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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.**

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.
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?

External credits: @lifeforstock

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=Length,Area,or Volume of total regionLength,Area,or Volume of success region $

In one-dimensional figures, the probability that a point $S,$ chosen at random from $AD,$ lies on $BC$ — the success region — is the ratio of the length of $BC$ to the length of $AD.$

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 $R.$

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 $C.$

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.{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:1em;vertical-align:-0.25em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.13889em;\">P<\/span><span class=\"mopen\">(<\/span><span class=\"mord\">\ud83c\udf6c<\/span><span class=\"mord text\"><span class=\"mord\">\u00a0<\/span><span class=\"mord Roboto-Regular\">on<\/span><span class=\"mord\">\u00a0<\/span><span class=\"mord Roboto-Regular\">a<\/span><span class=\"mord\">\u00a0<\/span><span class=\"mord Roboto-Regular\">gray<\/span><span class=\"mord\">\u00a0<\/span><span class=\"mord Roboto-Regular\">region<\/span><\/span><span class=\"mclose\">)<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["0.64"]}}

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.

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.
The probability of the candy falling on the gray area of the floor is approximately $0.64$ or $64%.$

$Yellow:Gray: 15⋅2=30ft_{2}18⋅3=54ft_{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=84ft_{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)=Area of total regionArea of success region $

SubstituteValues

Substitute values

$P(🍬on a gray region)=8454 $

ReduceFrac

$ba =b/6a/6 $

$P(🍬on a gray region)=149 $

CalcQuot

Calculate quotient

$P(🍬on a gray region)=0.642857…$

RoundDec

Round to $2$ decimal place(s)

$P(🍬on a gray region)≈0.64$

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 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.{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:1.01025390625em;vertical-align:-0.25em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.13889em;\">P<\/span><span class=\"mopen\">(<\/span><span class=\"mord text\"><span class=\"mord Roboto-Regular\">forest<\/span><\/span><span class=\"mclose\">)<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["0.32"]}}

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:1.01025390625em;vertical-align:-0.25em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.13889em;\">P<\/span><span class=\"mopen\">(<\/span><span class=\"mord text\"><span class=\"mord Roboto-Regular\">field<\/span><\/span><span class=\"mclose\">)<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["0.6"]}}

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

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 | $8$ | $1200$ | $9600∣∣∣∣ 8⋅1200=$ |

Fields | $24$ | $750$ | $18000∣∣∣∣ 24⋅750=$ |

Lakes | $76$ | $3$ | $228∣∣∣∣ 76⋅3=$ |

$30000−228=29772km_{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)=Area of total regionArea of success region $

SubstituteValues

Substitute values

$P(forest)=297729600 $

UseCalc

Use a calculator

$P(forest)=0.322450…$

RoundDec

Round to $2$ decimal place(s)

$P(forest)≈0.32$

$P(field)=Area of total regionArea of success region $

SubstituteValues

Substitute values

$P(field)=2977218000 $

UseCalc

Use a calculator

$P(field)=0.604594…$

RoundDec

Round to $2$ decimal place(s)

$P(field)≈0.60$

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. ### Hint

### Solution

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 $103 ,$ what is the total length of the fence?

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

The geometric probability formula can be used to find the total length of the fence.
*without* the fence. By substituting the length of total region with $170$ and $P$ with $103 ,$ this value can be calculated.
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.

$P=Length of total regionLength of success 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 $P=Length of total regionLength of success region $

SubstituteValues

Substitute values

$103 =170Length of success region $

MultEqn

$LHS⋅170=RHS⋅170$

$103 ⋅170=Length of success region$

MoveRightFacToNum

$ca ⋅b=ca⋅b $

$10510 =Length of success region$

CalcQuot

Calculate quotient

$51=Length of success region$

RearrangeEqn

Rearrange equation

$Length of success region=51$

$Length of Fence 170−51=119ft $

Therefore, the length of the fence is $119$ feet. 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.{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.68333em;vertical-align:0em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.13889em;\">P<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["7\/8"]}}

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.

The experiment uses a sphere, a three-dimensional object, so the geometric probability formula for volume should be used.
*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.
Now the volume of the success outcomes can be found.
Finally, the found volumes can be substituted into the probability formula.
It can be concluded that the probability of the bubble being closer to the outside of the sphere is $87 .$

$P=Volume of total regionVolume of success 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}=34π r_{3} $

Outcomes in which the bubble is situated closer to the $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 $2r .$ This means that the volume of that sphere is as follows.
$V_{center}=34π (2r )_{3}$

Simplify right-hand side

PowQuot

$(ba )_{m}=b_{m}a_{m} $

$V_{center}=34π (2_{3}r_{3} )$

CalcPow

Calculate power

$V_{center}=34π (8r_{3} )$

MultFrac

Multiply fractions

$V_{center}=244πr_{3} $

ReduceFrac

$ba =b/4a/4 $

$V_{center}=6πr_{3} $

$V_{success}=V_{sphere}−V_{center}$

SubstituteII

$V_{sphere}=34π r_{3}$, $V_{center}=6πr_{3} $

$V_{success}=34π r_{3}−6πr_{3} $

Simplify right-hand side

MoveRightFacToNum

$ca ⋅b=ca⋅b $

$V_{success}=34πr_{3} −6πr_{3} $

ExpandFrac

$ba =b⋅2a⋅2 $

$V_{success}=68πr_{3} −6πr_{3} $

SubFrac

Subtract fractions

$V_{success}=67πr_{3} $

$P=Volume of total regionVolume of success region $

Substitute values and evaluate

SubstituteValues

Substitute values

$P=34πr_{3} 67πr_{3} $

DivFracByFracSL

$ba /dc =ba ⋅cd $

$P=67πr_{3} ⋅4πr_{3}3 $

SplitIntoFactors

Split into factors

$P=3⋅27πr_{3} ⋅4πr_{3}3 $

CrossCommonFac

Cross out common factors

$P=3 ⋅27πr_{3} ⋅4πr_{3}3 $

CancelCommonFac

Cancel out common factors

$P=27 ⋅41 $

MultFrac

Multiply fractions

$P=87 $

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. ### Hint

### Solution

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

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:1em;vertical-align:-0.25em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.13889em;\">P<\/span><span class=\"mopen\">(<\/span><span class=\"mord text\"><span class=\"mord Roboto-Regular\">meeting<\/span><\/span><span class=\"mclose\">)<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["11\/36"]}}

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.

