11.7: Geometric Probability
Introduction
The Bike Trick
“Look at this,” Carey said showing Telly a picture of a stunt bike rider.
The picture was of a bicyclist riding down a piece of pipe. In fact, the pipe had been split into two separate pieces and the bicyclist was able to ride on both pieces of the pipe.
“I made one of those once,” Ms. Kelley said looking at the article.
“How did you do that?” Telly asked.
“Well, you start with a 10 foot piece of pipe. Then you metal saw it into two separate pieces. You hope that your work is accurate and that you end up with a piece that is about 7 feet or longer. That way you can really ride down before the quick turn,” Ms. Kelly said walking away.
“What are the chances that the pipe is split like that?” Telly asked.
This is another place for probability. This is geometric probability though, so pay close attention and you will be able to figure out this problem.
What You Will Learn
By the end of this lesson, you will understand the following concepts.
- Recognize probability of a point in the area of a specified region as a ratio compared to the area of the entire region.
- Find geometric probabilities involving rectangular regions.
- Find geometric probabilities involving circular regions.
- Make predictions involving geometric probability.
Teaching Time
I. Recognize Probability of a Point in One Area of a Specific Region as a Ratio Compared to the Area of the Entire Region
The best way to begin thinking about geometric probability is to look at an example. Let’s start there.
Example
A bus is traveling on Four-Color Road through the towns of Greenville, Red Hook, Yellow Town, and Mt. Blue. Each town takes up exactly 1 mile of distance on the Four-Color Road.
If a rider gets on the bus at some random point along the 4-mile route, what is the probability that she will board the bus in Greenville?
Let’s think about how we can solve this problem. To solve it, imagine that not a single rider gets on the bus, but 100 riders. If each rider enters at some random point, you would expect to collect the riders to be evenly distributed over the 4 different towns.
Day | Greenville | Red Hook | Yellow town | Mt.blue |
---|---|---|---|---|
expected riders | 25 | 25 | 25 | 25 |
Of course, in real life, the data might be slightly different, but overall you would expect 25 of 100, or \begin{align*}\frac{1}{4}\end{align*}
\begin{align*}P(\text{Greenville}) = \frac{1}{4}\end{align*}
This is because there is one spot out of four possible points that a person could be picked up on the bus.
In general:
The probability of a randomly selected point to be located in a given “favorable” section of a distance is equal to the ratio of the length of the favorable section to the entire distance.
\begin{align*}P (\text{point in section}) &= \frac{\text{Length of ``favorable'' section}}{\text {Total distance}} \end{align*}
Let’s look at another problem where we will use the same diagram of the four color towns.
Example
In the bus problem above, what is the probability that a rider will randomly board the bus in Red Hook or Yellow Town?
To solve this problem represent the rider as a point that can appear anywhere along the route.
\begin{align*}P (\text{Red Hook or Yellow Town}) &= \frac{\text{length of ``favorable'' section}}{\text {total distance}} \\
&= \frac{1 \ mile + 1 \ mile}{4 \ miles} \\
&= \frac{1}{2}\end{align*}
You would expect the rider to board the bus in Red Hook or Yellow Town about \begin{align*}\frac{1}{2}\end{align*}
These two examples are related to geometric probability-space and possible outcomes. The space of the road is four roads. There are four possible outcomes within the four roads.
II. Find Geometric Probabilities Involving Rectangular Regions
If we think about geometric probability and rectangles, we can use the following example to help us understand this concept.
Example
As a joke, a tennis player on the left side of the court who was attempting to place his serve into service box \begin{align*}A\end{align*}
What is the probability that the serve landed in the service box \begin{align*}A\end{align*}
Think back to the information that you just learned in the last section. In this section, you learned how to determine where one point can be randomly located within other given sections. Here we have possible outcomes and we have our favorable outcome identified. It is identified physically in space and not just in number form.
\begin{align*}P (\text{point in section}) &= \frac{\text{length of ``favorable'' section}}{\text{total distance}} \end{align*}
A similar rule applies for area.
