Unlocking Geometric Probability: the Duet of Probability and Geometry

This lesson is will use various real-life situations to explore the concepts of geometric probability. It will be shown how geometric probability can be calculated for one

-,

two

-,

and three

-

dimensional objects.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Probability Experiment
Probability of an Event
Union
Intersection
Complement
Independent and Dependent Events
Conditional Probability

Challenge

Calculating the Probability of Bumping Into a Person

Magdalena went to a sea resort with her family during her summer vacation. She noticed that when swimming in the pool with an area of

150

square meters, one person occupies approximately

2

square meters in the pool.

External credits: @lifeforstock

If there are already

30

people in the pool, what is the probability of Magdalena bumping into another person when she jumps into the pool?

Discussion

Geometric Probability

The geometric probability of an event is a ratio that involves geometric measures such as length, area, or volume. In geometric probability, points on a line segment, on a plane, or as part of a three-dimensional figure represent outcomes.

$P = \frac{Length , Area , or Volume of success region}{Length , Area , or Volume of total region}$

Probability and Length

In one-dimensional figures, the probability that a point $S,$ chosen at random from $\overline{A D},$ lies on $\overline{B C}$ — the success region — is the ratio of the length of $\overline{B C}$ to the length of $\overline{A D} .$

Probability and Area

In two-dimensional figures, the probability that a point $S,$ chosen at random from a region $R,$ lies in region $N$ — the success region — is the ratio of the area of region $N$ to the area of region $R .$

Triangular Region inside a quadrilateral

Probability and Volume

In three-dimensional figures, the probability that a point $S,$ chosen at random from a solid $C,$ lies inside a solid $B$ — the success region — is the ratio of the volume of solid $B$ to the volume of solid $C .$

Example

Probability of a Candy Falling on a Gray Square

Magdalena decorates a banana-honey cake she baked with candies. One of the candies fell on the floor and started to bounce. The floor consists of yellow and gray squares.

If there are

15

yellow squares of

2

square feet each and

18

gray squares of

3

square feet each, what is the probability of the candy falling on the gray region? Round the answer to two decimal places.

Hint

Calculate the combined areas of the yellow and gray squares of the kitchen floor. Identify the area of the success region and the area of total region.

Solution

It is given that there are

15

yellow squares of

2

square feet each and

18

gray squares of

3

square feet each. Therefore, by calculating the product of

15

and

2,

as well as

18

and

3,

the total areas of yellow and gray regions can be found.

Yellow : Gray : 15 \cdot 2 = 30 ft^{2} 18 \cdot 3 = 54 ft^{2}

The sum of these values is the total area of the kitchen floor. Note that the whole area of the kitchen floor is the sample space of all the possible outcomes where the candy can fall.

Total : 30 + 54 = 84 ft^{2}

Since the probability of a candy falling on gray region should be calculated, the area of success region is the area of gray region. Therefore, the area of success region can be substituted with

54

and the number of total region can be substituted with

84

into the Probability Formula.

P (🍬 on a gray region) = \frac{Area of success region}{Area of total region}

SubstituteValues

Substitute values

P (🍬 on a gray region) = \frac{5 4}{8 4}

ReduceFrac

$\frac{a}{b} = \frac{a / 6}{b / 6}$

P (🍬 on a gray region) = \frac{9}{1 4}

CalcQuot

Calculate quotient

P (🍬 on a gray region) = 0.642857 \dots

RoundDec

Round to $2$ decimal place(s)

P (🍬 on a gray region) \approx 0.64

The probability of the candy falling on the gray area of the floor is approximately

0.64

64 % .

Example

Calculating the Probability of Being Lost in a Forest

Magdalena is exploring a hidden and unknown country and gets lost. She knows that the makeup of the country consists of $8$ great forests of approximately $1200$ square kilometers each, $24$ fields of $750$ square kilometers each, and $76$ lakes of $3$ square kilometers each.

The land of the country with forests, fields and lakes

The total area of the country is

30000

square kilometers. Assuming that she is not in a lake, what is the probability of her being lost in a forest? In a field? Round the answer to two decimals.

