The Central Idea
In a finite equally likely sample space,
\[ P(A)=\frac{|A|}{|S|}. \]In a uniform geometric model, counting is replaced by geometric measure.
The general idea is
\[ \boxed{ P(A)= \frac{\text{geometric measure of favorable region}} {\text{geometric measure of total region}} }. \]Depending on the problem, the measure may be length, area, angle, surface measure, or volume.
Why Counting No Longer Works
A continuous interval contains infinitely many possible points.
If a number is selected uniformly from
\[ [0,10], \]we cannot count the individual real numbers in the interval.
Instead, probability is distributed according to interval length.
Uniform Selection on an Interval
Suppose a point \(X\) is selected uniformly from
\[ [a,b]. \]If the favorable event is the subinterval
\[ [c,d]\subseteq[a,b], \]then
\[ \boxed{ P(c\le X\le d)=\frac{d-c}{b-a} }. \]Probability is proportional to length.
Example: A Uniform Interval
A number is selected uniformly from
\[ [0,20]. \]Find the probability that it lies between \(5\) and \(12\).
The favorable interval has length
\[ 12-5=7. \]The total interval has length
\[ 20. \]Therefore,
\[ \boxed{ P(5\le X\le12)=\frac7{20} }. \]Endpoints Usually Do Not Affect Continuous Probability
In standard continuous uniform models, individual points have probability zero.
Therefore,
\[ P(5are equal under a continuous uniform model.
Individual Points Have Probability Zero
If \(X\) is selected uniformly from an interval such as
\[ [0,1], \]then
\[ \boxed{P(X=0.5)=0}. \]The point \(0.5\) is possible, but a single point has zero length.
Zero Probability Does Not Mean Impossible
Continuous probability models distinguish between an impossible event and a possible event with geometric measure zero.
A single point in an interval is possible, yet it has zero probability in the standard uniform model.
Thus,
\[ P(A)=0 \]does not always imply
\[ A=\varnothing. \]Probability by Length
If a point is selected uniformly from a line segment of total length \(L\), and the favorable portion has total length \(F\), then
\[ \boxed{ P(A)=\frac{F}{L} }. \]The favorable portion may consist of one interval or several disjoint intervals.
Example: Several Favorable Intervals
A point is selected uniformly from
\[ [0,10]. \]Suppose the favorable region is
\[ [1,3]\cup[7,9]. \]The total favorable length is
\[ (3-1)+(9-7)=2+2=4. \]Therefore,
\[ \boxed{ P(A)=\frac4{10}=\frac25 }. \]Complementary Length
If the unfavorable region is easier to measure, use the complement:
\[ P(A)=1-P(A^c). \]In length form,
\[ \boxed{ P(A) = 1- \frac{\text{unfavorable length}} {\text{total length}} }. \]Example: Avoiding a Central Interval
A point is selected uniformly from
\[ [0,100]. \]Find the probability that it does not lie in
\[ [40,60]. \]The excluded interval has length
\[ 20. \]Therefore,
\[ P(\text{outside }[40,60]) = 1-\frac{20}{100} = \boxed{\frac45}. \]Uniform Angular Probability
If a direction or angle is selected uniformly from a full rotation, then probability is proportional to angle measure.
For a favorable angular region of measure \(\theta\),
\[ \boxed{ P(A)=\frac{\theta}{2\pi} } \]when radians are used.
Equivalently,
\[ \boxed{ P(A)=\frac{\theta}{360^\circ} }. \]Example: Spinner Sector
A spinner is equally likely to stop at any angular position around a circle.
A favorable sector has angle
\[ 90^\circ. \]Therefore,
\[ P(A) = \frac{90^\circ}{360^\circ} = \boxed{\frac14}. \]Sector Area Gives the Same Result
If the spinner is a circular disk and probability is uniform by area, a sector with central angle \(\theta\) occupies the fraction
\[ \frac{\theta}{2\pi} \]of the disk.
Thus angular and area reasoning agree for sectors of a circle with common radius.
Probability by Area
If a point is selected uniformly from a planar region \(R\), and a subregion \(A\subseteq R\) is favorable, then
\[ \boxed{ P(A) = \frac{\operatorname{Area}(A)} {\operatorname{Area}(R)} }. \]The model assumes every equal-area part of the region is treated uniformly.
Example: Point in a Rectangle
A point is selected uniformly from a rectangle measuring
\[ 10\times6. \]A favorable subrectangle measures
\[ 4\times3. \]The total area is
\[ 60, \]and the favorable area is
\[ 12. \]Therefore,
\[ \boxed{ P(A)=\frac{12}{60}=\frac15 }. \]Example: Circle Inside a Square
A point is selected uniformly from a square of side length \(2r\).