The probability of a randomly selected point to be located in a section of an area is equal to the ratio of the area of the “favorable” section to the entire area.
\begin{align*}P (\text{point in section}) &= \frac{\mathbf{area} \ \text{of ``favorable'' section}}{\text{total} \ \mathbf{area}}\end{align*}
Write this probability ratio down in your notebook.
To find the probability that the ball will land in the box, just find the ratio of the area of box \begin{align*}A\end{align*} to the area of the entire green side of the court.
\begin{align*}P (\text{ball lands in box}) &= \frac{\text{area of favorable section}}{\text{total area}} \\ &= \frac{\text{area of box}\ A}{\text{area box} \ A + \text{box} \ B + \text{box} \ C}\end{align*}
Now calculate the area of box \begin{align*}A\end{align*}, box \begin{align*}B\end{align*}, and box \begin{align*}C\end{align*}. You can look at the diagram to determine this. Remember that the formula for the area of a rectangle is \begin{align*}A = lw\end{align*} and the units are measured in square units. In this case, it will be square feet.
\begin{align*} \text{Box} \ A &= 21 \ ft \cdot 13.5 \ ft \\ &= 283.5 \ sq \ ft \\ \text{Box} \ B &= 21 \ ft \cdot 13.5 \ ft \\ &= 283.5 \ sq \ ft \\ \text{Box} \ C &= 18 \ ft \cdot 27 \ ft \\ &= 486 \ sq \ ft \end{align*}
So:
\begin{align*}P(\text{ball lands in box}) &= \frac{\text{area of ``favorable'' section}}{\text{total area}}\\ &= \frac{\text{area} \ A}{\text{area} \ B + \text{area} \ B + \text{area} \ C} \\ &= \frac{283.5}{1053} \\ &= 26.9\% \end{align*}
The ball has a little more than a 1 in 4 chance of landing in the service box.
Good question. We got that because 25% would be the same as one-quarter or \begin{align*}\frac{1}{4}\end{align*}. Here we have a final percentage of 26.9%, so it is a little greater than 25%. We can therefore say that there is a little greater than a 1 in 4 chance of the ball landing in the service box.
III. Find Geometric Probabilities Involving Circular Regions
In the last section, we looked at geometric probability involving rectangular regions. To solve these problems, we used the formula for the area of a rectangle and then we used information tha we have already learned about ratios and percentages to determine our outcomes. We can also work with circular regions. This lesson will teach you how to figure out the geometric probability of circular regions. Let’s look at an example.
Example
The probability for a random point being located in the white circle is therefore:
\begin{align*}P(\text{white}) &= \frac{\text{Area of white}}{\text{Total area}}\end{align*}
Use the formula for the area of a circle, \begin{align*}A = \pi r^2\end{align*} to find the area of the white circle and the total area of the figure.
\begin{align*}A (\text{white}) &= \pi r^2 \\ &= (3.14) \cdot (4.5) \cdot(4.5) \\ &= 63.6 \ sq \ in \\ A \text{(large)} &= \pi r^2 \\ &= (3.14) \cdot(7.5) \cdot (7.5) \\ &= 176.6 \ sq \ in \end{align*}
So:
\begin{align*}P (\text{white}) &= \frac{\text{area of white}}{\text{total area}} \\ &= \frac{63.6}{176.6} \\ &= 36\% \end{align*}
To find the probability of a point being randomly located in the black region, first find the area of the black region.
\begin{align*}A (\text{black}) &= \text{total area} - \text{white area} \\ &= 176.6 - 63.6 \\ &= 113.0 \end{align*}
So:
\begin{align*}P (\text{black}) &= \frac{\text{area of black}}{\text{total area}} \\ &= \frac{113.0}{176.6} \\ &= 64\% \end{align*}
Notice that an easy way to find the \begin{align*}P(\text{black})\end{align*} is to recognize that \begin{align*}P(\text{black})\end{align*} and \begin{align*}P(\text{white})\end{align*} are complementary events.
