1. + Chapter 5 Probability: What Are the Chances? 5.1Randomness, Probability, and Simulation 5.2Probability Rules 5.3Conditional Probability and Independence Section 5.2 Probability Rules After this section, you should be able to… DESCRIBE chance behavior with a probability model DEFINE and APPLY basic rules of probability DETERMINE probabilities from two-way tables CONSTRUCT Venn diagrams and DETERMINE probabilities Learning Objectives

2. Sample Space - the collection of all possible outcomes of a chance experiment Roll a dieS = {1,2,3,4,5,6} Event - any collection of outcomes from the sample space Complement - Consists of all outcomes that are not in the event - Not rolling a even # = 1- P(rolling an even #)

3. Union - the event A or B happening consists of all outcomes that are in at least one of the two events - Rolling a prime # or even numberE={2,3,4,5,6} Intersection - the event A and B happening consists of all outcomes that are in both events - Drawing a red card and a “2” E = {2 hearts, 2 diamonds}

4. Addition rule: If A and B are disjoint events, then P(A or B ) = P(A) + P(B) If A and B are NOT disjoint events, then P(A or B ) = P(A) + P(B) – P (A ∩ B ) Mutually Exclusive (disjoint) -two events that have no outcomes in common - Roll a “2” or a “5”

5. At least one: The probability that at least one outcome happens is 1 minus the probability that no outcomes happen. P(at least 1) = 1 – P(none) Multiplication rule: If there are n ways to do a first event & m ways to do a second event. Then the number of all possible outcomes = n ∙ m

6. Disjoint example: A large auto center sells cars made by many different manufacturers. Three of these are Honda, Nissan, and Toyota. Suppose that P(H) =.25, P(N) =.18, P(T) =.14. Are these disjoint events? P(H or N or T) = P(not (H or N or T) = yes.25 +.18+.14 =.57 1 -.57 =.43

7. NonDisjoint example: Musical styles other than rock and pop are becoming more popular. A survey of college students finds that the probability they like country music is.40. The probability that they liked jazz is.30 and that they liked both is.10. P(C or J) =.4 +.3 -.1 =.6 What is the probability that they like country or jazz? Are these disjoint events? No

8. For a sales promotion the manufacturer places winning symbols under the caps of 10% of all Dr. Pepper bottles. You buy a six-pack. What is the probability that you win something? P(at least one winning symbol) = 1 – P(no winning symbols) 1 -.9 6 1 -.531441=.4686

9. Rules of probability SUMMARY What is the range of values for any given probability? What is the sum of the probability of all events? Compliment rule: For any event A, P(A c ) = Addition rule: If A and B are disjoint events, then If A and B are NOT disjoint events, then P(A or B ) =  Multiplication rule: If there are n ways to do a first event & m ways to do a second event. Then the number of all possible outcomes = n ∙ m At least one: P(at least 1) = 0 ≤ P(A) ≤ 1 1 – P(A) P(A or B ) =P(A) + P(B) P(A) + P(B) – P (A ∩ B ) Sum = 1 1 – P(none)

10. Simple Probability – Example # 1 License plates in the State of Altered require 2 letters followed by 4 digits. No two letters can be the same, but it is ok to repeat digits. How many different license plates can me made? 26 ∙ 25 ∙ 10 ∙ 10 ∙ 10 ∙ 10 = 6,500,000

11. + Simple Probability example #2 : Roll the Dice You are rolling two fair, six-sided dice – one that’s red and one that’s green.How many possible outcomes are there? Probability Rules Sample Space 36 Outcomes Sample Space 36 Outcomes Since the dice are fair, each outcome is equally likely. Each outcome has probability 1/36. Since the dice are fair, each outcome is equally likely. Each outcome has probability 1/36.

12. + Suppose event A is defined as the “sum of 5”. Find P(A). There are 4 outcomes that result in a sum of 5. Since each outcome has probability 1/36, P(A) = 4/36. P(B) = 1 – 4/36 = 32/36 Suppose event B is defined as the “sum is not 5”. Find P(B).

13. + Simple Probability Example #3: Distance Learning Distance-learning courses are rapidly gaining popularity among college students. Randomly select an undergraduate student who is taking distance-learningcourses for credit and record the student’s age. Here is the probability model: Age group (yr):18 to 2324 to 2930 to 3940 or over Probability:0.570.170.140.12 (a)Show that this is a legitimate probability model. (b)Find the probability that the chosen student is not in the traditional college age group (18 to 23 years). Each probability is between 0 and 1 and 0.57 + 0.17 + 0.14 + 0.12 = 1 P(not 18 to 23 years) = 1 – P(18 to 23 years) = 1 – 0.57 = 0.43

14. Simple Probability Example #4 Canada has 2 official languages, English and French. Choosing a Canadian at random in a recent survey gave the following distribution of responses: EnglishFrenchAsian/PacificOther 0.630.220.06? What is the probability that a Canadian’s mother tongue is either French or Asian Pacific? What is the probability of ‘Other’? Why? 0.09, because the entire P(sample space) = 1 What is the probability that a Canadian’s “mother tongue” is not English? 1 – P (English) = 1 - 0.63 =.37 P(Fr) + P(AP) = 0.22+0.06 =.28

15. + Example #5: Two-Way Tables When finding probabilities involving two events, a two-way table can display the sample space in a way that makes probability calculations easier. Consider the example on page 303. Suppose we choose a student at random. Find the probability that the student Probability Rules (a)has pierced ears. (b)is a male with pierced ears. (c)is a male or has pierced ears. (a) Each student is equally likely to be chosen. 103 students have pierced ears. So, P(pierced ears) = P(B) = 103/178. Define events A: is male and B: has pierced ears. (b) We want to find P(male and pierced ears), that is, P(A and B). Look at the intersection of the “Male” row and “Yes” column. There are 19 males with pierced ears. So, P(A and B) = 19/178. (c) We want to find P(male or pierced ears), that is, P(A or B). There are 90 males in the class and 103 individuals with pierced ears. However, 19 males have pierced ears – don’t count them twice! P(A or B) = (19 + 71 + 84)/178. So, P(A or B) = 174/178

16. Tree Diagrams - When there are many options, you may want to use a TREE DIAGRAM to help you find all the different combinations of outcomes. Simple Probability Example #6 – Four different people are each flipping a coin (4 distinct coin flips). What is the probability of getting... a)At least 2 heads? b) Either 3 heads or 3 tails?