Hint

Start by calculating the total area of the forests, fields, and lakes. Then use the Probability Formula.

Solution

First, the total areas of the forests, fields, and lakes should be calculated by multiplying the area of each geographical feature by the number of those features.

Object	Number	Area	Total Area
Forests	$8$	$1200$	$9600 ∣ ∣ ∣ ∣ 8 \cdot 1200 =$
Fields	$24$	$750$	$18000 ∣ ∣ ∣ ∣ 24 \cdot 750 =$
Lakes	$76$	$3$	$228 ∣ ∣ ∣ ∣ 76 \cdot 3 =$

It is given that the total area of the country is

30000

square kilometers. Since it is also given that Magdalena is not in a lake, the area of total region she could be in is the difference between the total area of the country and the total area of lakes.

30000 - 228 = 29772 km^{2}

Magdalena must be somewhere in a

29772

square kilometer area. To calculate the probability of the event of her being lost in a forest, substitute

9600,

the total area of land covered by forests, for the area of success region and

29772

for the area of total region.

P (forest) = \frac{Area of success region}{Area of total region}

SubstituteValues

Substitute values

P (forest) = \frac{9 6 0 0}{2 9 7 7 2}

UseCalc

Use a calculator

P (forest) = 0.322450 \dots

RoundDec

Round to $2$ decimal place(s)

P (forest) \approx 0.32

The probability of Magdalena being lost in a field can be calculated similarly by substituting

18000,

the total land area made up of fields, for the area of success region and the same number for the area of total region.

P (field) = \frac{Area of success region}{Area of total region}

SubstituteValues

Substitute values

P (field) = \frac{1 8 0 0 0}{2 9 7 7 2}

UseCalc

Use a calculator

P (field) = 0.604594 \dots

RoundDec

Round to $2$ decimal place(s)

P (field) \approx 0.60

Therefore, the probability of Magdalena being lost in a forest is approximately

0.32,

32 %,

while the probability of her being lost in a field is about

0.60,

60 % .

Example

Calculating the Probability of Turning Successfully

Earlier geometric probabilities were calculated for two-dimensional problems. This time it will be shown how geometric probability can be used in a one-dimensional problem. Magdalena's friend Tiffaniqua was walking along a

170

foot long path in a park. Portions of the path are bordered by a fence.

External credits: @macrovector-official, @pikisuperstar

In places where the path does not have a fence, Tiffaniqua can go and sit on the grass. She decides to close her eyes and randomly turn. If the probability of Tiffaniqua successfully turning and finding a place to lie on the grass to admire the sky is

\frac{3}{1 0},

what is the total length of the fence?

Hint

Use the formula for the geometric probability of length. Remember what the length of the success region represents.

Solution

The geometric probability formula can be used to find the total length of the fence.

P = \frac{Length of success region}{Length of total region}

Since the probability of the event of turning and successfully finding a place in the grass is given, the length of success region is the length of parts without the fence. By substituting the length of total region with

170

and

P

with

\frac{3}{1 0},

this value can be calculated.

P = \frac{Length of success region}{Length of total region}

SubstituteValues

Substitute values

\frac{3}{1 0} = \frac{Length of success region}{1 7 0}

MultEqn

$LHS \cdot 170 = RHS \cdot 170$

\frac{3}{1 0} \cdot 170 = Length of success region

MoveRightFacToNum

$\frac{a}{c} \cdot b = \frac{a \cdot b}{c}$

\frac{5 1 0}{1 0} = Length of success region

CalcQuot

Calculate quotient

51 = Length of success region

RearrangeEqn

Rearrange equation

Length of success region = 51

It can be concluded that the length of the parts without the fence is

51

feet. The parts of the path without the fence form the sample space of the event of turning successfully. By subtracting

51

from the length of the path, the total length of the fence can be found.

\underline{Length of Fence} 170 - 51 = 119 ft

Therefore, the length of the fence is

119

feet.

Example

Calculating the Probability of Different Positions of a Bubble

Next, the case in which the geometric probability of a three-dimensional object can be calculated will be presented. Two scientists are conducting an experiment in which they place a small bubble of water into a vacuum sphere.

