Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

Geometric Probability

Date: 05/23/2001 at 12:38:29
From: Olivia
Subject: geometric probability

I am supposed to figure out the probability of an arrow hitting the
bull's-eye of a target.  I know that the target is a circle that is 3
feet in diameter and the bull's-eye is 6 inches in diameter.  The
problem also says that the arrow is equally likely to hit any spot on
the target.  How do I solve this?

Date: 05/23/2001 at 12:41:55
From: Doctor Annie
Subject: Re: geometric probability

Hi, Olivia.  

Let's start with the basic idea of probability, which is
used to calculate the "likelihood" that something will happen in a
given situation. We find this by dividing the number of "favorable"
outcomes, meaning what you want to happen, by the number of total
outcomes, meaning all of the things that might happen.

                       number of favorable outcomes
        probability = ------------------------------
                        number of possible outcomes

For example, if we want to know the probability of rolling a 5 with a
standard six-sided die, we know that there is only one favorable
outcome (5), while there are 6 total outcomes (one for each side of
the die). This means that the probability of rolling a 5 is 1/6. If we
wanted to know the probability of rolling an odd number, we first
count the favorable outcomes - 1, 3, or 5. There are 3. There are
still 6 total possible outcomes, so the probability of rolling an odd
number is 3/6, or 1/2.

(If you don't know much about probability and want to learn more,
check out the Dr. Math Introduction to Probability at

  http://mathforum.org/dr.math/faq/faq.prob.intro.html .) 

"Geometric probability", which is what your target problem is about,
is exactly the same idea, except that we are dealing with the areas of
regions instead of the "number" of outcomes. The equation becomes

                       area of favorable region
        probability = ------------------------------
                         area of total region

A typical problem might be this: If you are throwing a dart at the
rectangular target below and are equally likely to hit any point on
the target, what is the probability that you will hit the small square?

                        25 cm
           |                               |
           |        5 cm                   |
           |       -----                   |
           |      |     |                  | 10 cm
           |      |     |                  |
           |       -----                   |
           |                               |

To solve this, we need to find the area of the favorable region, which
is the small square, and the area of the total region, which is the

    * The area of the square is (5 cm)^2, or 25 cm^2.

    * The area of the rectangle is 10 cm * 25 cm, or 250 cm^2.

                     favorable      25 cm^2
    * probability = ----------- = ---------- = 1/10.
                       total       250 cm^2

This means that there is a 1 in 10 chance that a dart thrown at the
rectangle will hit the small square.

Try using these ideas to tackle your problem, and be sure to write
back if you need more help!

- Doctor Annie, The Math Forum
Associated Topics:
Middle School Probability
Middle School Two-Dimensional Geometry

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.