1. Sections 3-3 and 3-4 Probability

2. PROBABILITY AND STATISTICS CHAPTER 3 NOTES WARM-UP Respondents in three cities were asked whether they would buy a new breakfast cereal that was being tested. Use the responses from the contingency table below to answer the questions. Fort WorthDallasChicagoRow Totals Yes (Y)100150150400 No (N)12513095350 Undecided (U)751705250 Column Totals3004502501000 A.What is the probability that the person responded yes? B.What is the probability that the person is from Dallas? C. What is the probability that the person responded yes, and is from Dallas? D.What is the probability that the person responded yes or is from Dallas? E.What is the probability the person responded yes, given that they are from Dallas? F. What is the probability that the person is from Fort Worth, given that they responded yes?

3. PROBABILITY AND STATISTICS CHAPTER 3 NOTES WARM-UP Respondents in three cities were asked whether they would buy a new breakfast cereal that was being tested. Use the responses from the contingency table below to answer the questions. Fort WorthDallasChicagoRow Totals Yes (Y)100150150400 No (N)12513095350 Undecided (U)751705250 Column Totals3004502501000 A.What is the probability that the person responded yes? The total in the Yes row is 400; the total number of respondents is 1000. P(Y) = 400/1000 =.400 B.What is the probability that the person is from Dallas? The total in the Dallas column is 450; the total number of respondents is 1000. P(Dallas) = 450/1000 =.450

4. PROBABILITY AND STATISTICS CHAPTER 3 NOTES WARM-UP Respondents in three cities were asked whether they would buy a new breakfast cereal that was being tested. Use the responses from the contingency table below to answer the questions. Fort WorthDallasChicagoRow Totals Yes (Y)100150150400 No (N)12513095350 Undecided (U)751705250 Column Totals3004502501000 C. What is the probability that the person responded yes, and is from Dallas? The categories intersect. There were 150 people who both said yes and were from Dallas P(Yes and Dallas) = 150/1000 =.150 D.What is the probability that the person responded yes or is from Dallas? The events are not mutually exclusive. P(Yes) + P(Dallas) – P(Yes and Dallas) =.4 +.45 -.15 =.7

5. PROBABILITY AND STATISTICS CHAPTER 3 NOTES WARM-UP Respondents in three cities were asked whether they would buy a new breakfast cereal that was being tested. Use the responses from the contingency table below to answer the questions. Fort WorthDallasChicagoRow Totals Yes (Y)100150150400 No (N)12513095350 Undecided (U)751705250 Column Totals3004502501000 E.What is the probability the person responded yes, given that they are from Dallas? Only refer to the Dallas column P(Yes, given Dallas) = 150/450 =.3333 F. What is the probability that the person is from Fort Worth, given that they responded yes? Only refer to the Yes row. P(Fort Worth, given yes) = 100/400 =.250

6. PROBABILITY AND STATISTICS CHAPTER 3 NOTES SECTION 3 A.The OR combination P(A or B) 1.The question concerns the probability of either or two (or more) events occurring. When we say “A or B”, we include the possibility of just A, just B, or both A and B. 2.Look for key words like OR, EITHER, or phrases like AT LEAST ONE. 3.To select the correct formula, we must first determine whether or not the events A and B are mutually exclusive. a)A and B are mutually exclusive if they cannot both occur within the same trial or experiment. 1)If A occurs, then B cannot, and vice-versa. 2)In other words, the occurrence of A excludes the occurrence of B. 3)Example:Draw a card from a deck. a.A = card is a ten b.B = card is a king 1.Once we know that the card is a ten, it cannot be a king. The two are mutually exclusive.