The point must be either in the black area or the white area, so the two probabilities must add up to 100 percent.
\begin{align*}P (\text{black}) + P (\text{white}) &= 100\% \\ P(\text{black}) + 36\% &= 100\% \\ {\color{red}64\%} + 36\% &= 100\% \end{align*}
This is a way that we can check our work. Notice that you will need to use the formula for the area of a circle to determine geometric probability related to area.
IV. Make Predictions Involving Geometric Probability
Now that you understand geometric probability, we can also make predictions involving geometric probability. This will help us to understand how to work with geometric probability. Let’s look at an example.
Example
The Ultra Company displays this giant logo on a downtown billboard. The logo is lit up by thousands of small high definition LCD pixels. The pixels are often damaged or burn out. Predict where the next 60 damaged pixels will be located.
Before making predictions, we must figure out the probability of each section. To find the probability of where a damaged pixel will be located, first find the area of each section. As the diagram shows, each side of green square measures 20 feet, while each side of the blue diamond measures 28.2 feet
\begin{align*}A (\text{green}) &= 20 \cdot 20 \\ &= 400 \ sq \ ft \\ A (\text{diamond}) &= 28.2 \cdot 28.2 \\ &= 795.2 \ sq \ ft \end{align*}
Now the total area of all 4 equal-sized triangles is equal to the area of the entire figure minus the central green square.
\begin{align*}A (\text{diamond}) - A (\text{green}) &= A (4 \ \text{triangles}) \\ 795.2 - 400 &= 395.2 \ sq \ ft \end{align*}
Since there are 4 equal-sized triangles, each triangle has the following areas.
\begin{align*} \text{Area of all triangles} \div 4 &= \text{area of single triangle} \\ 395.4 \div 4 &= 98.9 \ sq \ ft \end{align*}
So:
\begin{align*}A \text{(red)} &= 98.9 \ sq \ ft \\ A \text{(blue)} &= 3 \cdot 98.9\\ &= 296.6 \ sq \ ft \end{align*}
To find the probability of each area:
\begin{align*}P (\text{green}) &= \frac{\text{area of green}}{\text{area of diamond}} \\ &= \frac{400}{795.2} \\ &= 50.2\% \\ \\ P (\text{blue}) \ &= \frac{\text{area of blue}}{\text{area of diamond}} \\ &= \frac{296.6}{795.2} \\ &= 37.3\%\\ \\ P \text{(red)} \ \ &= \frac{\text{area of red}}{\text{area of diamond}} \\ &= \frac{98.9}{795.2} \\ &= 12.5\% \end{align*}
Now that we have all of the probabilities figured out, we can make predictions. Remember that before you make a prediction, you will need to figure out the probability first. The probability will help you to figure out each prediction.
Problem: Predict where the next 64 damaged pixels will be located in the figure above.
Step 1: Find the probabilities. (You already found them above.)
\begin{align*}P (\text{green}) &= 50.2\% \\ P (\text{blue}) &= 37.3\% \\ P (\text{green}) &= 12.5\% \end{align*}
Step 2: Multiply each probability by the number of events. In this case, the total number of events is 64, the number of damaged pixels. Round off where necessary
\begin{align*} \text{green} &= 0.502 \cdot 64 \\ &= 32 \\ \text{blue} &= 0.373 \cdot 64 \\ &= 24 \\ \text{red} &= 0.125 \cdot 64 \\ &= 8 \end{align*}
This is the answer.
Real-Life Example Completed
The Bike Trick
Here is the problem from the introduction. Use the information to figure out the probability that one piece of the pipe is greater than or equal to 7 feet in length.
“Look at this,” Carey said showing Telly a picture of a stunt bike rider.
The picture was of a bicyclist riding down a piece of pipe. In fact, the pipe had been split into two separate pieces and the bicyclist was able to ride on both pieces of the pipe.
“I made one of those once,” Ms. Kelley said looking at the article.
“How did you do that?” Telly asked.
“Well, you start with a 10 foot piece of pipe. Then you metal saw it into two separate pieces. You hope that your work is accurate and that you end up with a piece that is about 7 feet or longer.That way you can really ride down before the quick turn,” Ms. Kelly said walking away.