A circle of radius \(r\) is inscribed in the square.
The square has area
\[ (2r)^2=4r^2. \]The circle has area
\[ \pi r^2. \]Therefore,
\[ \boxed{ P(\text{point lies in circle}) = \frac{\pi r^2}{4r^2} = \frac{\pi}{4} }. \]Geometric Probability and Estimating \(\pi\)
The inscribed-circle example can be reversed.
If points are generated uniformly in the square, then the long-run proportion landing inside the circle should approximate
\[ \frac{\pi}{4}. \]Therefore,
\[ \pi \approx 4\left( \frac{\text{points inside circle}} {\text{total points}} \right). \]This is the basis of a simple Monte Carlo estimate of \(\pi\).
Example: Circular Ring
A point is selected uniformly from a disk of radius \(R\).
Find the probability that the point lies outside the concentric disk of radius \(r\), where
\[ 0Thus,
\[ \boxed{ P(A) = \frac{\pi R^2-\pi r^2}{\pi R^2} = 1-\frac{r^2}{R^2} }. \]Probability by Volume
If a point is selected uniformly from a three-dimensional region \(V\), and a subregion \(A\) is favorable, then
\[ \boxed{ P(A) = \frac{\operatorname{Vol}(A)} {\operatorname{Vol}(V)} }. \]The same ratio principle extends from length and area to volume.
Example: Smaller Cube Inside a Cube
A point is selected uniformly from a cube of side length \(10\).
A smaller cube of side length \(4\) lies entirely inside it.
The total volume is
\[ 10^3=1000. \]The smaller cube has volume
\[ 4^3=64. \]Therefore,
\[ \boxed{ P(\text{point lies in smaller cube}) = \frac{64}{1000} = \frac8{125} }. \]Scaling Matters by Dimension
Geometric probability reflects the dimension of the measure being used.
If a length scale is multiplied by \(k\), then:
- length scales by \(k\);
- area scales by \(k^2\);
- volume scales by \(k^3\).
Therefore, probability ratios involving similar shapes depend on the corresponding power of the scale factor.
Example: Concentric Disks
A point is selected uniformly from a disk of radius \(10\).
Find the probability that it lies within distance \(5\) of the center.
The favorable disk has radius half as large, so its area is
\[ \left(\frac12\right)^2=\frac14 \]of the total area.
Therefore,
\[ \boxed{P=\frac14}. \]Example: Concentric Spheres
A point is selected uniformly from a solid sphere of radius \(R\).
Find the probability that its distance from the center is at most
\[ \frac R2. \]Volume scales with the cube of radius, so
\[ P = \left(\frac12\right)^3 = \boxed{\frac18}. \]Coordinate Geometry Can Describe Favorable Regions
Many geometric probability problems define the favorable event using inequalities.
For example, if a point \((x,y)\) is selected uniformly from the unit square
\[ 0\le x\le1,\qquad0\le y\le1, \]then the event
\[ x+y\le1 \]describes a triangular region.
Example: Triangle in the Unit Square
In the unit square,
\[ 0\le x\le1,\qquad0\le y\le1, \]consider the event
\[ x+y\le1. \]The favorable region is a right triangle with legs of length \(1\).
Its area is
\[ \frac12. \]The square has area \(1\), so
\[ \boxed{ P(x+y\le1)=\frac12 }. \]Example: A Band in the Unit Square
A point \((x,y)\) is selected uniformly from the unit square.
Find the probability that
\[ |x-y|\le\frac12. \]The unfavorable regions are two corner triangles, each with leg length
\[ \frac12. \]Each triangle has area
\[ \frac12\left(\frac12\right)^2=\frac18. \]The total unfavorable area is
\[ \frac14. \]Therefore,
\[ \boxed{ P\left(|x-y|\le\frac12\right)=1-\frac14=\frac34 }. \]Geometric Complements
Complement reasoning is often especially efficient in geometric probability.
If a complicated favorable region is easier to describe by what it excludes, use
\[ \boxed{ P(A) = 1- \frac{\text{measure of excluded region}} {\text{measure of total region}} }. \]Geometric Unions
If a favorable event consists of several overlapping geometric regions, then area or length obeys inclusion–exclusion:
\[ \operatorname{Area}(A\cup B) = \operatorname{Area}(A) + \operatorname{Area}(B) - \operatorname{Area}(A\cap B). \]Dividing by the total area gives the corresponding probability rule.
Example: Two Overlapping Subregions
Suppose a point is selected uniformly from a region of area \(100\).
