Geometric Probability

We know that the probability of an event occurring is how likely it is to occur. We can apply this concept to geometry by asking, "What are the odds a randomly-selected point will be within a particular line or region?" This is called geometric probability. For example, consider line segment PQ:

If point X is placed at random, what are the odds that it will appear somewhere in line segment PR? In this article, we'll explore how to solve problems like that. Let's get started!

Working with geometric probability

The probability of point X being on line segment PR isn't that difficult to figure out. We know we're putting point X somewhere on line segment PQ, so we don't need to concern ourselves with any scenarios where no point X exists. That means that point X will either be on segment PR or segment RQ. However, we cannot assume $\frac{50}{50}$ odds since the two line segments in question aren't the same length. Therefore, we set up a ratio to determine the geometric probability:

Probability of X being on $\mathrm{PR}=\frac{\mathrm{Length\; of\; line\; segment\; PR}}{\mathrm{Length\; of\; line\; segment\; PQ}}$

It's important to read the question carefully when setting up this ratio. If we wanted the odds of point X being on line segment $\angle \mathrm{RQ}$ instead of $\angle \mathrm{PR}$ , we would need to use $\angle \mathrm{RQ}$ in our ratio.

Geometric probability works similarly if we're looking at figures instead of line segments. Consider the image below:

If we place a random point P somewhere in the image above, the geometric probability of it appearing in region S would be:

Probability of P in S $=\frac{\mathrm{Area\; of\; Region\; S}}{\mathrm{Area\; of\; Region\; R}}$

Again, we're assuming that point P is somewhere and ignoring the possibility that it wouldn't be in either region.

Let's look at a sample problem to see how this works in practice. Find the probability that a randomly-placed point will be on line segment AC using the image below:

We can see that segment AC measures 3 units and segment CB measures 9 units. Our formula for geometric probability places the length of AC (3) over the total length of line segment AB, $3+9=12$ , giving us $\frac{3}{12}$ . The numerator and denominator share a common factor of 3, so we can simplify our answer to $\frac{1}{4}$ or 25%.

An example of geometric probability

a. Find the probability that a randomly-placed point will be on line segment CB using the image below:

The image is the same as above, but the question has changed. Now, we're looking for the probability that the point is on segment CB, not AC. We can see that segment CB measures 9 units and the total line segment measures 12 units, so we get $\frac{9}{12}$ . Both the numerator and denominator share a common factor of 3, allowing us to simplify our answer to $\frac{3}{4}$ or 75%.

Topics related to the Geometric Probability

Geometric Mean

Probability Distribution

Using Probabilities to Make Fair Decisions

Flashcards covering the Geometric Probability

Statistics Flashcards

Probability Theory Flashcards

Practice tests covering the Geometric Probability

Probability Theory Practice Tests

Common Core: High School - Statistics and Probability Diagnostic Tests

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