# Geometric Probability

*Geometric probability* is a tool to deal with the problem of infinite outcomes by measuring the number of outcomes geometrically, in terms of length, area, or volume. In basic probability, we usually encounter problems that are "discrete" (e.g. the outcome of a dice roll; see probability by outcomes for more). However, some of the most interesting problems involve "continuous" variables (e.g., the arrival time of your bus).

Dealing with continuous variables can be tricky, but geometric probability provides a useful approach by allowing us to transform probability problems into geometry problems. If this sounds surprising, take a look at the following problem:

## Your bus is coming at a random time between 12pm and 1pm. If you show up at 12:30pm, how likely are you to catch the bus?

Intuitively, the answer seems to be \(\frac{1} {2} \). We can show this geometrically by considering a point chosen randomly on a 1-dimensional number line: the length of the number line between 12:30pm and 1pm is equal to the length from 12pm to 12:30pm.

While this example is fairly straightforward, many complicated problems can be solved simply by using geometric probability. On this page, we will start with \(1\text{D}\) examples, which are the simplest and easy to understand and then work our way up to \(2\text{D},\) \(3\text{D},\) and higher dimensions.

#### Contents

## Introduction

One of the main ideas in probability is to count the number of equally likely "desired" outcomes, and then divide that by the number of equally likely total outcomes: \[ P(X) = \frac{\mbox{desired outcomes}}{\mbox{total outcomes}} .\]

However, when a variable is continuous, it becomes impossible to "count" the outcomes in the traditional sense. For example, if \(X\) is a random real number between 0 and 1, it could be \(0.2\) or \(0.53\) or \(0.434662565465465\) or even something irrational like \(\frac{\pi}{4}.\) It is clear that there are infinite outcomes if we count in the traditional sense.

## 1-dimensional Geometric Probability

Let's look more at the situation where \(X\) is a random real number, as mentioned in the Introduction section.

## \(X\) is a random real number between 0 and 3. What is the probability \(X\) is closer to 0 than it is to 1?

Since there are infinitely many possible outcomes for the value of \(X,\) we will take the equally likely outcomes as random points along the number line from 0 to 3. It’s easy to see that \(X\) will be closer to 0 than it is to 1 if \(X<0.5.\)

Now, we can use the measures (lengths, in this 1D case) of our possible outcomes and apply the usual probability formula. Here,

\[P(X\text{ is closer to 0 than to 1}) = \frac{\mbox{length of segment where }0<X<0.5}{\mbox{length of segment where }0<X<3} = \frac{0.5}{3} = \frac{1}{6} \approx 17\%.\ _\square\]

To reiterate, the core idea in one-dimensional (1D) geometric probability is translating a probability question into a geometry problem on a number line, where we measure outcomes with *length*. To make sure you've got this concept down, try this problem related to rounding errors:

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.

## 2-dimensional Geometric Probability

Many probability problems include more than one variable, so 1D geometric probability won't be enough. For problems with two variables, it is often helpful to transform them into 2D geometric probability questions, where the outcomes are measured by *area*: \[P(X) = \dfrac{\text{area of desired outcomes}}{\text{area of total outcomes}}.\]

This is most easily understood when the problem at hand is explicitly a 2D geometry problem:

## A dart is thrown at a circular dartboard such that it will land randomly over the area of the dartboard. What is the probability that it lands closer to the center than to the edge?

The set of outcomes are all of the points on the dartboard, which make up an area of \(\pi r^2\) where \(r\) is the radius of the circle. The points that are closer to the center than to the edge are those that lie within the circle of radius \(\frac{r}{2}\) around the center, so the area of the "success" outcomes is \(\pi \left(\frac{r}{2}\right)^2 = \frac{\pi r^2}{4}.\)

Thus,

\[P(\text{closer to center than edge}) = \dfrac{\text{area of desired outcomes}}{\text{area of total outcomes}} = \frac{\frac{\pi r^2}{4}}{\pi r^2} = \frac{1}{4} = 25\%.\ _\square\]

## 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

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

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.

To make sure you've got down the basic ideas of 2D geometric probability, try this similar question. Note that many 2D geometry problems, such as the one below, use the ideas of composite figures. If you are not familiar with that concept, you may want to take a look at composite figures first.