Region \(A\) has area \(40\), region \(B\) has area \(35\), and their overlap has area \(10\).
Then
\[ \operatorname{Area}(A\cup B) = 40+35-10 = 65. \]Therefore,
\[ \boxed{ P(A\cup B)=\frac{65}{100}=0.65 }. \]Geometric Intersection
The event \(A\cap B\) corresponds to the portion of the geometric sample space satisfying both conditions.
If the model is uniform by area,
\[ \boxed{ P(A\cap B) = \frac{\operatorname{Area}(A\cap B)} {\operatorname{Area}(S)} }. \]Example: Two Coordinate Conditions
A point is selected uniformly from the unit square.
Let
\[ A=\{x\le0.6\} \]and
\[ B=\{y\le0.5\}. \]Their intersection is a rectangle measuring
\[ 0.6\times0.5. \]Thus,
\[ \boxed{ P(A\cap B)=0.30 }. \]Geometric Independence Can Appear as Area Factorization
In the unit-square example,
\[ P(A)=0.6 \]and
\[ P(B)=0.5. \]Since
\[ P(A\cap B)=0.30=0.6(0.5), \]the events are independent.
Geometrically, the rectangular intersection has area equal to the product of the corresponding one-dimensional proportions.
Random Arrival-Time Models
Time can be modeled geometrically when arrival times are treated as uniform over an interval.
If two independent arrival times are represented by coordinates
\[ (x,y), \]then the sample space is often a square.
Conditions involving waiting time become geometric regions in that square.
Example: Meeting Within Ten Minutes
Two people independently arrive uniformly during a \(60\)-minute interval.
Let their arrival times be \(x\) and \(y\).
They meet if their arrival times differ by at most \(10\) minutes:
\[ |x-y|\le10. \]The total sample space is a
\[ 60\times60 \]square.
The unfavorable region consists of two right triangles with leg length
\[ 50. \]Their total area is
\[ 2\left(\frac12\cdot50^2\right)=2500. \]The total area is
\[ 3600. \]Therefore,
\[ P(\text{meet}) = 1-\frac{2500}{3600} = \boxed{\frac{11}{36}}. \]Geometric Probability with Distance
Distance conditions often produce circles, disks, strips, annuli, or other geometric regions.
For example, the condition
\[ x^2+y^2\le r^2 \]describes a disk of radius \(r\).
The probability can then be found by comparing the disk's area with the area of the full sampling region.
Example: Distance from the Origin
A point is selected uniformly from the square
\[ [-1,1]\times[-1,1]. \]Find the probability that its distance from the origin is at most \(1\).
The favorable region is the unit disk
\[ x^2+y^2\le1. \]Its area is
\[ \pi. \]The square has area
\[ 4. \]Therefore,
\[ \boxed{ P=\frac{\pi}{4} }. \]Piecewise Geometric Regions
Favorable regions do not need to be single familiar shapes.
They may be decomposed into rectangles, triangles, sectors, circles, or other pieces whose measures are known.
If the pieces are disjoint, add their measures.
Example: Rectangle Plus Triangle
Suppose a point is selected uniformly from a region of total area \(80\).
The favorable region consists of a rectangle of area \(18\) and a disjoint triangle of area \(10\).
The favorable area is
\[ 18+10=28. \]Therefore,
\[ \boxed{ P(A)=\frac{28}{80}=\frac7{20} }. \]Geometric Probability Requires Uniformity
The ratio
\[ \frac{\text{favorable measure}} {\text{total measure}} \]is valid only when the model is uniform with respect to the measure being used.
If some locations are more likely than others, geometric measure alone does not determine probability.
Example: Nonuniform Selection
Suppose points near the center of a disk are more likely to be selected than points near the boundary.
Then the probability of landing inside a smaller concentric disk is not necessarily
\[ \frac{\text{small disk area}} {\text{large disk area}}. \]The area ratio assumes uniformity over the larger disk.
Uniform by Length Is Different from Uniform by Area
The phrase uniform must match the object being sampled.
Choosing a radius uniformly from
\[ [0,R] \]is not the same experiment as choosing a point uniformly from a disk of radius \(R\).
The first is uniform in radius. The second is uniform in area.
Example: Half the Radius Is Not Half the Disk Area
In a disk of radius \(R\), the region within radius
\[ \frac R2 \]has area
\[ \pi\left(\frac R2\right)^2 = \frac14\pi R^2. \]Therefore, a point chosen uniformly by area lies within half the radius with probability
\[ \boxed{\frac14}, \]not \(1/2\).
