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Geometric Probability

Use geometric properties to evaluate probability

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Understanding Geometric Probability

Emma and Abby were looking at 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,” Emma’s dad said looking at the article. “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”.

“What are the chances that the pipe is split like that?” Abby asked.

In this concept, you will learn to use geometric probability in problem solving.

Geometric Probability

Geometric probability is the probability associated with a geometric problem. The best way to think about geometric probability is through a real-world situation.

Let’s look at an example.

If a circle with a radius of 10 inches is placed inside a square with a side length of 20 inches, what is the probability that a dart thrown will land inside the circle?

First, find the area of the circle.

\begin{align*}\begin{array}{rcl} A & = & \pi r^2\\ A & = & \pi (10)^2\\ A & = & 314 \ in^2 \end{array}\end{align*}

Next, find the area of the square.

\begin{align*}\begin{array}{rcl} A & = & s^2\\ A & = & 20^2\\ A & = & 400 \end{array}\end{align*}

Then, use the formula for probability to solve the problem.

For this problem a favorable outcome would be hitting the circle. Therefore the number of favorable outcomes would be the area of the circle. The total outcomes would be the area of the square since the square covers the circle.

\begin{align*}\begin{array}{rcl} P (\text{event}) &=& \frac{\# \ \text{of favorable outcomes}}{\text{total} \ \# \ \text{of outcomes}} \\ P (\text{landing in circle}) &=& \frac{A_{\text{circle}}}{A_{\text{square}}} \\ P (\text{landing in circle}) &=& \frac{314}{400} \\ P (\text{landing in circle}) &=& 0.785 \end{array}\end{align*}

The probability of a dart being thrown and hitting the circle is 0.785 or 78.5%.

Let’s look at another 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*} hit a serve straight up in the air as high as he could. The ball landed on the green side of the court in some random location.

What is the probability that the serve landed in the service box \begin{align*}A\end{align*}?

First, 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*}\begin{array}{rcl} Area_{Box \ A} &=& l \times w \\ &=& 21 \ ft \times 13.5 \ ft \\ &=& 283.5 \ ft^2 \end{array}\end{align*}

\begin{align*}\begin{array}{rcl} Area_{Box \ B} &=& l \times w \\ &=& 21 \ ft \times 13.5 \ ft \\ &=& 283.5 \ ft^2 \end{array}\end{align*}

\begin{align*}\begin{array}{rcl} Area_{Box \ C} &=& l \times w \\ &=& 18 \ ft \times 27 \ ft \\ &=& 486 \ ft^2 \end{array}\end{align*}

Next, 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*}\begin{array}{rcl} P (\text{point in section}) &=& \frac{\text{area of favourable section}}{\text{total area}} \\ \\ P (\text{ball lands in Box}\ A) &=& \frac{\text{area of Box}\ A}{\text{area Box} \ A + \text{area Box} \ B + \text{area Box} \ C} \\ \\ &=& \frac{283.5}{283.5 + 283.5 + 486} \\ \\ &=& \frac{283.5}{1053} \\ \\ &=& 0.269 \end{array}\end{align*}

There is a 0.269 or 26.9% chance the ball will land in the service box \begin{align*}A\end{align*}.

You can also use geometric probability with circular regions.

Look at the image below. What is the probability that a random point selected would be in the white circle?

First, calculate the area of the inner white circle and the larger circle.

\begin{align*}\begin{array}{rcl} A_{\text{white circle}} &=& \pi r ^2 \\ &=& \pi (4.5)^2 \\ &=& \pi (20.25) \\ &=& 63.6 \ in^2 \end{array}\end{align*}

\begin{align*}\begin{array}{rcl} A_{\text{large circle}} &=& \pi r^2 \\ &=& \pi (7.5)^2 \\ &=& \pi (56.25) \\ &=& 176.7 \ in^2 \end{array}\end{align*}

Next, to find the probability that a random point will be located in the white circle, just find the ratio of the area of white circle to the area of the entire large circle.

\begin{align*}\begin{array}{rcl} P (\text{point in section}) &=& \frac{\text{area of favorable section}}{\text{total area}} \\ P (\text{point in white circle}) &=& \frac{\text{area of white circle}}{\text{total area}} \\ &=& \frac{63.6}{176.7} \\ &=& 0.3599 \end{array}\end{align*}

There is a 0.3599 or 36% chance the point selected will be in the white circle.