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.
### 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 $14:30$ $P.M.$ to $16:00$ $P.M..$
### 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.
Now, the area of the $15-$minute leeway zone can be found.
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.
It can be concluded that the probability that Magdalena and Tiffaniqua will meet at the library is $3611 .$

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=Area of total regionArea of success region $

The area of the total region can be calculated by substituting $90$ for $s$ into the area of a square formula.

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.

$A_{triangle}=21 bh$

Substitute values and evaluate

SubstituteII

$b=75$, $h=75$

$A_{triangle}=21 (75)(75)$

MoveRightFacToNum

$ca ⋅b=ca⋅b $

$A_{triangle}=2(75)(75) $

Multiply

Multiply

$A_{triangle}=25625 $

CalcQuot

Calculate quotient

$A_{triangle}=2812.5$

$A_{success}=A_{total}−2A_{triangle}$

SubstituteII

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

$A_{success}=8100−2(2812.5)$

Multiply

Multiply

$A_{success}=8100−5625$

SubTerm

Subtract term

$A_{success}=2475$

$P=Area of total regionArea of success region $

SubstituteValues

Substitute values

$P=81002475 $

ReduceFrac

$ba =b/25a/25 $

$P=32499 $

ReduceFrac

$ba =b/9a/9 $

$P=3611 $

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.

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b What is the probability of the dart hitting a red or green region? Round the answer to $3$ decimals.

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:1em;vertical-align:-0.25em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.13889em;\">P<\/span><span class=\"mopen\">(<\/span><span class=\"mord text\"><span class=\"mord Roboto-Regular\">red<\/span><span class=\"mord\">\u00a0<\/span><span class=\"mord Roboto-Regular\">or<\/span><span class=\"mord\">\u00a0<\/span><span class=\"mord Roboto-Regular\">green<\/span><\/span><span class=\"mclose\">)<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":[".282"]}}

c What is the probability of the dart hitting a black or white region? Round the answer to $3$ decimals.

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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.

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

$P=Area of total regionArea of success 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 $22 =1$ centimeter.

$r_{C_{1}}=1cm $

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 $C_{2}$ is $2+2=4$ 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}} =2cm=10cm $

Next, by adding the width of the rectangles to the radius of $C_{3},$ it can be concluded that the radius of $C_{4}$ is $11.5$ centimeters.
$r_{C_{4}}=11.5cm $

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 $C_{6}$ is $1.5$ centimeters greater than $r_{C_{5}}.$
$r_{C_{5}}r_{C_{6}} =16cm=17.5cm $

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.
$A_{C_{1}}=πr_{2}$

Substitute

$r=1$

$A_{C_{1}}=π(1)_{2}$

CalcPow

Calculate power

$A_{C_{1}}=π(1)$

MultByOne

$a⋅1=a$

$A_{C_{1}}=π$

Circle | Radius | $πr_{2}$ | Area |
---|---|---|---|

$C_{1}$ | $1$ | $π(1)_{2}$ | $π$ |

$C_{2}$ | $2$ | $π(2)_{2}$ | $4π$ |

$C_{3}$ | $10$ | $π(10)_{2}$ | $100π$ |

$C_{4}$ | $11.5$ | $π(11.5)_{2}$ | $132.25π$ |

$C_{5}$ | $16$ | $π(16)_{2}$ | $256π$ |

$C_{6}$ | $17.5$ | $π(17.5)_{2}$ | $306.25π$ |

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

Since there are $20$ rectangles and they all have the same size, the area of each rectangle can be calculated by dividing $23.25πcm_{2}$ by $20.$
$A_{small rectangle}=2032.25π ⇓A_{small rectangle}=1.6125π $

$10$ of the $20$ rectangles are red, so the total area of the red rectangles is $16.125π$ 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}}=306.25π−256π⇓A_{C_{6}}−A_{C_{5}}=50.25π $

The area of each bigger rectangle can be obtained by dividing this value by $20.$
$A_{big rectangle}=2050.25π ⇓A_{big rectangle}=2.5125π $

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+16.125π+25.125π⇓A_{red}=42.25π $

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)=Area of total regionArea of red region $

SubstituteValues

Substitute values

$P(red)=306.25π42.25π $

CancelCommonFac

Cancel out common factors

$P(red)=306.25π 42.25π $

SimpQuot

Simplify quotient

$P(red)=306.2542.25 $

UseCalc

Use a calculator

$P(red)=0.137959…$

RoundDec

Round to $3$ decimal place(s)

$P(red)≈0.138$

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 $32.25π$ and $50.25π$ square centimeters, respectively. Also, $A_{C_{2}}=4π$ square centimeters is the area of the center circles.

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

The total area of the success region is $86.5π$ square centimeters. The area of total region is the same as before, $306.25π$ square centimeters. Now enough information has been calculated to find the probability of hitting a red or green region.