1. Addition Rule for Probability Vicki Borlaug Walters State Community College Morristown, Tennessee Spring 2006

2. Are the statements TRUE or FALSE? Rita is playing the violin and soccer. Rita is playing the violin or soccer. This is Rita. “ or ” means one or the other (or both) are true “ and ” means both must be true FALSE TRUE

3. Elm St. Maple St. Elm St. Maple St. Which one is “Elm and Maple”? Which one is “Elm or Maple”? Elm and Maple Elm or Maple This is called UNION. This is called INTERSECTION. Like when two streets cross. Like when you put the North and the South together.

4. Next we will look at Venn Diagrams. In a Venn Diagram the box represents the entire sample space. A B Members that fit Event A go in this circle. Members that fit Event B go in this circle.

5. A B A B Event A and B Event A or B Which is “A and B”? Which is “A or B”? This is called INTERSECTION. This is called UNION.

6. A B A B A B A B + _ = The Addition Rule for Probability P(A or B) =P(A)+ P(B)- P(A and B) A B But we have added this piece twice! That is one extra time! We need to subtract off the extra time!

7. Example #1) Given the following probabilities: P(A)=0.8 P(B)=0.3 P(A and B)=0.2 Find the P(A or B). This can be solved two ways. 1. Using Venn Diagrams 2. Using the formula We will solve it both ways.

8. Example #1 (continued) P(A)=0.8 P(B)=0.3 P(A and B)=0.2 Find the P(A or B). Solution using Venn Diagrams: A B In this example we will fill up the Venn Diagram with probabilities.

9. Solution using Venn Diagrams: A B First fill in where the events overlap. The probability that a student fits the event A and B is 0.2. That means the entire A circle must add up to 0.8. 0.2 0.6 0.1 The probability that a student fits the event B is 0.3. The box represents the entire sample space and must add up to 1. 0.2 0.1 0.6 The probability that a student fits the event A is 0.8. That means the entire B circle must add up to 0.3. Example #1 (continued) P(A)=0.8 P(B)=0.3 P(A and B)=0.2 Find the P(A or B).

10. Then find the probability of A or B. A B 0.2 0.6 0.10.2 0.1 0.6 P(A or B) =0.6 + 0.2 + 0.1 I will start by shading A or B. Then I will add up the probabilities in the shaded area. = 0.9 Answer

11. Solution using the formula: P(A or B) = P(A) + P(B) - P(A and B) = 0.8 + 0.3 - 0.2 = 0.9 Example #1 (continued) P(A)=0.8 P(B)=0.3 P(A and B)=0.2 Find the P(A or B). Answer

12. Example #2.) There are 50 students. 18 are taking English. 23 are taking Math. 10 are taking English and Math. If one is selected at random, find the probability that the student is taking English or Math. E = taking English M = taking Math

13. Solution using Venn Diagrams: E M In this example we will fill up the Venn Diagram with the number of students. Example #2 (continued) There are 50 students. 18 are taking English. 23 are taking Math. 10 are taking English and Math. If one is selected at random, find the probability that the student is taking English or Math.

14. Solution using Venn Diagrams: E M First fill in where the events overlap. The number of students taking English and Math is 10. That means the number of students taking English must add up to 18. 10 8 13 19 The number of students taking Math is 23. The box represents the entire sample space and must add up to 50. 10 19 13 8 The number of students taking English is 18. That means the number of students taking Math must add up to 23. Example #2 (continued) There are 50 students. 18 are taking English. 23 are taking Math. 10 are taking English and Math. If one is selected at random, find the probability that the student is taking English or Math.

15. Then find the probability of English or Math. E M 10 8 1310 19 13 8 P(E or M) = I will start by shading E or M. Then I will find the probability in the shaded area. = 0.62 Answer

16. Solution using the formula: P(E or M) = P(E) + P(M) - P(E and M) = 0.62 Example #2 (continued) There are 50 students. 18 are taking English. 23 are taking Math. 10 are taking English and Math. If one is selected at random, find the probability that the student is taking English or Math. Answer

17. Class Activity #1) There are 1580 people in an amusement park. 570 of these people ride the rollercoaster. 700 of these people ride the merry-go-round. 220 of these people ride the roller coaster and merry-go-round. If one person is selected at random, find the probability that that person rides the roller coaster or the merry-go-round. a.) Solve using Venn Diagrams. b.) Solve using the formula for the Addition Rule for Probability.

18. Example #3) Population of apples and pears. Each member of this population can be described in two ways. 1. Type of fruit 2. Whether it has a worm or not We will make a table to organize this data.

19. Example #3) Population of apples and pears. 5 59 4 8 62 3 apple pear no wormworm grand total 14 ?? ? ?? ? ? ?

20. Ex. #3 (continued) 5 59 4 8 62 3 apple pear no wormworm grand total 14 Experiment: One is selected at random. Find the probability that... a.)... it is a pear and has a worm. b.)... it is a pear or has a worm.

21. Ex. #3 (continued) 5 59 4 8 62 3 apple pear no wormworm grand total 14 Solution to #3a.) P(pear and worm) = Answer

22. Ex. #3 (continued) 5 59 4 8 62 3 apple pear no wormworm grand total 14 Solution to #3b.) P(pear or worm) = Answer

23. Ex. #3 (continued) 5 59 4 8 62 3 apple pear no wormworm grand total 14 Alternate Solution to #3b.) P( pear or worm )=P( pear ) + P( worm ) – P (pear and worm ) Answer

24. Class Activity #2) There are our modes of transportation – horse, bike, & canoe. Each has a person or does not have a person. 1.) Make a table to represent this data. 2.) If one is selected at random find the following: b.) P( horse and has a person) a.) P( horse or has a person) c.) P( bike or does not have a person)

25. The end!