17. Example #6 – Four different people are each flipping a coin (4 distinct coin flips). What is the probability of getting... a) At least 2 heads? b) Either exactly 3 heads or 3 tails? H H T T T H H H H T T T T H H T H T T T H H H H T T T T H H HHHH HHHT HHTH HHTT HTHH HTHT HTTH HTTT THHH THHT THTH THTT TTHH TTHT TTTH TTTT

18. Simple Probability Example #7 – In a family of 3 children… a)What is the probability exactly two are boys? b)What is the probability at least one is a girl? G B G G G G B B B G GGG GGB GBG B B G B GBB BGG BGB BBG BBB

19. NOT Disjoint: P(A ∩ B) Disjoint: P(A ∪ B) Venn Diagrams

20. + There are 174 total people. You have 90 total males. 19 of them have their ears pierced. You have 103 total people with their ears pierced. Define event A as males Define event B as pierced ears.

21. In an apartment complex, 40% of residents read the USA Today, while 25% of residents read the New York Times. Five percent of residents read both. Suppose we select an apartment resident at random and record which of the two papers the person reads. Find the probability the person reads at least 1 of the 2 papers. Find the probability the person doesn’t read either paper. Venn Diagrams Example #1 Read USA TodayRead NY Times 5% 20%35%

22. Find the probability the person reads at least 1 of the 2 papers. Find the probability the person doesn’t read either paper. Topic D: Venn Diagrams Example #1 Read USA TodayRead NY Times 5% 20%35% P(A or B) = P(A) + P(B) - P(A ∩ B) =.40 +.25 –.05 =.60 P(only A or only B or both) =.35 +.05 +.20 =.60 1 – P( at least one) = P (none) 1 -.60 =.40.40

23. After observing several pizza orders, I discover the following pizza topping requests: 67 people wanted sausage on their pizza, 55 wanted pepperoni, 50 wanted mushrooms, 22 wanted sausage and mushrooms, 18 wanted pepperoni and mushrooms, 15 wanted sausage and pepperoni, and 10 wanted all three. What is the probability that a randomly selected person wants a)only sausage? b)pepperoni and mushrooms, but no sausage? c)pepperoni or sausage? d) Neither pepperoni nor sausage nor mushrooms? Venn Diagrams Example #2

24. only sausage? pepperoni and mushrooms, but no sausage? pepperoni or sausage? Sausage Pepperoni Mushrooms 40 32 5 12 10 8 20 d) Neither pepperoni nor sausage nor mushrooms? 23

25. + Section 5.2 Probability Rules In this section, we learned that… A probability model describes chance behavior by listing the possible outcomes in the sample space S and giving the probability that each outcome occurs. An event is a subset of the possible outcomes in a chance process. For any event A, 0 ≤ P(A) ≤ 1 P(S) = 1, where S = the sample space If all outcomes in S are equally likely, P(A C ) = 1 – P(A), where A C is the complement of event A; that is, the event that A does not happen. Summary

26. + Section 5.2 Probability Rules In this section, we learned that… Events A and B are mutually exclusive (disjoint) if they have no outcomes in common. If A and B are disjoint, P(A or B) = P(A) + P(B). A two-way table or a Venn diagram can be used to display the sample space for a chance process. The intersection (A ∩ B) of events A and B consists of outcomes in both A and B. The union (A ∪ B) of events A and B consists of all outcomes in event A, event B, or both. The general addition rule can be used to find P(A or B): P(A or B) = P(A) + P(B) – P(A and B) Summary

27. + Homework: # 39, 40, 45, 46, 49, 50, 51–54, 56 We’ll learn how to calculate conditional probabilities as well as probabilities of independent events. We’ll learn about Conditional Probability Independence Tree diagrams and the general multiplication rule Special multiplication rule for independent events Calculating conditional probabilities In the next Section…

28. + Additional notes on Venn Diagrams (UsefulResource) Note, the previous example illustrates the fact that we can’t use the addition rule for mutually exclusive events unless theevents have no outcomes in common. The Venn diagram below illustrates why. Probability Rules

29. + Additional Notes on Venn Diagrams (usefulresource) Because Venn diagrams have uses in other branches ofmathematics, some standard vocabulary and notation havebeen developed. Probability Rules The complement A C contains exactly the outcomes that are not in A. The events A and B are mutually exclusive (disjoint) because they do not overlap. That is, they have no outcomes in common.

30. + Additional notes on Venn Diagrams (usefulresource) Probability Rules The intersection of events A and B (A ∩ B) is the set of all outcomes in both events A and B. The union of events A and B (A ∪ B) is the set of all outcomes in either event A or B. Hint: To keep the symbols straight, remember ∪ for union and ∩ for intersection.

31. + Additional Notes on Venn Diagrams (usefulresource) Recall the example on gender and pierced ears. We can use a Venn diagram to display the information and determine probabilities. Probability Rules Define events A: is male and B: has pierced ears.