1. Geometry Geometric Probability

2. October 25, 2015 Goals  Know what probability is.  Use areas of geometric figures to determine probabilities.

3. October 25, 2015 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. October 25, 2015 Probability Example A ball is drawn at random from the box. What is the probability it is red? P(red) = ? ? 2 9

5. October 25, 2015 Probability A ball is drawn at random from the box. What is the probability it is green or black? P( green or black ) = ? ? 3 9

6. October 25, 2015 Probability A ball is drawn at random from the box. What is the probability it is green or black? P( green or black ) = 1 3

7. October 25, 2015 Geometric Probability Based on lengths of segments and areas of figures. Random: Without plan or order. There is no bias.

8. October 25, 2015 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. October 25, 2015 Example 1 Find the probability that a point chosen at random on RS is on JK. JK = 3 RS = 9 Probability = 1/3 123456789101112 R SJK

10. October 25, 2015 Your Turn Find the probability that a point chosen at random on AZ is on the indicated segment. 123456789101112 A Z B CD E

11. October 25, 2015 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 M J K

12. October 25, 2015 Example 2 Find the probability that a randomly chosen point in the figure lies in the shaded region. 8 8

13. October 25, 2015 Example 2 Solution 8 8 Area of Square = 8 2 = 64 Area of Triangle A=(8)(8)/2 = 32 Area of shaded region 64 – 32 = 32 Probability: 32/64 = 1/2 8

14. October 25, 2015 Example 3 Find the probability that a randomly chosen point in the figure lies in the shaded region. 5

15. October 25, 2015 Example 3 Solution 55 10 Area of larger circle A = (10 2 ) = 100 Area of one smaller circle A = (5 2 ) = 25 Area of two smaller circles A = 50 Shaded Area A = 100 - 50 = 50 Probability

16. October 25, 2015 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. October 25, 2015 Solution 6 ? 6 ? 3 ? Find the area of the hexagon:

18. October 25, 2015 Solution 6 6 3 Find the area of the circle: A = r 2 A=36  113.1 Shaded Area Circle Area – Hexagon Area 113.1 – 93.63 =19.57 113.1 19.57 93.53

19. October 25, 2015 Solution 6 6 3 Probability: Shaded Area ÷ Total Area 19.57/113.1 = 0.173 17.3% 113.1 19.57

20. October 25, 2015 Example 4 If 20 darts are randomly thrown at the target, how many would be expected to hit the red zone? 10

21. October 25, 2015 Example 4 Solution 10 Radius of small circles: 5 Area of one small circle: 25 Area of 5 small circles: 125

22. October 25, 2015 Example 4 Solution continued 10 Radius of large circle: 15 Area of large circle: (15 2 ) = 225 Red Area: (Large circle – 5 circles) 225  125 = 100 10 5

23. October 25, 2015 Example 4 Solution continued 10 Red Area:100 Total Area: 225 Probability: This is the probability for each dart.

24. October 25, 2015 Example 4 Solution continued 10 Probability: For 20 darts, 44.44% would likely hit the red area. 20  44.44%  8.89, or about 9 darts.

25. October 25, 2015 Your Turn 500 points are randomly selected in the figure. How many would likely be in the green area?

26. October 25, 2015 Solution 500 points are randomly selected in the figure. How many would likely be in the green area? 10 Area of Hexagon: A = ½ ap A = ½ (53)(60) A = 259.81 Area of Circle: A = r 2 A = (53)2 A= 235.62 60 30 5 10

27. October 25, 2015 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. October 25, 2015 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. October 25, 2015 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. October 25, 2015 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. October 25, 2015 Practice Problems