However, one of the most powerful uses of geometric probability is applying it to problems that are not inherently geometric. Identifying when and how to use geometric probability is never obvious, but a good sign is that you are dealing with probabilities in a situation with continuous variables. Let's take a look at a modified example of the bus problem mentioned at the beginning of this wiki.

## Both the bus and you get to the bus stop at random times between 12pm and 1pm. When the bus arrives, it waits for 5 minutes before leaving. When you arrive, you wait for 20 minutes before leaving if the bus doesn't come. What is the probability that you catch the bus?

We have two continuous variables here: \(b,\) the time in minutes past 12pm that the bus arrives, and \(y,\) the time in minutes past 12pm that you arrive. Since there are 2 independent variables, we will convert this into a 2-dimensional geometry problem. Specifically, we can think of the set of all outcomes as the points in a square:

Then, we need to determine the region of "success"; that is, the points where we catch the bus. Since the bus will wait for 5 minutes, you need to arrive within 5 minutes of the bus' arrival, or \(y \le b+5:\)

However, you only wait for 20 minutes, so you can't arrive more than 20 minutes before the bus, so \(y \ge b-20:\)

Combining our two conditions, we have a region of success as shown below: Now, we just need to find the area of this success region. A simple method is to find the area of the non-success region, and then subtract that from the total area: Thus, the probability of catching the bus is \[P(\text{catching the bus}) = \dfrac{\text{area of desired outcomes}}{\text{area of total outcomes}} = \frac{60^2 - \frac{55^2}{2}-\frac{40^2}{2}}{60^2} = \frac{103}{288} \approx 36\%.\ _\square\]

Now that we have changed our problem into a geometric one, we can easily answer other questions about the situation, such as:

1) What is the probability that the bus does not have to wait for you?

2) What is the probability that you had to wait less than 10 minutes, given that you were able to catch the bus?

3) What is the probability that the bus came and went before you, given that you were not able to board the bus?

To practice these ideas, let's try a similar question:

Dave and Kathy both arrive at Pizza Palace at two random times between 10:00 p.m. and midnight. They agree to wait exactly 15 minutes for each other to arrive before leaving. What is the probability that Dave and Kathy see each other?

If the probability is \( \frac ab\) for coprime positive integers, give the answer as \(a+b\).

## 3-dimensional Geometric Probability

At this point, you can probably guess where this is headed! 3D geometric probability is when we are dealing with 3 continuous variables, and we measure the *volume* of the various outcomes; that is,

\[P(X) = \dfrac{\text{volume of desired outcomes}}{\text{volume of total outcomes}}.\]

To get started, let's look at an example which is analogous to the first problem we solved in the 2D geometric probability section.

## An atom is inside a sphere and it is equally likely to be anywhere within the sphere. What is the probability that it lands closer to the center of the sphere than the outside?

The set of outcomes are all of the points in the sphere, which make up a volume of \(\frac{4\pi}{3} r^3\) where \(r\) is the radius of the sphere. The points that are closer to the center than to the edge are those that lie within the sphere of radius \(\frac{r}{2}\) around the center, so the volume of the "success" outcomes is \(\frac{4\pi}{3} \left(\frac{r}{2}\right)^3 = \frac{\pi}{6}r^3.\) Thus,

\[P(\text{closer to center than outside}) = \dfrac{\text{volume of desired outcomes}}{\text{volume of total outcomes}} = \frac{\frac{\pi}{6}r^3}{\frac{4\pi}{3} r^3} = \frac{1}{8} = 12.5\%.\ _\square\]

Of course, not all problems will be so explicitly geometric in nature. As usual, one of the signs that we might want to apply geometric probability is that we are dealing with continuous variables. Let's see how we can approach the following example:

## Alex, Bob, and Charlie each randomly pick a real number between 0 and 1. What is the probability that the sum of the squares of their numbers does not exceed 1?