7. PROBABILITY AND STATISTICS CHAPTER 3 NOTES SECTION 3 4)When the two events are mutually exclusive, use the simple addition rule: a.P(A or B) = P(A) + P(B). 1.P(A) = 4/52, or 1/13. 2.P(B) = 4/52, or 1/13. 3.P(A or B) = 1/13 + 1/13 = 2/13. b)A and B are not mutually exclusive if they can both occur within the same trial or experiment. 1)Example:Draw a card from a deck a.A = card is a ten b.B = card is a diamond. 1.The card can be both a ten and a diamond; the two events do not exclude each other. 2)When the events are not mutually exclusive, use the following addition rule: a.P(A or B) = (P(A)+P(B))-P(A and B). 1.Calculate P(A), and P(B). Add them together. 2.Calculate P(A and B).

8. PROBABILITY AND STATISTICS CHAPTER 3 NOTES SECTION 3 3.Subtract P(A and B) from the total you got in step 1. a)This is to prevent counting an event twice. 1)If the card happened to be the ten of diamonds, it would be counted as both a ten and as a diamond. Subtracting P(A and B) prevents this. b.P(A) = 4/52 There are four cards in the deck that are tens c.P(B) = 13/52 There are 13 cards in the deck that are diamonds. d.P(A and B) = 1/52. There is only one card in the deck that is both a ten and a diamond. e.P(A or B) = (4/52 + 13/52) - 1/52 = 16/52. 1.There are 16 cards in the deck that are either a ten, or a diamond, or both. One card, the ten of diamonds, is both and is counted twice. We subtract it once to keep the count accurate. f.When the two events are mutually exclusive, P(A and B) = 0. 1.The two events cannot happen together.

9. PROBABILITY AND STATISTICS CHAPTER 3 NOTES EXAMPLES Steps to solving problems 1.Determine what the events are. 2.Decide whether to use the “and” combination, the “or” combination, or the complement rule. 3.Before choosing the correct formula, ask one of the following questions: a)For the AND combination: ARE THE EVENTS INDEPENDENT? b)For the OR combination: ARE THE EVENTS MUTUALLY EXCLUSIVE? 4.After answering question #3, write down the appropriate formula. 5.Finally, answer the probability question.

10. PROBABILITY AND STATISTICS CHAPTER 3 NOTES EXAMPLES Two marbles are drawn without replacement from a bowl containing 7 red, 4 blue, and 9 yellow marbles. A.What is the probability that both are red? Events are:A = First is Red B = Second is Red This is an “AND” combination (the first is Red AND the second is Red) To determine the correct formula, we need to know “are the events independent or dependent?” The events are dependent since the first marble is not replaced before the second marble is drawn. Using the formula P(A and B) = P(A) * P(B, given A), we get: P(A) = 7/20 P(B, given A) = 6/19 (7/20)(6/19) = 42/380 = 21/190 =.1105. There is an 11.05% chance of getting two red marbles.

11. PROBABILITY AND STATISTICS CHAPTER 3 NOTES EXAMPLES Two marbles are drawn without replacement from a bowl containing 7 red, 4 blue, and 9 yellow marbles. B.What is the probability that NEITHER is red? This is an AND combination as well (The first marble is not red, AND the second marble is not red) The events are dependent, again because the first marble is not replaced before drawing the second P(First not Red) = 13/20 P(Second not Red, given First not Red) = 12/19. (13/20)(12/19) = 156/380 = 39/95 =.4105 C.What is the probability that at least one is red? The complement to “at least one” is “neither”. We can use the complement rule and get: P(at least one is red) = 1 – P(neither is red) = 1 – 39/95 = 56/95 =.5895 = 1 -.4105 =.5895

12. PROBABILITY AND STATISTICS CHAPTER 3 NOTES EXAMPLES One card is drawn from a standard deck of 52 cards (no jokers) A.What is the probability that it is a 4 or a 5? The events are:A = card is a 4 B = card is a 5. This is an OR combination: Either the card is a 4, OR it is a 5. The events are mutually exclusive, since a card cannot be both a 4 and a 5 at the same time. Use the simple addition rule. P(A or B) = P(A) + P(B) P(4 or 5) = 4/52 + 4/52 = 8/52 = 2/13 =.1538

