Unit 12 Probability Homework 3 Geometric Probability Answer Key

Geometric probability, a fascinating branch of probability theory, allows us to calculate the likelihood of events based on geometric measures such as length, area, or volume. It offers a visual and intuitive way to understand probabilities, moving beyond discrete counts to continuous spaces. Mastering geometric probability requires practice and a solid understanding of its underlying principles, which is why homework assignments, like those found in Unit 12 Probability Homework 3, are crucial. This comprehensive guide will delve into the concepts, provide step-by-step solutions, and offer insights to help you confidently tackle geometric probability problems.

Understanding Geometric Probability

Geometric probability leverages geometric concepts to determine the likelihood of an event. Instead of counting discrete outcomes, it focuses on measuring continuous spaces. The fundamental principle is that the probability of an event occurring is the ratio of the "favorable" geometric measure to the "total" geometric measure.

Key Concepts:

• Sample Space: The entire geometric region representing all possible outcomes.
• Event Space: The geometric region representing the specific event we're interested in.
• Probability: The ratio of the area (or length, volume) of the event space to the area (or length, volume) of the sample space.

Formula:

P(event) = (Measure of Event Space) / (Measure of Sample Space)

Common Geometric Shapes:

• Lines and Line Segments: Probabilities related to points falling on specific segments.
• Circles: Probabilities related to points falling within or on specific regions of a circle.
• Squares, Rectangles, and other Polygons: Probabilities related to points falling within these shapes.

Unit 12 Probability Homework 3: A Deep Dive with Answer Key

Let's analyze some typical geometric probability problems that might appear in Unit 12 Probability Homework 3, providing detailed solutions and explanations.

Problem 1: The Dartboard Dilemma

A circular dartboard has a radius of 10 inches. The bullseye has a radius of 1 inch. Assuming a dart lands randomly on the dartboard, what is the probability that it lands in the bullseye?

Solution:

1. Identify the Sample Space: The entire dartboard is the sample space, which is a circle with a radius of 10 inches.
2. Calculate the Area of the Sample Space: The area of the sample space (the entire dartboard) is π * (10 inches)^2 = 100π square inches.
3. Identify the Event Space: The bullseye is the event space, which is a circle with a radius of 1 inch.
4. Calculate the Area of the Event Space: The area of the event space (the bullseye) is π * (1 inch)^2 = π square inches.
5. Calculate the Probability: P(bullseye) = (Area of Bullseye) / (Area of Dartboard) = π / 100π = 1/100 = 0.01

Answer: The probability of hitting the bullseye is 1/100 or 1%.

Problem 2: The Random Meeting

Two friends agree to meet at a specific location between 2:00 PM and 3:00 PM. They agree that each will wait for 15 minutes, and if the other doesn't arrive within that time, they will leave. What is the probability that they will actually meet?

Solution:

1. Define the Sample Space: Let 'x' be the time the first friend arrives, and 'y' be the time the second friend arrives. Both x and y are between 0 and 60 minutes (representing the hour between 2:00 PM and 3:00 PM). The sample space is a square with sides of length 60.
2. Calculate the Area of the Sample Space: The area of the sample space (the square) is 60 * 60 = 3600 square minutes.
3. Define the Event Space: The friends will meet if the difference between their arrival times is less than or equal to 15 minutes. In other words, |x - y| <= 15. This inequality can be rewritten as two inequalities: x - y <= 15 and y - x <= 15. Graphically, this represents the area between the lines y = x - 15 and y = x + 15 within the square.
4. Calculate the Area of the Event Space: To find the area of the region where they don't meet, we consider the two triangles formed outside the region defined by |x - y| <= 15. Each triangle has a base and height of 45 (60 - 15). The area of each triangle is (1/2) * 45 * 45 = 1012.5. The combined area of the two triangles is 2 * 1012.5 = 2025. The area of the event space (where they do meet) is the total area of the square minus the area of the two triangles: 3600 - 2025 = 1575 square minutes.
5. Calculate the Probability: P(meeting) = (Area of Meeting Region) / (Area of Sample Space) = 1575 / 3600 = 7/16 = 0.4375

Answer: The probability that the two friends will meet is 7/16 or 43.75%.

Problem 3: Point on a Line Segment

A line segment AB has a length of 10 cm. A point C is randomly chosen on the line segment. What is the probability that the length of AC is less than 4 cm?

Solution:

1. Identify the Sample Space: The sample space is the entire line segment AB, which has a length of 10 cm.
2. Calculate the Length of the Sample Space: The length of the sample space is 10 cm.
3. Identify the Event Space: The event space is the portion of the line segment where the distance from A to C is less than 4 cm. This is a segment of length 4 cm starting at point A.
4. Calculate the Length of the Event Space: The length of the event space is 4 cm.
5. Calculate the Probability: P(AC < 4) = (Length of AC < 4) / (Total Length of AB) = 4 / 10 = 2/5 = 0.4

Answer: The probability that the length of AC is less than 4 cm is 2/5 or 40%.

Problem 4: The Waiting Game

You arrive at a bus stop knowing that a bus arrives every 20 minutes. What is the probability that you will have to wait more than 8 minutes for the bus?

Solution:

1. Identify the Sample Space: The sample space is the interval of time between bus arrivals, which is 20 minutes.
2. Calculate the Length of the Sample Space: The length of the sample space is 20 minutes.
3. Identify the Event Space: You'll wait more than 8 minutes if you arrive in the first 12 minutes after the previous bus (because the next bus will arrive in 20 minutes). So, the event space is the first 12 minutes of the interval.
4. Calculate the Length of the Event Space: For waiting less than 8 minutes, the event space would be 8 minutes long. However, the problem asks for the probability of waiting more than 8 minutes. Since the bus arrives every 20 minutes, you will wait more than 8 minutes if you arrive in the first 20-8 = 12 minutes.
5. Calculate the Probability: P(wait > 8) = (Length of Time You Wait More Than 8 Minutes) / (Total Interval) = 12 / 20 = 3/5 = 0.6

Answer: The probability of waiting more than 8 minutes is 3/5 or 60%.

