5. Probability in Length, Area, and Volume
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Chapter 6
5.
Probability in Length, Area, and Volume
| Student Learning Objectives: |
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Why
Geometric probability merges the realms of geometry and probability to address tangible situations. This study delves into how points on line segments, planes, or within three-dimensional figures can represent potential outcomes. By examining scenarios like the likelihood of a candy landing on a specific floor tile or the chances of two friends meeting at a library within a given time frame, one can grasp the practical applications of geometric probability. Whether it's determining the odds of a dart hitting a particular region on a dartboard or calculating the probability of being lost in a vast landscape, geometric probability offers a unique lens to view and solve these problems. It's not just about numbers; it's about understanding the spatial relationships and dimensions that influence outcomes.
Lesson Settings & Tools
| | 10 Theory slides |
| | 11 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Probability in Length, Area, and Volume
Slide of 10
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.
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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
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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= 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 AD, lies on BC — the success region — is the ratio of the length of BC to the length of AD.
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.
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.
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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.
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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:& 15* 2=30 ft^2 Gray:& 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=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)=Area of success region/Area of total region
SubstituteValues
Substitute values
P(🍬on a gray region)=54/84
ReduceFrac
a/b=.a /6./.b /6.
P(🍬on a gray region)=9/14
CalcQuot
Calculate quotient
P(🍬on a gray region)=0.642857...
RoundDec
Round to 2 decimal place(s)
P(🍬on a gray region)≈ 0.64
The probability of the candy falling on the gray area of the floor is approximately 0.64 or 64 %.
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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 total area of the country is 30 000 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.
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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 | 8* 1200= 9600 |
| Fields | 24 | 750 | 24* 750= 18 000 |
| Lakes | 76 | 3 | 76* 3= 228 |
It is given that the total area of the country is 30 000 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. 30 000-228=29 772 km^2 Magdalena must be somewhere in a 29 772 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 29 772 for the area of total region.
P(forest)=Area of success region/Area of total region
SubstituteValues
Substitute values
P(forest)=9600/29 772
UseCalc
Use a calculator
P(forest)=0.322450...
RoundDec
Round to 2 decimal place(s)
P(forest)≈ 0.32
The probability of Magdalena being lost in a field can be calculated similarly by substituting 18 000, 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)=Area of success region/Area of total region
SubstituteValues
Substitute values
P(field)=18 000/29 772
UseCalc
Use a calculator
P(field)=0.604594...
RoundDec
Round to 2 decimal place(s)
P(field)≈ 0.60
Therefore, the probability of Magdalena being lost in a forest is approximately 0.32, or 32 %, while the probability of her being lost in a field is about 0.60, or 60 %.
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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.
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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=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 310, this value can be calculated.
P=Length of success region/Length of total region
SubstituteValues
Substitute values
3/10=Length of success region/170
MultEqn
LHS * 170=RHS* 170
3/10* 170=Length of success region
MoveRightFacToNum
a/c* b = a* b/c
510/10=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. Length of Fence 170-51=119 ft Therefore, the length of the fence is 119 feet.
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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.
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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=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.
\begin{gathered}
V_\text{sphere}=\dfrac{4\pi}{3}r^3
\end{gathered}
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. \begin{gathered} V_\text{success}=V_\text{sphere}-V_\text{center} \end{gathered} 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 r2. This means that the volume of that sphere is as follows.
V_\text{center}=\dfrac{4\pi}{3}\bigg(\dfrac{r}{2}\bigg)^3
▼
Simplify right-hand side
PowQuot
(a/b)^m=a^m/b^m
V_\text{center}=\dfrac{4\pi}{3}\bigg(\dfrac{r^3}{2^3}\bigg)
CalcPow
Calculate power
V_\text{center}=\dfrac{4\pi}{3}\bigg(\dfrac{r^3}{8}\bigg)
MultFrac
Multiply fractions
V_\text{center}=\dfrac{4\pi r^3}{24}
ReduceFrac
a/b=.a /4./.b /4.
