be closer to 0 than it is to 1 if the point is less than \( \frac{1}{2} \). So what does that make the probability? Tricks that we can use with finite sets don’t work here. For example, I can map the interval \( (0,.5) \) to the interval \( (.5,3) \) by the function \( f(x) = 5x + .5 \). In the finite case, given this bijection, we would say that the events are equally likely. Does this then mean that the probability is \( \frac {1}{2} \)?

However, I can also map the interval \( (0,.5) \) to each of the sets \( (.5,.75),(.75,1), \ldots, (2.75,3) \) by using similar transformations. So I can show that the number of elements in my target region is the same as the number in the complement region, or that the complement region has 10 times as many elements as the target region. Does this mean that the probability is \( \frac { 1} { 1+10} = \frac {1}{ 11} \)? If not, what is the correct answer?

The problem here is that we are working with sets of infinite cardinality, and so we cannot use the ideas of finite sets to help us. What we can do is compare the “areas” of the sets to get the correct answer. The area of our target interval (since we are in one dimension, the area is the length) is \( 0.5 \), and the area of our whole interval is 3, so the probability that a point is in the target interval is \( \frac{0.5}{3} = \frac{1}{6} \).

The reason as to why this works is a more advanced topic, which deals with the idea of **Measure Theory**. Measure Theory gives a rigorous framework for probability theory, including probabilities on finite sets. Measure Theory is also the key idea behind integration in calculus, and can be used to find integrals of functions that seem non-integrable using “standard” methods. These two ideas are not unrelated, as at a fundamental level, probability theory is just a special case of integration.

We will do a few more examples on working with geometric probabilities in higher dimensions to get a better feel for how to work with the concept. It is often helpful to use a figure to help with understanding and solving these types of problems.

## 1. A toothpicked is dropped to the floor and it breaks in two places, creating three pieces. The position of the breaks are uniformly random along the entire length of the toothpick. What is the probability that these 3 pieces can form a triangle?

We can parameterize the toothpick along its length from \( 0 \) to \( 1 \). Let \( x \) be the position of the first break and \( y \) be the position of the second break. We have two possibilities, either \( x < y \) or \( y < x \). We need not consider \( x = y \), since the question states the toothpick was broken into three pieces. However, even if this was not stated, we do not need to consider \( x = y \), since the probability of this happening is \( 0 \).

Let us first consider the case when \( x < y \). For the three pieces to form a triangle, the length of the longest pieces must be at most \( 0.5 \), so that the three pieces will satisfy the triangle inequality. For this to occur, we must have \( x < .5\), \( y - x < .5 \) and \( y > 0.5\). We can plot these in two dimensions on the \( xy \)-plane.

The shaded area is the area that satisfies all three of the constraints. If we consider the case where \( y < x \), we will get the same things reflected along the line \( y = x \). The total area of the square is 1, and the sum of the areas of the two small triangles is \( \frac{1}{4} \), so the probability is \( \frac{1}{4} \).

## 2. A point is chosen uniformly at random from the interior of a sphere. What is the probability that it is closer to center of the sphere than it is to the surface of the sphere?

For any point in the interior of the sphere, there is a radius from the center of the sphere to the surface that goes through that point. This radius gives the shortest distances from the point to the center and the point to the surface. So the point will be closer to the center if it is at most half way along the radius. In other words, the set of points that are closer to the center of the sphere will itself be a sphere with half the radius of the original sphere. Since the volume of a sphere is \( V = \frac{4}{3}\pi r^3 \), the ratio of the volumes of the spheres will be \( \frac{1}{8} \), so that is our probability.

## 3. A square \( S \) has side length 30. A standard 20-sided die is rolled, and a square \( t \) is constructed inside \( S \) with side length equal to the roll. Then, a dart is thrown and lands randomly somewhere inside square \( S \). What is the probability that the dart also lands inside square \( T \)?

Suppose the die rolls \( i \). Then the probability that the dart will land inside square \( T \) is the ratio of the area of square \( T \) to the area of square \( S \). This is \( \frac{i^2}{900} \). For each \( i \), the probability that the die will roll \( i \) is \( \frac{1}{20}, \) so the probability that the dart lands inside \( T \) will be

\[ \sum\limits_{i=1}^{20} \frac{1}{20}\cdot \frac{i^2}{900} = \frac{1}{20} \cdot \frac{20 \times 21 \times 41}{900 \times 6} = \frac{287}{1800} .\]

The difficulty associated with geometric probability usually comes from one of two areas, the first is finding a good way to model the problem geometrically, and the second is in trying to determine the areas/volumes of particular regions in order to calculate the relative probabilities. As in finite probability, it is sometimes simpler to find the probability of the complement.

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