Geometric Probability Notes

Geometric Probability

  • Geometric probability is similar to regular probability: Probability=Area of what we wantTotal possible area\text{Probability} = \frac{\text{Area of what we want}}{\text{Total possible area}}
  • There will be two geometric probability questions on the assessment.
  • Show all work, even for simple calculations, to potentially earn partial credit.

Showing Work Examples

  • Rectangle: If a rectangle is 8 by 10, write down A=80A = 80 around the rectangle.
  • Circle: If a circle has a radius of 2, write down A=4πA = 4\pi. Even if obvious, this shows the area has been calculated.

Homework

  • The instructor does not recommend doing the geometric probability homework problems from 9.4.
  • Instead, focus on the problems from the review (written by the instructor) and the quizzes.

Using Angle Measures to Find Geometric Probability

  • Circles total 360 degrees.
  • When using angle measures: Probability=Desired angle measure360 degrees\text{Probability} = \frac{\text{Desired angle measure}}{\text{360 degrees}} as total area.
Examples
  • Spinner with different colored sections.
  • Red (80 degrees): Probability =803600.22= \frac{80}{360} \approx 0.22 (22%).
  • Purple (75 degrees) or Blue (60 degrees): Probability =75+60360=135360=0.375= \frac{75 + 60}{360} = \frac{135}{360} = 0.375 (37.5% or approximately 0.38).
  • Not landing on Yellow (100 degrees): Probability =360100360=2603600.72= \frac{360 - 100}{360} = \frac{260}{360} \approx 0.72.
  • Assessment questions will be more like the example in the bottom half of the slide (using areas of shapes within a rectangle).

Example Problem: Shapes in a Rectangle

  • Find the probability that a point chosen randomly inside a rectangle lands within a given shape.
  • Rectangle dimensions: 45 x 20.
  • Shapes inside: triangle, circle, and trapezoid.
Step 1: Find Areas
  • Rectangle Area: A=45×20=900A = 45 \times 20 = 900. Write A=900A = 900 in the corner.
  • Triangle Area: A=12×base×height=12bh=40A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} bh = 40.
  • Circle Area: radius r=6r = 6, A=πr2=36π113.1A = \pi r^2 = 36\pi \approx 113.1.
  • Trapezoid Area: A=12×height×(base1+base2)A = \frac{1}{2} \times \text{height} \times (\text{base1} + \text{base2}). The height is 10, and the sum of the bases is 20, thus: A=12×10×20=100A = \frac{1}{2} \times 10 \times 20 = 100.
Step 2: Calculate Probabilities
  • Triangle: 409000.04\frac{40}{900} \approx 0.04 (4%)
  • Circle: 36π9000.13\frac{36\pi}{900} \approx 0.13
  • Trapezoid: 100900=190.11\frac{100}{900} = \frac{1}{9} \approx 0.11.
Additional Question: Probability of Landing in the White Area
  • Calculate the white area by subtracting the areas of all the shapes (triangle, circle, trapezoid) from the total area of the rectangle.
  • Area of white region: 9004010036π900 - 40 - 100 - 36\pi.
  • Probability of landing in the white area: 9004010036π9000.72\frac{900 - 40 - 100 - 36\pi}{900} \approx 0.72.
Alternative Method
  • Add the probabilities of landing in each of the colored shapes: 0.04+0.13+0.11=0.280.04 + 0.13 + 0.11 = 0.28.

  • Subtract this sum from 1: 10.28=0.721 - 0.28 = 0.72. The probability of landing in the white area is 0.72.

  • The sum of probabilities for all regions (white, yellow, green, blue) should equal 1 or 100%.

  • The assessment may not provide individual probabilities, so calculating the white area directly might be necessary. It may not be set up in a way to do it the easier way, as the assessment itself might not ask for the individual probabilities so you won't already have them calculated.