Problem 5: The Random Chord

A circle has a radius of r. What is the probability that a randomly chosen chord is longer than the side of an equilateral triangle inscribed in the circle?

Solution:

This problem is a classic example of Bertrand's Paradox, highlighting that the answer depends on how "randomly chosen chord" is defined. Let's explore one common interpretation:

• Method 1: Random Endpoint Choose two points randomly on the circumference of the circle.

  1. Visualize the equilateral triangle: Imagine an equilateral triangle inscribed in the circle. A side of this triangle will subtend an angle of 120 degrees at the center of the circle.
  2. Fix one point: Fix one endpoint of the chord. The chord will be longer than the side of the equilateral triangle if the other endpoint lies within the 120-degree arc on either side of the fixed point's antipodal point (the point directly opposite the fixed point). This creates a 240-degree arc (120 degrees on either side).
  3. Calculate the probability: The probability is the ratio of the favorable arc length (240 degrees) to the total circumference (360 degrees): 240/360 = 2/3.

Answer (Method 1): The probability that a randomly chosen chord (chosen by randomly selecting two endpoints on the circumference) is longer than the side of an inscribed equilateral triangle is 2/3.

Important Note on Bertrand's Paradox: Other methods of choosing a "random chord" lead to different answers. For instance:

• Method 2: Random Midpoint Choose a random point inside the circle as the midpoint of the chord. The chord will be longer than the side of the inscribed equilateral triangle if the midpoint falls within a smaller concentric circle. This leads to a probability of 1/4.

Understanding Bertrand's Paradox is critical in geometric probability. Always carefully consider how "random" is defined in the problem.

Strategies for Solving Geometric Probability Problems

Here are some helpful strategies for tackling geometric probability problems:

1. Visualize the Problem: Draw a diagram! Visual representation is essential for understanding the sample space and event space.
2. Define the Sample Space: Clearly identify the set of all possible outcomes. What are the boundaries? What geometric shape represents all possibilities?
3. Define the Event Space: Identify the region that corresponds to the event you're interested in. What conditions must be met?
4. Calculate Measures: Determine the relevant geometric measures (length, area, volume) of both the sample space and the event space. Use appropriate formulas.
5. Calculate the Probability: Divide the measure of the event space by the measure of the sample space.
6. Simplify: Simplify the resulting fraction or decimal to express the probability in its simplest form.
7. Consider Different Interpretations of "Random": Be aware of potential ambiguities in the problem statement, especially when the term "random" is used. Consider different ways of interpreting "random" and how they might affect the solution (as illustrated by Bertrand's Paradox).

Common Mistakes to Avoid

• Incorrectly Identifying the Sample Space: The sample space must encompass all possible outcomes.
• Incorrectly Identifying the Event Space: Make sure the event space accurately reflects the conditions of the event you're analyzing.
• Using the Wrong Formulas: Double-check that you're using the correct formulas for calculating area, length, or volume.
• Forgetting Units: Keep track of units (e.g., square inches, centimeters) to ensure consistent calculations.
• Not Simplifying: Simplify your final answer to its simplest form.
• Ignoring Hidden Assumptions: Be mindful of hidden assumptions or constraints that might not be explicitly stated in the problem.
• Assuming Uniform Distribution: Geometric probability often assumes a uniform distribution – that is, that all points within the sample space are equally likely to occur. Make sure this assumption is valid.

Frequently Asked Questions (FAQ)

Q: What is the difference between geometric probability and discrete probability?

A: Discrete probability deals with countable outcomes (e.g., the number of heads in 3 coin flips), while geometric probability deals with continuous spaces and measures (e.g., the probability of a point falling within a specific area).

Q: Can geometric probability be used in higher dimensions?

A: Yes, geometric probability can be extended to three dimensions (using volume) and even higher dimensions, although visualization becomes more challenging.

Q: What are some real-world applications of geometric probability?

A: Geometric probability has applications in various fields, including:

• Operations Research: Optimizing resource allocation.
• Physics: Modeling particle collisions.
• Computer Graphics: Generating random points and distributions.
• Statistics: Spatial statistics and point process modeling.

Q: How do I know which geometric formulas to use?

A: The geometric formulas you need will depend on the shapes involved in the problem. Common formulas include:

• Area of a circle: πr^2
• Area of a square: s^2
• Area of a rectangle: lw
• Area of a triangle: (1/2)bh
• Length of a line segment: Distance formula
• Volume of a sphere: (4/3)πr^3
• Volume of a cube: s^3
• Volume of a rectangular prism: lwh

Q: What if the problem involves an irregular shape?

A: If the problem involves an irregular shape, you may need to use calculus (integration) to find the area or volume. Alternatively, you might be able to approximate the shape using simpler geometric figures.

Conclusion

Geometric probability provides a powerful and intuitive framework for understanding probabilities in continuous spaces. By mastering the fundamental concepts, practicing problem-solving techniques, and being aware of potential pitfalls, you can confidently tackle even the most challenging geometric probability problems in Unit 12 Probability Homework 3 and beyond. Remember to visualize the problem, carefully define the sample space and event space, and apply the appropriate geometric formulas. With consistent effort and a solid understanding of the underlying principles, you'll unlock a deeper appreciation for the beauty and versatility of geometric probability.