V_\text{center}=\dfrac{\pi r^3}{6}
Now the volume of the success outcomes can be found.
V_\text{success}=V_\text{sphere}-V_\text{center}
SubstituteII
V_\text{sphere}={\color{#0000FF}{\dfrac{4\pi}{3}r^3}}, V_\text{center}={\color{#009600}{\dfrac{\pi r^3}{6}}}
V_\text{success}={\color{#0000FF}{\dfrac{4\pi}{3}r^3}}-{\color{#009600}{\dfrac{\pi r^3}{6}}}
▼
Simplify right-hand side
MoveRightFacToNum
a/c* b = a* b/c
V_\text{success}=\dfrac{4\pi r^3}{3}-\dfrac{\pi r^3}{6}
ExpandFrac
a/b=a * 2/b * 2
V_\text{success}=\dfrac{8\pi r^3}{6}-\dfrac{\pi r^3}{6}
SubFrac
Subtract fractions
V_\text{success}={\color{#FF0000}{\dfrac{7\pi r^3}{6}}}
Finally, the found volumes can be substituted into the probability formula.
P=Volume of success region/Volume of total region
▼
Substitute values and evaluate
SubstituteValues
Substitute values
P=7π r^36/4π r^33
DivFracByFracSL
.a/b /c/d.=a/b*d/c
P=7π r^3/6* 3/4π r^3
SplitIntoFactors
Split into factors
P=7π r^3/3* 2* 3/4π r^3
CrossCommonFac
Cross out common factors
P=7 π r^3/3* 2* 3/4 π r^3
CancelCommonFac
Cancel out common factors
P=7/2* 1/4
MultFrac
Multiply fractions
P=7/8
It can be concluded that the probability of the bubble being closer to the outside of the sphere is 78.
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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.
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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=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. to 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_\text{total}=s^2
Substitute
s= 90
A_\text{total}={\color{#0000FF}{90}}^2
CalcPow
Calculate power
A_\text{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_\text{triangle}=\dfrac{1}{2}bh
▼
Substitute values and evaluate
SubstituteII
b= 75, h= 75
A_\text{triangle}=\dfrac{1}{2}({\color{#0000FF}{75}})({\color{#009600}{75}})
MoveRightFacToNum
a/c* b = a* b/c
A_\text{triangle}=\dfrac{(75)(75)}{2}
Multiply
Multiply
A_\text{triangle}=\dfrac{5625}{2}
CalcQuot
Calculate quotient
A_\text{triangle}=2812.5
Now, the area of the 15-minute leeway zone can be found.
A_\text{success}=A_\text{total}-2A_\text{triangle}
SubstituteII
A_\text{total}={\color{#0000FF}{8100}}, A_\text{triangle}={\color{#009600}{2812.5}}
A_\text{success}={\color{#0000FF}{8100}}-2({\color{#009600}{2812.5}})
Multiply
Multiply
A_\text{success}=8100-5625
SubTerm
Subtract term
A_\text{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=Area of success region/Area of total region
SubstituteValues
Substitute values
P=2475/8100
ReduceFrac
a/b=.a /25./.b /25.
P=99/324
ReduceFrac
a/b=.a /9./.b /9.
P=11/36
It can be concluded that the probability that Magdalena and Tiffaniqua will meet at the library is 1136.
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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.
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b What is the probability of the dart hitting a red or green region? Round the answer to 3 decimals.
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c What is the probability of the dart hitting a black or white region? Round the answer to 3 decimals.
