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Geometry

Geometry. Geometric Probability. Goals. Know what probability is. Use areas of geometric figures to determine probabilities. Probability. A number from 0 to 1 that represents the chance that an event will occur. P(E) means “the probability of event E occuring”.

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Geometry

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  1. Geometry Geometric Probability

  2. Goals • Know what probability is. • Use areas of geometric figures to determine probabilities.

  3. Probability • A number from 0 to 1 that represents the chance that an event will occur. • P(E) means “the probability of event E occuring”. • P(E) = 0 means it’s impossible. • P(E) = 1 means it’s certain. • P(E) may be given as a fraction, decimal, or percent.

  4. Probability Example A ball is drawn at random from the box. What is the probability it is red? ? 2 P(red) = ? 9

  5. Probability A ball is drawn at random from the box. What is the probability it is green or black? ? 3 P(green or black) = ? 9

  6. Probability A ball is drawn at random from the box. What is the probability it is green or black? 1 P(green or black) = 3

  7. Geometric Probability Based on lengths of segments and areas of figures. Random: Without plan or order. There is no bias.

  8. Probability and Length Let AB be a segment that contains the segment CD. If a point K on AB is chosen at random, then the probability that it is on CD is

  9. J K S R 1 2 3 4 5 6 7 8 9 10 11 12 Example 1 Find the probability that a point chosen at random on RS is on JK. JK = 3 RS = 9 Probability = 1/3

  10. B E Z A C D 1 2 3 4 5 6 7 8 9 10 11 12 Your Turn Find the probability that a point chosen at random on AZ is on the indicated segment.

  11. Probability and Area Let J be a region that contains region M. If a point K in J is chosen at random, then the probability that it is in region M is J M K

  12. Example 2 Find the probability that a randomly chosen point in the figure lies in the shaded region. 8 8

  13. Example 2 Solution Area of Square = 82 = 64 Area of Triangle A=(8)(8)/2 = 32 Area of shaded region 64 – 32 = 32 Probability: 32/64 = 1/2 8 8 8

  14. Example 3 Find the probability that a randomly chosen point in the figure lies in the shaded region. 5

  15. Example 3 Solution Area of larger circle A = (102) = 100 Area of one smaller circle A = (52) = 25 Area of two smaller circles A = 50 Shaded Area A = 100 - 50 = 50 10 5 5 Probability

  16. Your Turn A regular hexagon is inscribed in a circle. Find the probability that a randomly chosen point in the circle lies in the shaded region. 6

  17. 6 Solution Find the area of the hexagon: ? 6 ? ? 3

  18. 6 Solution 19.57 Find the area of the circle: A = r2 A=36  113.1 Shaded Area Circle Area – Hexagon Area 113.1 – 93.63 =19.57 93.53 6 113.1 3

  19. 6 Solution 19.57 Probability: Shaded Area ÷ Total Area 19.57/113.1 = 0.173 17.3% 6 113.1 3

  20. Example 4 If 20 darts are randomly thrown at the target, how many would be expected to hit the red zone? 10

  21. Example 4 Solution Radius of small circles: 5 Area of one small circle: 25 Area of 5 small circles: 125 10

  22. Example 4 Solution continued Radius of large circle: 15 Area of large circle: (152) = 225 Red Area: (Large circle – 5 circles) 225  125 = 100 10 5 10

  23. Example 4 Solution continued Red Area:100 Total Area: 225 Probability: 10 This is the probability for each dart.

  24. Example 4 Solution continued Probability: 10 For 20 darts, 44.44% would likely hit the red area. 20  44.44%  8.89, or about 9 darts.

  25. Your Turn 500 points are randomly selected in the figure. How many would likely be in the green area?

  26. 30 10 60 5 Solution 500 points are randomly selected in the figure. How many would likely be in the green area? Area of Hexagon: A = ½ ap A = ½ (53)(60) A = 259.81 Area of Circle: A = r2 A = (53)2 A= 235.62 10

  27. Solution 500 points are randomly selected in the figure. How many would likely be in the green area? Area of Hexagon: A = 259.81 Area of Circle: A= 235.62 Green Area: 259.81 – 235.62 24.19

  28. Solution 500 points are randomly selected in the figure. How many would likely be in the green area? Area of Hexagon: A = 259.81 Green Area: 24.19 Probability: 24.19/259.81 = 0.093 or 9.3%

  29. Solution 500 points are randomly selected in the figure. How many would likely be in the green area? Probability: 0.093 or 9.3% For 500 points: 500  .093 = 46.5 47 points should be in the green area.

  30. Summary • Geometric probabilities are a ratio of the length of two segments or a ratio of two areas. • Probabilities must be between 0 and 1 and can be given as a fraction, percent, or decimal. • Remember the ratio compares the successful area with the total area.

  31. Practice Problems

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