What if you wanted to find the probability of a point being randomly located in the black region.

First, you know the area of the white circle is \begin{align*}63.6 \ in^2\end{align*} and the total area of the circle is \begin{align*}176.7 \ in^2\end{align*}. From these you can find the area of just the black region.

\begin{align*}\begin{array}{rcl} A_{\text{black region}} &=& \text{total area} - A_{\text{white circle}} \\ &=& 176.7 - 63.6 \\ &=& 113.1 \end{array}\end{align*}

Next, calculate the probability of a randomly selected point being found in the black region.

\begin{align*}\begin{array}{rcl} P(\text{black region}) &=& \frac{\text{area of black region}}{\text{total area}} \\ &=& \frac{113.1}{176.7} \\ &=& 0.64 \end{array}\end{align*}

There is a 0.64 or 64% chance the point selected will be in the black region.

Notice that an easy way to find the \begin{align*}P\end{align*} (black) is to recognize that \begin{align*}P\end{align*} (black) and \begin{align*}P\end{align*} (white) 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*}\begin{array}{rcl} P (\text{black}) + P (\text{white}) &=& 100\% \\ P (\text{black}) + 36\% &=& 100\% \\ 64 \% + 36\% &=& 100\% \end{array}\end{align*}

This is a way that you can check your work.

Notice that you will need to use the formula for the area of a circle to determine geometric probability related to area.

Examples

Example 1

Earlier, you were given a problem about the pipe for the bike trick.  Emma’s dad was talking about building the pipe.

First, 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.

Next, 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.

Then, 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*}\begin{array}{rcl} P (> \ 7 \ ft) &=& \frac{\text{length of favorable section}}{\text{total length}} \\ &=& \frac{3 \ ft + 3 \ ft}{10 \ ft} \\ &=& \frac{6}{10} \\ &=& \frac{3}{5} \end{array}\end{align*}

The answer is \begin{align*}\frac{3}{5}\end{align*}.

The probability is \begin{align*}\frac{3}{5}\end{align*} or 60% that one piece of the pipe will be 7 feet long or greater.

Example 2

The Ultra Company displays this giant logo on a downtown billboard. The sides of the green rectangle measure 16 feet by 20 feet, while the sides of the blue square measure 28.2 feet. The logo is lit up by thousands of small high definition LCD pixels to light the area. What is the probability that one of the pixels will be burned out in the blue region?

First, find the area of each section.

\begin{align*}\begin{array}{rcl} A_{\text{blue diamond}} &=& s^2 \\ &=& (28.2)^2 \\ &=& 795.24 \ ft^2 \end{array}\end{align*}

\begin{align*}\begin{array}{rcl} A_{\text{green rectangle}} &=& l \times W \\ &=& 16 \times 20 \\ &=& 320 \ ft^2 \end{array}\end{align*}

Next, find the total area of the 4 equal-sized triangles. This area is equal to the area of the entire figure minus the central green rectangle.

\begin{align*}\begin{array}{rcl} A_{4 \ \text{triangles}} &=& A_{\text{blue diamond}} - A_{\text{green diamond}} \\ A_{4 \ \text{triangles}} &=& 795.24 - 320 \\ A_{4 \ \text{triangles}} &=& 475.24 \ ft^2 \end{array}\end{align*}

Then, find the probability that the next pixel will be burned out in the blue region.

\begin{align*}\begin{array}{rcl} P(\text{blue region}) &=& \frac{\text{area of blue region}}{\text{total area}} \\ &=& \frac{475.24}{795.24} \\ &=& 0.5976 \end{array}\end{align*}

There is a 0.5976 or 59.76% chance the next pixel will burn out in the blue region.