“What are the chances that the pipe is split like that?” Telly asked.
Now work through the solution.
Solution to Real – Life Example
Step 1: The minimum size for the larger piece would be 7 feet. That would make the smaller piece 3 feet in length. Mark this off on your diagram.
Step 2: Now notice that the pipe can also break on the other side. This is shown in red.
With 3 feet on either end, that leaves a 4 foot length in the center of the pipe.
Step 3: Now think of the pipe as separate areas. The shaded areas show regions where one of the pieces will be greater than 7 feet in length. These are your favorable sections.
So the probability of breaking into a piece that is 7 feet in length or greater is:
\begin{align*}P (> 7 \ ft) &= \frac{\text{length of favorable sections}}{\text{total distance}} \\ &= \frac{3 \ ft + 3 \ ft}{10 \ ft} \\ &= \frac{6}{10} \\ &= \frac{3}{5}\end{align*}
The probability is 60% that one piece of the pipe will be 7 feet long or greater.
Time to Practice
Directions: Draw diagrams to solve the problems.
- Geoff rode his bike along an 8 mile path and lost his cell phone at some random location somewhere along the way. What is the probability that Geoff’s phone dropped during the first mile of the path?
- In the phone problem above, what is the probability that Geoff dropped his phone during the first mile or the last 2 miles?
- In the phone problem above, Geoff searched from mile 4.5 to mile 7. What is the phone probability that he found the phone?
- In the phone problem above, a cell phone tower is located at mile 4 in the exact center of the path. The tower has a range of 2.75 miles. If Geoff uses a second cell phone to call his lost phone, what is the probability that the lost phone will ring?
- In the phone problem above, what range would the tower need to have to be sure that Geoff’s lost phone would ring?
Directions: Use this information on the football field to answer the following questions.
A football field is 120 yards long – 100 yards (green) plus two end zones (shown in red and blue) and 53 yards wide. The two hash mark lines that run across the center of the field are 13 yards apart and 20 yards from the sidelines. A pigeon flies over the stadium and lands at some random location on the football field.
- What is the probability that the pigeon will land in one of the end zones?
- What is the probability that the pigeon will land between the goal line and the 20 yard line on either side of the field?
- What is the probability that the pigeon will land within 5 yards of the 50-yard line?
- What is the probability that the pigeon will land between the two hashmarks somewhere on the green part of the field?
- What is the probability that the pigeon will land somewhere on the green part of the field outside of the hashmarks (not between both hashmarks)?
- What is the probability that the pigeon will land between the two hashmarks and between the two 40-yard lines?
Directions: Look at the diagram and then answer each question as it is related to geometric probabilities.
- The radius of circle 1 (the inner-most yellow circle) is 1 meter. Each radius thereafter increases by 1 m, as shown. What is the probability of a randomly thrown dart landing on circle 1?
- What is the probability that the dart will land in circle 2?
- What is the probability that the dart will land in circle 3?
- What is the probability that the dart will land in circle 4?
- What is the probability that the dart will land in circle 5?
- What pattern describes how the probability changes for each circle?
- What is the probability that the dart will land in a yellow area?
- What is the probability that the dart will land in an orange area?
Directions: Now use what you have learned to make predictions.
Measurements of the be green, not mean sign are:
\begin{align*}\text{black circle radius} &= 13.8 \ ft \\ \text{pink circle radius} &= 10.0 \ ft \\ \text{white square side} &= 20.0 \ ft \\ \text{green square side} &= 14.1 \ ft \end{align*}
- Of the next 240 pixels in the be green, not mean sign to be damaged, predict how many will be green.
- Of the next 500 pixels in the be green, not mean sign to be damaged, predict how many will be white.
- Of the next 60 pixels in the be green, not mean sign to be damaged, predict how many will be black.
- Of the next 200 pixels in the be green, not mean sign to be damaged, predict how many will be black or white.
- Of the next 81 pixels in the be green, not mean sign to be damaged, predict how many will not be green.