Boundary Effects
In ordinary uniform geometric models, boundaries such as lines inside a region or surfaces inside a volume usually have lower dimension than the full sample space.
Their probability is therefore often zero.
For example, a point selected uniformly from a square has probability zero of landing on one specific diagonal.
Example: Landing on a Diagonal
A point is selected uniformly from the unit square.
The diagonal
\[ y=x \]has area zero.
Therefore,
\[ \boxed{ P(y=x)=0 }. \]The diagonal is possible as a location, but it occupies no area.
Geometric Probability and Coordinate Transformations
Simple changes of coordinates can make a region easier to describe, but one must preserve the probability model correctly.
Length-, area-, and volume-preserving transformations keep uniform geometric ratios unchanged.
More advanced transformations may require adjustment for how area or volume changes.
Simulation Can Approximate Geometric Probability
When geometric regions are difficult to measure exactly, random simulation can estimate their probability.
Generate points uniformly in the total region and compute
\[ \frac{\text{number of simulated points in favorable region}} {\text{total simulated points}}. \]With many trials, this empirical proportion can approximate the geometric probability.
Monte Carlo Reasoning
Monte Carlo methods use repeated random sampling to estimate quantities that may be difficult to compute directly.
Geometric probability provides a natural setting:
\[ \text{geometric ratio} \approx \text{long-run simulated proportion}. \]Later lessons on simulation and empirical probability develop this idea more systematically.
Common Error: Counting Points in a Continuous Region
A continuous interval or area contains infinitely many points, so direct point-counting is not meaningful for ordinary uniform geometric probability.
Use length, area, volume, or another appropriate geometric measure instead.
Common Error: Using the Wrong Dimension
In a two-dimensional uniform region, probability is based on area, not perimeter length.
In a three-dimensional solid, probability is based on volume, not surface area.
Match the measure to the dimension of the sample space.
Common Error: Assuming Uniformity Without Justification
A geometric region alone does not determine a uniform probability model.
The experiment must state or justify that equal-measure portions are equally likely.
Without uniformity, area or length ratios may be wrong.
Common Error: Confusing Radius Ratio with Area Ratio
If two similar disks have radii in ratio
\[ \frac rR, \]then their area ratio is
\[ \boxed{ \left(\frac rR\right)^2 }. \]Probability under uniform area sampling follows the area ratio, not the radius ratio.
Common Error: Confusing Length Ratio with Volume Ratio
For similar solids, a linear scale factor \(k\) produces a volume scale factor
\[ k^3. \]Thus, halving every linear dimension gives only
\[ \frac18 \]of the original volume.
Common Error: Ignoring Overlap
If favorable geometric regions overlap, adding their areas directly counts the overlap twice.
Use inclusion–exclusion:
\[ \operatorname{Area}(A\cup B) = \operatorname{Area}(A) + \operatorname{Area}(B) - \operatorname{Area}(A\cap B). \]Common Error: Treating Boundary Inclusion as Important in a Continuous Model
For ordinary continuous geometric probability, including or excluding a finite collection of boundary points usually does not change probability.
The difference has measure zero.
A Reliable Procedure for Geometric Probability
- Identify the total geometric sample space.
- Confirm that the model is uniform with respect to the relevant measure.
- Determine whether the measure is length, angle, area, or volume.
- Translate the event into a geometric region.
- Decompose the region into measurable pieces if necessary.
- Use complements or inclusion–exclusion when they simplify the geometry.
- Compute the favorable measure.
- Compute the total measure.
- Form the ratio favorable measure divided by total measure.
- Check that the probability lies in \([0,1]\).
Example: Full Area Workflow
A point is selected uniformly from a \(12\times8\) rectangle.
A \(4\times5\) subrectangle is favorable.
The total area is
\[ 12\cdot8=96. \]The favorable area is
\[ 4\cdot5=20. \]Therefore,
\[ \boxed{ P(A)=\frac{20}{96}=\frac5{24} }. \]Example: Full Coordinate Workflow
A point is selected uniformly from the unit square.
Find the probability that
\[ y\le x. \]The line
\[ y=x \]divides the unit square into two congruent triangles.
The favorable triangle has area
\[ \frac12. \]Therefore,
\[ \boxed{ P(y\le x)=\frac12 }. \]Practice 1: Uniform Interval
A number is selected uniformly from
\[ [0,30]. \]Find the probability that it lies in
\[ [8,20]. \]Practice 2: Complement on an Interval
A point is selected uniformly from
\[ [0,50]. \]Find the probability that it lies outside
\[ [15,35]. \]Practice 3: Spinner Sector
A spinner is uniform in angle.