Example 3

Which of the following geometric probabilities is impossible?

1. 0.95
2. 1.25
3. 0.00
4. 1.00

Probabilities always lie between 0 and 1. You can have a 0% probability of snow in the desert and you can have a 100% probability that that there is a sun. Geometric probabilities cannot be greater 1 or be less than 0. Therefore answer b is incorrect.

Example 4

In the figure below, a point is selected at random in the square. What is the probability that this point will be found in the isosceles right triangle \begin{align*}CEF\end{align*}?

First, calculate the area of the yellow square and the green triangle.

\begin{align*}\begin{array}{rcl} A_{\text{square}} &=& s^2 \\ &=& (10)^2 \\ &=& 100 \ in^2 \end{array}\end{align*}

\begin{align*}\begin{array}{rcl} A_{\text{triangle}} &=& \frac{1}{2} \ bh \\ &=& \frac{1}{2} \times 5 \times 5 \\ &=& \frac{1}{2} \times 25 \\ &=& 12.5 \ in^2 \end{array}\end{align*}

Next, to find the probability that a random point will be located in the triangle, just find the ratio of the area of triangle to the area of the square.

\begin{align*}\begin{array}{rcl} P (\text{point in triangle}) &=& \frac{\text{area of triangle}}{\text{total area}} \\ &=& \frac{12.5}{100} \\ &=& 0.125 \end{array}\end{align*}

There is a 0.125 or 12.5% chance the point selected will be in the green triangle.

Example 5

A dart thrown at a line segment \begin{align*}\overline{AC}\end{align*} can land at any point on it. What is the probability that it lands on \begin{align*}\overline{BC}\end{align*}? [Given \begin{align*}a=19, b=9\end{align*}.]

First, find the length of \begin{align*}AC\end{align*}.

\begin{align*}\begin{array}{rcl} \overline{AC} &=& a+b \\ &=& 9 + 19 \\ &=& 28 \end{array}\end{align*}

Next, fill in the probability formula to find the ratio of the length of \begin{align*}BC\end{align*} to the length of \begin{align*}AC\end{align*}.

\begin{align*}\begin{array}{rcl} P (\text{dart landing on} \ \overline{BC}) &=& \frac{\text{length of} \ \overline{BC}}{\text{length of} \ \overline{AC}} \\ &=& \frac{9}{28} \\ &=& 0.321 \end{array}\end{align*}

There is a 0.321 or 32.1% chance the dart landing on \begin{align*}\overline{BC}\end{align*}.

Review

Draw diagrams to solve the problems.

1. 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?

2. In the phone problem above, what is the probability that Geoff dropped his phone during the first mile or the last 2 miles?

3. In the phone problem above, Geoff searched from mile 4.5 to mile 7. What is the probability that he found the phone?

4. 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?

5. In the phone problem above, what range would the tower need to have to be sure that Geoff’s lost phone would ring?

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.

6. What is the probability that the pigeon will land in one of the end zones?

7. What is the probability that the pigeon will land between the goal line and the 20 yard line on either side of the field?

8. What is the probability that the pigeon will land within 5 yards of the 50-yard line?

9. What is the probability that the pigeon will land somewhere on the green part of the field?

Look at the diagram and then answer each question as it is related to geometric probabilities.

10. 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?

11. What is the probability that the dart will land in circle 2?

12. What is the probability that the dart will land in circle 3?

13. What is the probability that the dart will land in circle 4?

14. What is the probability that the dart will land in circle 5?

15. What is the probability that the dart will land in a yellow area?

16. What is the probability that the dart will land in a red area?

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Vocabulary Language: English

Geometric Probability

Probability is calculated through the use of geometry.

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