1. UNIT 5 PROBABILITY IN OUR DAILY LIVES Lesson 5-2: Finding Probabilities Essential Question: How can we calculate the probability of various events and combinations of events?

2. LEARNING OBJECTIVES 1. Sample Space 2. Events 3. Probability of Events 4. Rules for Finding Probabilities 5. Unions 6. Intersections 7. Contingency Tables 8. Probabilities Not Equal 9. Independent or Not

3. LEARNING OBJECTIVE 1: SAMPLE SPACE For a random phenomenon, the sample space is the set of all possible outcomes. When you roll a die, there are ___ outcomes: { } When you flip two coins at once, there are ___ outcomes: { } 1, 2, 3, 4, 5, 6 6 4 HH, HT, TH, TT

5. LEARNING OBJECTIVE 1: SAMPLE SPACE Often a tree diagram is helpful for establishing a sample space. For example, if your teacher is giving you a three question T/F test, your answer you each question is either correct (C) or incorrect (I). C I C C C C C C I I I I I CCC III IIC ICI ICC CII I CIC CCI We see there are ___ outcomes in our sample space. 8

6. LEARNING OBJECTIVE 1: SAMPLE SPACE Tree diagrams will be helpful in the near future when the probability of C and I are not equal. We can’t construct tree diagrams for problems with too many outcomes and trials, so let’s make sure we can represent it mathematically. OUTCOMES TRIALS 2 3 = 8 (CORRECT/INCORRECT) THREE QUESTIONS

7. LEARNING OBJECTIVE 2: EVENTS An event is a subset of the sample space and corresponds to a particular outcome or a group of possible outcomes. Events are typically denoted by letters (Event A, B, C, etc.) For example, (see tree diagram above): Event A = student answers all 3 questions correctly = { } = ____ outcome Event B = student passes (with at least 2 correct) = { } = ____ outcomes CCC1 CCICICICC 1/8 =.125 = 13% 4/8 =.5 = 50% 4 CCC

8. LEARNING OBJECTIVE 3: PROBABILITIES OF EVENTS Each outcome in a sample space has a probability.  The probability of each individual outcome is between 0 and 1  The total of all the individual probabilities equals 1.  The probability of an event A, denoted by P(A), is obtained by adding the probabilities of the individual outcomes in the event.  When all the possible outcomes are equally likely:

9. LEARNING OBJECTIVE 3: PROBABILITIES OF EVENTS EXAMPLE: Jamal, Ken, Linda, and Mary are participants in a herbal remedy/placebo experiment. Two are chosen to receive the herbal remedy. a) Identify the sample space: b) What is the probability of any single individual receiving the herbal treatment? c) What is the probability that the treatment is given to a one man and one woman pair? JK, JL, JM, KL, KM, LM 3 out of 6 = 3/6 =.5 = 50% JK or LM = 2/6 =.3333 = 33%

10. LEARNING OBJECTIVE 4: RULES FOR FINDING PROBABILITIES The COMPLEMENT of an Event The complement of an event A consists of all outcomes in the sample space that are not in A. The probabilities of A and of A c add to 1 P(A c ) = 1 – P(A) Why does this matter? Sometimes it is easier to calculate the chance something will not happen, and then take the complement for the chance it will (or vice versa). We will do this a lot in lesson 5-4. EXAMPLE: A pro basketball player’s free throw percentage is.832. Find the complement and state it’s meaning. A ACAC 1 -.832 =.168 The chance he will miss his next free throw is 16.8%.

11. LEARNING OBJECTIVE 4: RULES FOR FINDING PROBABILITIES UNION of Two Events  The union of A and B consists of outcomes that are in A or B or in both A and B. KEY WORD:SYMBOL: INTERSECTION of Two Events The intersection of A and B consists of outcomes that are in both A and B  KEY WORD:SYMBOL: When Two Events are DISJOINT Events that do not share any outcomes in common are said to be disjoint  OR AND A B B A A B A or B or both A and B only

12. LEARNING OBJECTIVE 5: PROBABILITY OF THE UNION OF TWO EVENTS When you see the word ‘or’, you are working with the Addition Rule : If the events are disjoint, P(A and B) = 0 P(A or B) = P(A) + P(B) For the union of two events: P(A or B) = P(A) + P(B) – P(A and B) B A A or B or both

13. Example: Given a standard 52-card deck… (26 black, 26 red, 13 hearts, 4 kings, 4 queens, 4 aces) What is the probability of drawing…. a black card or a hearts?a king or a hearts card? a red card or a hearts card? an ace or a queen?