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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=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 22=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 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)&=2 cm r_(C_3)&=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 is 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 is 1.5 centimeters greater than r_(C_5). r_(C_5)&=16 cm r_(C_6)&=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 * 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 | 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 |
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)=132.25π-100π ⇓ 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 23.25π cm^2 by 20. \begin{gathered} A_\text{small rectangle}=\dfrac{32.25\pi}{20}\\[0.5em] \Downarrow\\ A_\text{small rectangle}= 1.6125\pi \end{gathered} 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. \begin{gathered} A_\text{big rectangle} =\dfrac{50.25\pi}{20} \\[0.5em] \Downarrow \\ A_\text{big rectangle}=2.5125\pi \end{gathered} 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. \begin{gathered} A_\text{red}={\color{#0000FF}{\pi}}+{\color{#FD9000}{16.125\pi}}+{\color{#A800DD}{25.125\pi}}\\ \Downarrow\\ A_\text{red}=42.25\pi \end{gathered} 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 red region/Area of total region
SubstituteValues
Substitute values
P(red)=42.25π/306.25π
CancelCommonFac
Cancel out common factors
P(red)=42.25π/306.25π
SimpQuot
Simplify quotient
P(red)=42.25/306.25
UseCalc
Use a calculator
P(red)=0.137959...
RoundDec
Round to 3 decimal place(s)
P(red)≈ 0.138
The probability of the dart hitting a red region is about 0.138, or 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 32.25π and 50.25π square centimeters, respectively. Also, A_(C_2)=4π square centimeters is the area of the center circles.
\begin{gathered} A_\text{red and green}=32.25\pi\!+\!50.25\pi\!+\!4\pi \\ \Downarrow \\ A_\text{red and green}=86.5\pi \end{gathered} 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.
P(red or green)=Area of red and green regions/Area of total region
SubstituteValues
Substitute values
P(red or green)=86.5π/306.25π
CancelCommonFac
Cancel out common factors
P(red or green)=86.5π/306.25π
SimpQuot
Simplify quotient
P(red or green)=86.5/306.25
UseCalc
Use a calculator
P(red or green)=0.282448...
RoundDec
Round to 3 decimal place(s)
P(red or green)≈ 0.282
The probability of hitting a red or green region is approximately 0.282, or 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, or 71.8 %.
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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.
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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* 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=Area of success region/Area of total region
SubstituteValues
Substitute values
P=60/150
ReduceFrac
a/b=.a /30./.b /30.
P=2/5
It can be concluded that the probability of Magdalena bumping into a person in the pool is 25.
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Probability in Length, Area, and Volume
Exercises
Consider the following polygons.
Pair the polygons with the correct probability of selecting the shaded part of a polygon. i. 1/2 ii. 1/3 iii. 1/4 iv. 3/8 v. 3/4
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Let's start with the rectangles. To determine how much of each rectangle is shaded, let's number each of the parts in every rectangle until there are no more parts to number.
For A, we see that 1 of 3 parts are shaded, which gives a ratio of 13. For D, we have 3 of 8 shaded parts, so the shaded part covers 38 of the whole area. Finally, for E, there are 2 shaded parts out of 8, which reduces to 14. i.& 1/2 → ? [1em] ii.& 1/3 → A [1em] iii.& 1/4 → E [1em] iv.& 3/8 → D [1em] v.& 3/4 → ? Finally, we will identify the ratios that correspond to B and C. For C, we can see that the shaded part forms a semicircle. Therefore, this corresponds to a ratio of 12. Since we only have one option left, we can pair v with 34. To prove this, let's divide the circle into quarters and then count how many of them are shaded.
There are indeed 3 shaded parts out a total of 4 quarters. Now we can complete the pairing. i.& 1/2 → C [1em] ii.& 1/3 → A [1em] iii.& 1/4 → E [1em] iv.& 3/8 → D [1em] v.& 3/4 → B
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What is the probability that a ship lands at a random point along the shoreline of the Gulf of Mexico in the following state? Round your answer to the nearest percent.