Find the probability of landing in a sector measuring
\[ 120^\circ. \]Practice 4: Rectangle
A point is selected uniformly from a rectangle measuring
\[ 15\times10. \]A favorable subrectangle measures
\[ 6\times5. \]Find the probability that the point lies in the subrectangle.
Practice 5: Circle in a Square
A circle of radius \(r\) is inscribed in a square of side length \(2r\).
A point is selected uniformly from the square.
Find the probability that it lies inside the circle.
Practice 6: Concentric Disk
A point is selected uniformly from a disk of radius \(12\).
Find the probability that it lies within distance \(3\) of the center.
Practice 7: Cube
A point is selected uniformly from a cube of side length \(8\).
A smaller cube of side length \(2\) lies inside it.
Find the probability that the point lies in the smaller cube.
Practice 8: Unit Square Triangle
A point \((x,y)\) is selected uniformly from the unit square.
Find
\[ P(x+y\le1). \]Practice 9: Coordinate Rectangle
A point \((x,y)\) is selected uniformly from the unit square.
Find the probability that
\[ x\le0.4 \]and
\[ y\le0.7. \]Practice 10: Annulus
A point is selected uniformly from a disk of radius \(10\).
Find the probability that its distance from the center is at least \(6\).
Practice 11: Boundary Probability
A point is selected uniformly from the unit square.
What is the probability that it lies exactly on the line
\[ y=x? \]Practice 12: Identify the Error
A student says:
“A point chosen uniformly from a disk is equally likely to be at any radius from \(0\) to \(R\), so the probability of being within radius \(R/2\) is \(1/2\).”
Explain the error and give the correct probability.
Practice Answers
Practice 1
The favorable length is
\[ 20-8=12. \]Therefore,
\[ \boxed{ P=\frac{12}{30}=\frac25 }. \]Practice 2
The excluded interval has length
\[ 35-15=20. \]Therefore,
\[ \boxed{ P(\text{outside}) = 1-\frac{20}{50} = \frac35 }. \]Practice 3
\[ \boxed{ P=\frac{120^\circ}{360^\circ}=\frac13 }. \]Practice 4
The total area is
\[ 15\cdot10=150, \]while the favorable area is
\[ 6\cdot5=30. \]Therefore,
\[ \boxed{ P=\frac{30}{150}=\frac15 }. \]Practice 5
\[ \boxed{ P = \frac{\pi r^2}{(2r)^2} = \frac{\pi}{4} }. \]Practice 6
\[ P = \frac{\pi(3)^2}{\pi(12)^2} = \frac9{144} = \boxed{\frac1{16}}. \]Practice 7
\[ P = \frac{2^3}{8^3} = \frac8{512} = \boxed{\frac1{64}}. \]Practice 8
The favorable region is half of the unit square:
\[ \boxed{ P=\frac12 }. \]Practice 9
The favorable rectangle has area
\[ 0.4(0.7)=0.28. \]Therefore,
\[ \boxed{P=0.28}. \]Practice 10
The favorable annulus has area
\[ \pi(10)^2-\pi(6)^2 = 100\pi-36\pi = 64\pi. \]Therefore,
\[ \boxed{ P=\frac{64\pi}{100\pi}=\frac{16}{25} }. \]Practice 11
The diagonal has area zero, so
\[ \boxed{P(y=x)=0}. \]Practice 12
Uniform selection from a disk is uniform by area, not by radius.
The disk of radius \(R/2\) has area
\[ \pi\left(\frac R2\right)^2 = \frac14\pi R^2. \]Therefore,
\[ \boxed{ P\left(r\le\frac R2\right)=\frac14 }. \]What to Remember
In a uniform geometric model,
\[ \boxed{ P(A)= \frac{\text{measure of favorable region}} {\text{measure of total region}} }. \]Depending on the sample space, the relevant measure may be:
- length for intervals and line segments;
- angle for uniform directions;
- area for planar regions;
- volume for three-dimensional solids.
Key habits include:
- confirm uniformity before using a geometric ratio;
- match the measure to the dimension of the sample space;
- translate algebraic inequalities into geometric regions;
- use complements when the excluded region is easier to measure;
- use inclusion–exclusion for overlapping regions;
- remember that lower-dimensional boundaries often have probability zero.
The most important habit is: measure the favorable region in the same geometric language as the total sample space.
Where We Go Next
Geometric probability computes probabilities by restricting a uniform continuous sample space to regions of interest. A different but closely related idea asks how probability changes when we already know that some event has occurred.
The next lesson, Conditional Probability, develops probability under restricted information and explains how conditioning changes the effective sample space.