Continuous Uniform Distributions

Up to now, we have considered discrete probability spaces, namely spaces that contain a countable number of points, that is points that can be counted by using natural numbers, 1, 2, 3,… For example, in the case of a die, we considered a space made of 6 points. Then, under the assumption of equiprobability (each point has the same probability), we determined the probability of an event by dividing the number of favorable occurrences by the number of possible occurrences. So, in the case of a fair die, Pr(2v5)=2/6.

However, often we need to deal with sample spaces that are not discrete but continuous. For example, suppose that we randomly choose a point within a circle of radius 4. What is the probability that it is within distance 3 from the center?

Obviously, all the points within the inner circle will do (these are the favorable cases). However, there is an infinity of them, and the number of possible points (that is, all the points in the larger circle) is also infinite. The problem is that we cannot divide an infinite number by an infinite number. The solution is clear; what we need to do is to divide the area of the inner circle by that of the outer circle. So,

Pr(point is within distance 3 from center)=9π/16π, that is, 9/16.

In general, when a point is randomly chosen in an interval (length), an area, or a volume, we have a continuous uniform distribution, and the probability of an event is obtained by dividing the favorable length, area, or volume by the total length, area, or volume. Many real life situations can be modeled in this way.

Example 1

You arrive randomly between 1:30 and 1:45. What’s the probability that you’ll arrive before 1:35? Obviously, the favorable length is 5 minutes, while the total length is 15 minutes. Hence

Pr(you arrive before 1:35)=favorable length/total length=5/15=1/3.

Example 2

You arrive randomly between 2:00 and 2:30. What’s the probability that you arrive between 2:15 and 2:20? The favorable length is 5 minutes and the total length is 30 minutes. Hence,

Pr(you arrive between 2:15 and 2:20)= 1/6.

What is the probability you arrive at 2:15? It is zero, because 2:15 has length zero. In this model only intervals of time (lengths) have probability greater than zero.

Example 3

A movie starts at 5:00, 5:15, and 5:30, and that you get to the movie theater at random between 5:00 and 5:30. What is the probability that you’ll have to wait 5 minutes or less before the movie starts? We can model the situation with a length of 30 and note that if you get there between 5:10 and 5:15 you’ll wait 5 minutes or less, and similarly if you get there between 5:25 and 5:30. So, the favorable length is 10 minutes, while the total length is 30 minutes. Hence

Pr(you wait less then 5 minutes)=10/30, that is, 1/3.

To deal with more complex cases, we need to learn how to graph inequalities. Consider the inequality

x-y<2.

In order to graph it, first we graph the corresponding equality

x-y=2.

When x=0, y=-2, and when y=0, x=2. Hence the graph is

The graph of x-y<2 is given by those points in the plane that satisfy it. Obviously, the points on the line satisfy x-y=2.

So, our problem is to determine whether the points above that line or below that line satisfy x-y<2. The easiest way to determine this is by inspection. Consider the point P:(5,0). Obviously, it does not satisfy x-y<2, as 5-0>2. Hence, the points below the line x-y=2 do not satisfy the inequality, and consequently those above it do. In short the graph of the inequality is given by the area above the line x-y=2.

We are now ready to model more complex problems.

Example 4

Joe and Jill agree to meet and both randomly arrive between 10:00 and 10:30. What’s the probability that Jill arrives at least 10 minutes before Joe? Let x be Joe’s arrival time and y be Jill’s arrival time. Then, the probability space of their meeting time is given by the square below, and (x,y) is uniformly distributed in it. In other words, each point is the square has the same probability of being the time at which they meet.

The requirement that Jill arrives at least 10 minutes before Joe is given by

x-y≥10.

So, first we need to determine which part of the square (the probability space) has points satisfying x-y≥10. As before, we draw the line x-y=10, and by inspection we find out that the graph of the inequality is the area below the line x-y=10. Hence, the favorable area of the probability space is given by the triangle below line x-y=10; by simple geometry, its area is F=(20ּ20)/2=200. The area of the total probability space is

T=30ּ30 =900. Hence,

Pr(x-y≥10)=2/9.

Note that if what is picked is one-dimensional (a number, the time of arrival of one person, for example) then the model is one-dimensional, that is, a line. So, if the probability is uniformly distributed within an interval [a,b], then one models what is picked by a variable x such that a≤x≤b and determines the probability by dividing the favorable length by the total length. However, if what is picked is two-dimensional (two numbers, the time of meeting of two persons, for example), then the model is two-dimensional, that is a rectangle, for example. So, if the probability is uniformly distributed within two (possibly identical) intervals [a,b] and [c,d], then one models the event by two numbers x and y such that a≤x≤b and c≤y≤d, and the probability space by the rectangle abcd. The relevant probability is obtained by dividing the favorable area (typically obtained by determining the graph of an inequality) by the total area. More generally, an n-dimensional problem is modeled in an n-dimensional probability space.