14. You need to be able to use the union formula to solve for a missing probability (a little bit of algebraic reasoning is needed). Example: The probability a student going to the vending machines and getting a snack is.356. The probability they get a snack or a drink is.423. The probability they get a snack and a drink is.253. Calculate the probability of getting just a drink. P(S or D) = P(S) + P(D) – P(S and D) LEARNING OBJECTIVE 5: PROBABILITY OF THE UNION OF TWO EVENTS.423 =.356 + P(D) –.253.423 = P(D) +.103.423 –.103 = P(D) +.103 -.103 P(D) =.320 or 32%

15. LEARNING OBJECTIVE 6: PROBABILITY OF THE INTERSECTION OF TWO EVENTS When you see the word ‘and’, you are working with the Multiplication Rule : For the intersection of two independent events, A and B: P(A and B) = P(A) x P(B) Example: Given two standard six-sided dice, what is the probability of rolling snake eyes? Example: Given two decks of cards, what is the probability of an ace card being drawn from each one? This rule extends to more than two events also… Example: 3 flips of tails in a row = A B A and B only

16. A national survey was conducted for 235 individuals over age 30, asking about marital status and whether the individual ever attended college. What is the probability someone went to college? P(YC) What is the probability someone was not married or never went to college? P(NM or NC) What is the probability someone was married and went to college? P(YM and YC) LEARNING OBJECTIVE 7: APPLYING PROBABILITY RULES TO CONTINGENCY TABLES 155 235 113 + 80 - 58 = 135 235 100 235 =.659, 66% =.574, 57% =.425, 43%

17. Looking back to our first example about the 3 question test. There are two outcomes, C or I, but in reality the way we set up the problem before assumed that each outcome was equal, a T/F has a 50/50 chance. Imagine the test actually has 3 questions but each is multiple choice with 4 possible answers. The chance you pick correctly (assuming you are purely guessing is actually ___ and the chance you are wrong is____. LEARNING OBJECTIVE 8: WHEN PROBABILITIES AREN’T EQUAL 25%75%

18. Set up the tree diagram again with this information and calculate the probabilities of every outcome in the sample space. I(.75) C(.25) CCC III IIC ICI ICC CII CIC CCI C (.25) C (.25) I (.75) I (.75) I(.75) C(.25) LEARNING OBJECTIVE 8: WHEN PROBABILITIES AREN’T EQUAL

19. Now let’s answer the first two questions from objective 1 again: Event A: What is the probability of a perfect score? Event B: What is the probability of passing the test by getting at least two correct answers? LEARNING OBJECTIVE 8: WHEN PROBABILITIES AREN’T EQUAL CCC.25 x.25 x.25 =.0156 = 1.6% = very small CCI CIC ICC CCC.25 x.25 x.75 =.0469.25 x.75 x.25 =.0469.75 x.25 x.25 =.0469.25 x.25 x.25 =.0156 =.1563 = 16%

20. In reality, events are not always independent, for example, taking a 3 question quiz you probably aren’t randomly ever randomly guessing on all the questions and you can eliminate possible answers. Example: Playing Duck Hunt (with two ducks) on the classing NES, Mario records how he did on each round H=hit and M=miss, and determines his experimental probability. He finds that he is likely to either hit both ducks or miss both. Sometimes he misses the first but hits the second, but if he hits the first, he rarely misses the second. LEARNING OBJECTIVE 9: EVENTS OFTEN ARE NOT INDEPENDENT Outcome: MM MH HM HH Probability: 0.26 0.11 0.05 0.58.58.05.11.26.69.31.37.63 1.00

21. P(A and B) = P(HH) = If A and B are independent, then we should get.58 by applying the probability rule with the total percentages as well. P(A) = first duck hit P(B) = second duck hit P(A and B) = We see that the events _________ independent. Explain why or why not. LEARNING OBJECTIVE 9: EVENTS OFTEN ARE NOT INDEPENDENT Outcome: MM MH HM HH Probability: 0.26 0.11 0.05 0.58.58.05.11.26.69.31.37.63 1.00 58%.63 x.69 =.4347 or 43% ARE NOT The given HH does not match the calculated HH.

22. EXAMPLE: Catalog sales: You are the marketing director for a museum that raises money by selling gift items from a mail order catalog. For each catalog sent to a potential customer the customer’s entry in the data file is Y if they ordered something and N if they did not. After you have mailed the fall and the winter catalogs, you estimate the probabilities of the buying patterns based on those who received the catalog as follows: LEARNING OBJECTIVE 9: EVENTS OFTEN ARE NOT INDEPENDENT Outcome (Fall/Winter)YYYNNYNN Probability0.300.100.050.55 Let F denote buying from the fall catalog and W denote buying from the winter catalog. P(F) = P(W) = P(F or W) = P(F and W) = Are F and W independent events? Explain why or why not..30+.10 =.40 or 40%.05+.55 =.60 or 60%.40+.60 -.30 =.70 or 70% (YY) =.30 or 30%.40 x.60 =.24, which is not.30, so not independent.