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To determine the probability of the ship landing in Alabama, we must divide the length of Alabama's shoreline by the total length of the Gulf of Mexico's shoreline. P=length of Alabama's shoreline/length of Gulf of Mexico's shoreline Let's first add the length of the shorelines. 367+397+44+53+770=1631 Finally, we divide the length of Alabama's shoreline by the length of the Gulf of Mexico's shoreline. P=53/1631≈ 3 % There is a 3 % probability of randomly landing in Alabama.
As in part A, we must divide the length of the given state's shoreline by the total length of the Gulf of Mexico's shoreline. We already calculated the length of the shoreline in part A. With this information, we can determine the probability of a ship landing in Texas. P=367/1631≈ 23 %
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Every day she throws a tennis ball out her window and onto the garden. Depending on where the tennis ball lands, she uses that ingredient in her cooking that day. What is the probability in percent of the tennis ball landing on the area containing the following ingredient?
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The probability of the tennis ball landing on the plot of land where the watermelons grow is the ratio of the area of this piece of land to the area of the entire garden.
Examining the diagram, we see that the watermelons occupy a square piece of land with a side length of 5 meters. The garden itself is also a square, with a side length of 10 meters. Now we can determine the areas that we need. 5* 5&=25 m^2 10* 10&=100 m^2 Finally, we can determine the probability of the tennis ball landing where the watermelons grow. area of watermelons/area of entire garden=25/100 =0.25 Therefore, the probability is 0.25, or 25 %.
As in the previous exercise, we must first find the area of the land where the strawberries are grown. We can see that the strawberry region can be divided into two parts, a square and an isosceles triangle that is also a right triangle.
We have already determined the area of the whole garden. To find the area of the part with strawberries, we should add the area of the square and the area of the right isosceles triangle. \begin{aligned} {\color{#0000FF}{A_\text{square}}}+{\color{#009600}{A_\text{triangle}}}&= \\[0.2cm] {\color{#0000FF}{5\cdot 5}}+{\color{#009600}{\dfrac{1}{2}(5)(5)}}&=37.5 \end{aligned} Now we have everything we need to determine the probability of the tennis ball landing in the area were strawberries are grown. area of strawberries/area of entire garden=37.5/100=0.375 Therefore, the probability is 0.375 or 37.5 %.
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Find the probability of the following events. Answer with a fraction in its simplest form.
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To determine the probability that the point falls somewhere along AC, we must divide the length of this segment by the length of AB. Let's identify the lengths of these segments.
Now we can determine the probability of the point falling on AC. P(X is on AC)=1/9
As in the previous exercise, we need to identify the length of the given segment CE. We already know that AB is 9 units.
With this information, we can determine the probability of the point falling on CE. P(X is on CE)=7/9
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Find the probability in percent that a point chosen at random lies in the shaded region. Round to the nearest percent.
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To determine the probability that a point randomly chosen lies in the shaded region, we have to divide the area of the shaded region by the area of the circle. The area of the shaded region can be calculated as the difference of the area of the circle and areas of the white triangles.
Area of the Circle A_C
The circle has a diameter of 8 inches, which means it has a radius of 4 inches. With this information, we can calculate the area of the circle.
Areas of the Triangles A_S and A_L
To calculate the areas of the triangles, we multiply the triangle's base and height and then divide the product by 2. A=1/2bh Notice that the smaller triangle has legs that correspond to the circle's radius. As for the big triangle, it has a base and height that correspond to the diameter and radius, respectively. Let's add this information to the diagram.
Now we can determine the areas of each triangle. Small:& A_S=1/2( 4)( 4)=8 in.^2 [1em] Large:& A_L=1/2( 8)( 4)=16 in.^2
Area of Shaded Region A_(SR)
The area of the shaded region is the circle's area less the areas of the two triangles. A_(SR)=16π -8-16=16π -24
Probability
Now that we know the area of the shaded region and the area of the circle, we can determine the probability of picking a point that is on the shaded area by dividing the area of the shaded region by the area of the circle.
There is a 52 % probability that the point lies in the shaded region.
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Probability in Length, Area, and Volume
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