Assuming that the bubble is equally likely to be anywhere within the sphere, what is the probability that it lands closer to the outside of the sphere than its center? Give an exact answer as a fraction in its simplest form.

Hint

Think of how the outcomes in which the bubble is closer to the outside than the center of the sphere can be described. Try to relate them to the radius of the sphere.

Solution

The experiment uses a sphere, a three-dimensional object, so the geometric probability formula for volume should be used.

P = \frac{Volume of success region}{Volume of total region}

The sample space in this case is the whole region inside the sphere. The volume of total region is the same as the volume of the sphere. Let

r

be the radius of the sphere. Its volume is then determined by the following formula.

V_{sphere} = \frac{4 π}{3} r^{3}

Outcomes in which the bubble is situated closer to the outside than the center of the sphere are said to be the success outcomes.

They can be calculated as the difference between the total volume of the sphere and the outcomes in which the bubble is situated closer to the center of the sphere.

V_{success} = V_{sphere} - V_{center}

The bubble is closer to the center if it is in the sphere whose radius is half the radius of the whole sphere. The radius of this smaller sphere is

\frac{r}{2} .

This means that the volume of that sphere is as follows.

V_{center} = \frac{4 π}{3} (\frac{r}{2})^{3}

▼

Simplify right-hand side

PowQuot

$(\frac{a}{b})^{m} = \frac{a ^{m}}{b ^{m}}$

V_{center} = \frac{4 π}{3} (\frac{r ^{3}}{2 ^{3}})

CalcPow

Calculate power

V_{center} = \frac{4 π}{3} (\frac{r ^{3}}{8})

MultFrac

Multiply fractions

V_{center} = \frac{4 π r ^{3}}{2 4}

ReduceFrac

$\frac{a}{b} = \frac{a / 4}{b / 4}$

V_{center} = \frac{π r ^{3}}{6}

Now the volume of the success outcomes can be found.

V_{success} = V_{sphere} - V_{center}

SubstituteII

$V_{sphere} = \frac{4 π}{3} r^{3}$ , $V_{center} = \frac{π r ^{3}}{6}$

V_{success} = \frac{4 π}{3} r^{3} - \frac{π r ^{3}}{6}

▼

Simplify right-hand side

MoveRightFacToNum

$\frac{a}{c} \cdot b = \frac{a \cdot b}{c}$

V_{success} = \frac{4 π r ^{3}}{3} - \frac{π r ^{3}}{6}

ExpandFrac

$\frac{a}{b} = \frac{a \cdot 2}{b \cdot 2}$

V_{success} = \frac{8 π r ^{3}}{6} - \frac{π r ^{3}}{6}

SubFrac

Subtract fractions

V_{success} = \frac{7 π r ^{3}}{6}

Finally, the found volumes can be substituted into the probability formula.

P = \frac{Volume of success region}{Volume of total region}

▼

Substitute values and evaluate

SubstituteValues

Substitute values

P = \frac{\frac{7 π r ^{3}}{6}}{\frac{4 π r ^{3}}{3}}

DivFracByFracSL

$\frac{a}{b} / \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c}$

P = \frac{7 π r ^{3}}{6} \cdot \frac{3}{4 π r ^{3}}

SplitIntoFactors

Split into factors

P = \frac{7 π r ^{3}}{3 \cdot 2} \cdot \frac{3}{4 π r ^{3}}

CrossCommonFac

Cross out common factors

P = \frac{7 π r ^{3}}{3 \cdot 2} \cdot \frac{3}{4 π r ^{3}}

CancelCommonFac

Cancel out common factors

P = \frac{7}{2} \cdot \frac{1}{4}

MultFrac

Multiply fractions

P = \frac{7}{8}

It can be concluded that the probability of the bubble being closer to the outside of the sphere is

\frac{7}{8} .

Example

Calculating the Probability of the Girls Meeting at the Library

Magdalena and Tiffaniqua decided to meet at the school library before going home after school. Since the girls take different classes, they could arrive at two random times between

14 : 30

P . M .

and

16 : 00

P . M .