First, if we let their 3 numbers be \(x,\) \(y,\) and \(z,\) it is easy to see that the outcomes can be represented as points within a unit cube (the cube that encloses the region \(x,y,z \in [0,1]\)), which has volume \(1^3 = 1.\)

Then, the region where the sum of the squares of their numbers does not exceed 1 is given by \(x^2+y^2+z^2 \le 1,\) which (without restriction) is the sphere of radius 1 centered around the origin \((0,0,0).\) However, since \(x,y,z\ge 0,\) exactly \(\left(\frac{1}{2}\right)^3 = \frac{1}{8}\) of this sphere (one "octant") lies within the unit cube of possible outcomes. Hence, the volume of this "success" region is \(\frac{1}{8} \cdot \left(\frac{4\pi}{3} \cdot 1^3\right) = \frac{\pi}{6}.\)

Thus, \[P(\text{sum of squares does not exceed 1}) = \dfrac{\text{volume of desired outcomes}}{\text{volume of total outcomes}} = \frac{\frac{\pi}{6}}{1} = \frac{\pi}{6} = 52\%.\ _\square\]

If you'd like to test your skills at turning probability problems into 3D geometry problems, take a shot at this challenging problem which is similar to the example above:

## Applications

In addition to being a useful mathematical problem-solving tool, geometric probability can also be applied in other scientific fields. Let's start with an example from Mechanics, using the ideas of velocity and acceleration.

## We are playing shuffleboard on the table below, where the lengths of the regions are labelled below (in meters).

## You push the puck with initial velocity \(v,\) where \(v\) is randomly chosen between 5 and 15 meters/second. Since the table is rough, the puck decelerates at a constant rate of 5 \(m/s^2.\) What is the probability that you slide the puck off of the table? (You may assume that the puck is negligibly small.)

In this problem we have only one variable - the initial speed of the puck so this is going to be a 1-dimensional geometry problem. Recall the kinematics formula \( v_{f} ^{2} = v_{i}^{2} + 2 a s \). The final speed \(v_ {f }\) is zero because the puck comes to a rest. The initial speed \( v_{i} = v\). The distance traveled \(s\) will decide how much points we get. After plugging in the values, we get

\[s = \frac{ v^{ 2 } }{ 10 }, \]

so \(s>8+4+2+1=15\) occurs when \(v>\sqrt{150}.\) If we think of \(v\) as being a point on a number line between 5 and 15, then we can find our probability as

\[\frac{15-\sqrt{150}}{15-5} \approx 28\%.\ _\square\]

Awesome! Who would have expected that geometric probability would allow us to solve a Physics problem? Let's check out a few more examples. If you'd like to contribute to this wiki, add a solution to one of these examples!

## A block performs simple harmonic motion with time period \(T\) and maximum speed \(v\). The speed of the block is measured at a random time. What is the probability that the measured speed is more than \(v/2\)?

## Johannes Kepler worked out that all planets revolve around the sun in elliptical orbits with the sun as one of the foci. He also deduced that planets revolve around the sun with constant areal velocity.

## Let's model a small star system in which a planet revolves around the star in an elliptical orbit. It has semi-major axis \(a\) and semi-minor axis \(b\). During one revolution, the minimum speed of the planet is \(u\) and the maximum speed is \(v\). In a complete revolution, an instant of time is randomly and uniformly chosen. What is the probability that the distance between the planet and the star at that instant is more than \(b\)?

## Extra Challenges

There are many great problems in geometric probability. If you'd like some extra challenges, check out these problems. If you'd like to contribute to this wiki, you can add a solution to one of the examples!

## Two numbers are chosen randomly and uniformly from \([-a, a]\). What is the probability that the absolute value of the smaller number is greater than two times the absolute value of the larger number? Does the final answer depend on the value of \(a\)?

Two points are randomly and uniformly selected on the circumference of a circle. The center of the circle and the two points are joined together. What is the probability that we obtain:\(\begin{array}{rrl} &\text{i)} &\text{a line segment,} \\

&\text{ii)} &\text{an acute angled triangle,} \\ &\text{iii)} &\text{a right angled triangle,} \\ &\text{iv)} &\text{an obtuse angled triangle?} \end{array}\)

Two points are randomly and uniformly selected from the interior of a circle. The centre of the circle and the two points are joined together. What is the probability that we obtain:\(\begin{array}{rrl} &\text{i)} &\text{a line segment,} \\

&\text{ii)} &\text{an acute angled triangle,} \\ &\text{iii)} &\text{a right angled triangle,} \\ &\text{iv)} &\text{an obtuse angled triangle?} \end{array}\)

**Cite as:**Geometric Probability.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/1-dimensional-geometric-probability/