Magdalena and Tiffaniqua agreed to wait exactly

15

minutes for each other to arrive before leaving.

External credits: @pikisuperstar, @katemangostar

What is the probability that Magdalena and Tiffaniqua will see each other? Give the answer as a fraction in the simplest form.

Hint

Draw a graph in which the $x -$ and $y -$ axes represent Tiffaniqua's and Magdalena's timelines. Think of how the success region can be identified.

Solution

First, draw a graph that represents the possible times in which Magdalena and Tiffaniqua could meet. If Tiffaniqua arrives at

14 : 30

P . M .,

then Magdalena must arrive no later than

14 : 45

P . M . .

If Tiffaniqua arrives at

14 : 31

P . M .,

Magdalena must arrive no later than

14 : 46

P . M .,

and so on.

Therefore, the

15 -

minute leeway shown on the graph represents their opportunity of meeting. The geometric probability formula for area will be used to calculate the probability of both girls being in that

15 -

minute span.

P = \frac{Area of success region}{Area of total region}

Area of the Total Region

The total time available for the girls to meet or a sample space is represented by the square shown in the diagram. Its sides represent the

90 -

minute time span from

14 : 30

P . M .

16 : 00

P . M . .

The area of the total region can be calculated by substituting

90

for

s

into the area of a square formula.

A_{total} = s^{2}

Substitute

$s = 90$

A_{total} = 90^{2}

CalcPow

Calculate power

A_{total} = 8100

Area of the Success Region

Analyzing the diagram, it can be noted that the area of the

15 -

minute time allowance is equal to the difference between the area of the square and the areas of the top and bottom triangles. Also, the areas of the top and bottom triangles are the same, so calculating only one of them would be enough.

As can be seen, these are right triangles whose legs represent

75

minutes. By using the area of a triangle formula, their areas can be calculated.

A_{triangle} = \frac{1}{2} b h

▼

Substitute values and evaluate

SubstituteII

$b = 75$ , $h = 75$

A_{triangle} = \frac{1}{2} (75) (75)

MoveRightFacToNum

$\frac{a}{c} \cdot b = \frac{a \cdot b}{c}$

A_{triangle} = \frac{( 7 5 ) ( 7 5 )}{2}

Multiply

A_{triangle} = \frac{5 6 2 5}{2}

CalcQuot

Calculate quotient

A_{triangle} = 2812.5

Now, the area of the

15 -

minute leeway zone can be found.

A_{success} = A_{total} - 2 A_{triangle}

SubstituteII

$A_{total} = 8100$ , $A_{triangle} = 2812.5$

A_{success} = 8100 - 2 (2812.5)

Multiply

A_{success} = 8100 - 5625

SubTerm

Subtract term

A_{success} = 2475

Therefore, the area of success region is

2475 .

Calculating Probability

Finally, by substituting

2475

for the area of the success region and

8100

for the area of the total region, the probability of the event of the girls meeting can be calculated.

P = \frac{Area of success region}{Area of total region}

SubstituteValues

Substitute values

P = \frac{2 4 7 5}{8 1 0 0}

ReduceFrac

$\frac{a}{b} = \frac{a / 2 5}{b / 2 5}$

P = \frac{9 9}{3 2 4}

ReduceFrac

$\frac{a}{b} = \frac{a / 9}{b / 9}$

P = \frac{1 1}{3 6}

It can be concluded that the probability that Magdalena and Tiffaniqua will meet at the library is

\frac{1 1}{3 6} .

Example

Calculating Probability of Hitting Different Regions on a Dartboard

On a standard dartboard, the diameter of the center red circle is $2$ centimeters and the diameter of the green circle around it is $2$ centimeters greater. Each rectangle has the width of $1.5$ centimeters. Some other lengths are given on the diagram.

A dart is thrown at the dartboard.

a What is the probability of the dart hitting a red region? Round the answer to

3

decimals.

b What is the probability of the dart hitting a red or green region? Round the answer to

3

decimals.

c What is the probability of the dart hitting a black or white region? Round the answer to

3

decimals.

Hint

a Calculate the radii of all of the circles on the dartboard and then use them to find the areas of the circles. What is the total area of all the red regions? Calculate the areas in terms of pi.

b Use the geometric probability formula.

c Note that if a dart did not hit neither red nor green region, then it must have hit white or black region. Use the Complement Rule.

Solution

a In order to find the probability of hitting a red region with a dart, the geometric probability formula can be used.

P = \frac{Area of success region}{Area of total region}

To use the formula, first the areas of the success region — the red region — and total regions should be found. For the purposes of the solution the circles on the board can be named.

First, the radii of all the circles can be found. It is given that the diameter of the small red circle

C_{1}

at the center of the dartboard is

2

centimeters, so its radius is

\frac{2}{2} = 1

centimeter.

r_{C_{1}} = 1 cm

The diameter of the green circle

C_{2}

is said to be

2

centimeters greater than the diameter of

C_{1}

. This implies that the diameter of

C_{2}

2 + 2 = 4

centimeters and its radius is

2

centimeters. It is given that the length between

C_{2}

and the next ring is

8

centimeters, so the radius of

C_{3}

is obtained by adding

2

and

8 .

r_{C_{2}} r_{C_{3}} = 2 cm = 10 cm

Next, by adding the width of the rectangles to the radius of

C_{3},

it can be concluded that the radius of

C_{4}

11.5

centimeters.

r_{C_{4}} = 11.5 cm

The radius of

C_{5}

is equal to the sum of

14

and radius of

C_{2},

which is

2

centimeters. Finally, the radius of

C_{6}

1.5

centimeters greater than

r_{C_{5}} .

r_{C_{5}} r_{C_{6}} = 16 cm = 17.5 cm

Now that all the radii are known, they can be substituted into the circle area formula. First, the area of

C_{1}

can be found. Leave the areas in terms of pi to get a precise result at the end.

A_{C_{1}} = π r^{2}

Substitute

$r = 1$

A_{C_{1}} = π (1)^{2}

CalcPow

Calculate power

A_{C_{1}} = π (1)

MultByOne

$a \cdot 1 = a$

A_{C_{1}} = π

Similarly, the areas of the rest of the circles can be calculated.

Circle	Radius	$π r^{2}$	Area
$C_{1}$	$1$	$π (1)^{2}$	$π$
$C_{2}$	$2$	$π (2)^{2}$	$4 π$
$C_{3}$	$10$	$π (10)^{2}$	$100 π$
$C_{4}$	$11.5$	$π (11.5)^{2}$	$132.25 π$
$C_{5}$	$16$	$π (16)^{2}$	$256 π$
$C_{6}$	$17.5$	$π (17.5)^{2}$	$306.25 π$

By calculating the difference between the areas of

C_{4}

and

C_{3},

the total area of smaller red and green rectangles can be found.

A_{C_{4}} - A_{C_{3}} = 132.25 π - 100 π ⇓ A_{C_{4}} - A_{C_{3}} = 32.25 π

Since there are

20

rectangles and they all have the same size, the area of each rectangle can be calculated by dividing

23.25 π cm^{2}

20 .

A_{small rectangle} = \frac{3 2 . 2 5 π}{2 0} ⇓ A_{small rectangle} = 1.6125 π

10

of the

20

rectangles are red, so the total area of the red rectangles is

16.125 π

square centimeters. Finally, the total area of the greater green and red rectangles can be found as the difference between the areas of

C_{6}

and

C_{5} .

A_{C_{6}} - A_{C_{5}} = 306.25 π - 256 π ⇓ A_{C_{6}} - A_{C_{5}} = 50.25 π

The area of each bigger rectangle can be obtained by dividing this value by

20 .

A_{big rectangle} = \frac{5 0 . 2 5 π}{2 0} ⇓ A_{big rectangle} = 2.5125 π

Just as in case with smaller rectangles, there are

10

bigger red rectangles, so the total area of the bigger red rectangles is

25.125

square centimeters. The total red region on the board consists of

1

small

circle,

10

small

rectangles

and

10

big

rectangles .

Therefore, the total area of the red region is the sum of these areas.

A_{red} = π + 16.125 π + 25.125 π ⇓ A_{red} = 42.25 π

The area of the success region was found to be

42.25

square centimeters. The area of the total region is the area of the largest circle

C_{6} .

With this information in mind, the probability of the dart hitting a red region can be found.

P (red) = \frac{Area of red region}{Area of total region}

SubstituteValues

Substitute values

P (red) = \frac{4 2 . 2 5 π}{3 0 6 . 2 5 π}

CancelCommonFac

Cancel out common factors

P (red) = \frac{4 2 . 2 5 π}{3 0 6 . 2 5 π}

SimpQuot

Simplify quotient

P (red) = \frac{4 2 . 2 5}{3 0 6 . 2 5}

UseCalc

Use a calculator

P (red) = 0.137959 \dots

RoundDec

Round to $3$ decimal place(s)

P (red) \approx 0.138

The probability of the dart hitting a red region is about

0.138,

13.8 % .

b Since the dart can hit the red or the green regions, the success region is the total area of both colors. Earlier it was found that the total areas of the small and big rectangles are

32.25 π

and

50.25 π

square centimeters, respectively. Also,

A_{C_{2}} = 4 π

square centimeters is the area of the center circles.

A_{red and green} = 32.25 π + 50.25 π + 4 π ⇓ A_{red and green} = 86.5 π

The total area of the success region is

86.5 π

square centimeters. The area of total region is the same as before,

306.25 π

square centimeters. Now enough information has been calculated to find the probability of hitting a red or green region.

P (red or green) = \frac{Area of red and green regions}{Area of total region}

SubstituteValues

Substitute values

P (red or green) = \frac{8 6 . 5 π}{3 0 6 . 2 5 π}

CancelCommonFac

Cancel out common factors

P (red or green) = \frac{8 6 . 5 π}{3 0 6 . 2 5 π}

SimpQuot

Simplify quotient

P (red or green) = \frac{8 6 . 5}{3 0 6 . 2 5}

UseCalc

Use a calculator

P (red or green) = 0.282448 \dots

RoundDec

Round to $3$ decimal place(s)

P (red or green) \approx 0.282

The probability of hitting a red or green region is approximately

0.282,

28.2 % .

c Note that if a dart did not hit neither red nor green region, then it must have hit white or black region. In other words, hitting a white or black region is a complement of the event of hitting red or green region. Therefore, using the Complement Rule, the probability of hitting white or black region can be found.

P (black or white) = 1 - P (red or green)

Substitute

$P (red or green) = 0.282$

P (black or white) = 1 - 0.282

SubTerm

Subtract term

P (black or white) = 0.718

The probability of hitting a black or white region is approximately

0.718,

71.8 % .

Closure

Finding Probability Using the Values of Area

Magdalena went to a sea resort with her family during her summer vacation. She noticed that when swimming in the pool with an area of $150$ square meters, one person occupies approximately $2$ square meters in the pool.

External credits: @lifeforstock

If there are already

30

people in the pool, what is the probability of the event of Magdalena bumping into another person when she jumps into the pool? Give the answer as a fraction in the simplest form.

Hint

Use the Probability Formula. What is the area of the success region?

Solution

It is given that there are

30

people in the pool, each occupying

2

square meters. By multiplying these values, the area occupied by all the people in the pool can be calculated.

30 \cdot 2 = 60

The area of the pool occupied by other swimmers is

60

square meters. The Probability Formula can be used to calculate the probability of Magdalena bumping into another person. The area of the success region and the area of the total region should be substituted with

60

and

150,

respectively.

P = \frac{Area of success region}{Area of total region}

SubstituteValues

Substitute values

P = \frac{6 0}{1 5 0}

ReduceFrac

$\frac{a}{b} = \frac{a / 3 0}{b / 3 0}$

P = \frac{2}{5}

It can be concluded that the probability of Magdalena bumping into a person in the pool is

\frac{2}{5} .

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Catch-Up and Review

Probability and Length

Probability and Area

Probability and Volume

Hint

Solution

Hint

Solution

Hint

Solution

Hint

Solution

Hint

Solution

Area of the Total Region

Area of the Success Region

Calculating Probability

Hint

Solution

Hint

Solution