13. PROBABILITY AND STATISTICS CHAPTER 3 NOTES EXAMPLES One card is drawn from a standard deck of 52 cards (no jokers) B.What is the probability that the card is neither a 4 nor a 5? This is the complement to A. The card is NOT a “4 or 5” 1 – 2/13 = 11/13 =.8461

14. PROBABILITY AND STATISTICS CHAPTER 3 NOTES EXAMPLES One card is drawn from a standard deck of 52 cards (no jokers) C.Find the probability that it is either a 4 or it is red. This is an OR combination. Are the events mutually exclusive? NO – a card can be both a 4 and red at the same time. Use the addition rule to compute the probability P(A or B) = P(A) + P(B) – P(A and B) P(A) = 4/52 and P(B) = 26/52 (There are 4 cards that are 4’s, and 26 red cards) P( A and B) = 2/52 (two red 4’s) P(4 or Red) = 4/52 + 26/52 – 2/52 = 28/52 =.5385

15. PROBABILITY AND STATISTICS CHAPTER 3 NOTES EXAMPLES Two dice are rolled and the sum of the dice is noted. A.What is the probability that the sum is a 7 or 11? This is an OR combination, and the events are mutually exclusive. Use the simple addition rule: P(A or B) = P(A) + P(B) P(7 or 11) = 6/36 + 2/36 = 8/36 =.2222

16. PROBABILITY AND STATISTICS CHAPTER 3 NOTES EXAMPLES Two dice are rolled and the sum of the dice is noted. B. What is the probability that the sum is at least ten? This is an OR combination. At least ten means 10 OR 11 OR 12. These are mutually exclusive. P(at least 10) = P(10) + P(11) + P(12) = 3/36 + 2/36 + 1/36 = 6/36 =.1667 C.What is the probability that the sum is less than 10? This is the complement to at least ten. P(less than 10) = 1 – P(at least 10) = 1 – 6/36 = 30/36 =.8333

17. PROBABILITY AND STATISTICS CHAPTER 3 NOTES EXAMPLES At Studymore University, 46% of the students are female and 13% of the females are psychology majors. 9% of all the students are psychology majors. A student is picked at random. A.Find the probability that the person is a female psychology major. This is an AND combination (a female AND a psychology major) Are the events independent? No – the probability of a female majoring in psychology is different than the probability of a student at large majoring in psychology (13% to 9%) P(female and a psychology major) = P(female) * P(psychology major, given female) P(female and a psychology major) =.46 *.13 =.0598

18. PROBABILITY AND STATISTICS CHAPTER 3 NOTES EXAMPLES At Studymore University, 46% of the students are female and 13% of the females are psychology majors. 9% of all the students are psychology majors. A student is picked at random. B.Find the probability that the student is a female or a psychology major This is an OR combination The events are not mutually exclusive: one can be both female and majoring in psychology at the same time. P(female or psychology major) = P(female) + P(psychology major) – P(female and psychology major).46 +.09 -.0598 =.4902

19. PROBABILITY AND STATISTICS CHAPTER 3 NOTES EXAMPLES ASSIGNMENTS: Classwork: Page 165, #1-12 All Homework:Pages 166-169, #14-26 Evens

20. PROBABILITY AND STATISTICS CHAPTER 3 NOTES SECTION 3-4 WARM-UP The percent distribution of the number of occupants in vehicles crossing the Tacoma Narrows Bridge in Washington is as follows: One --55.5% Two --29.8% Three -7.6% Four--4.7% Five - 1.4% Six or More --1.0% Find each probability: A)Randomly selecting a car with two occupants B)Randomly selecting a car with two or more occupants C)Randomly selecting a car with between two and five occupants, inclusive.

21. PROBABILITY AND STATISTICS CHAPTER 3 NOTES SECTION 3-4 WARM-UP The percent distribution of the number of occupants in vehicles crossing the Tacoma Narrows Bridge in Washington is as follows: One --55.5% Two --29.8% Three -7.6% Four--4.7% Five - 1.4% Six or More --1.0% Find each probability: A)Randomly selecting a car with two occupants Look at the given information!! 29.8% of cars have 2 occupants, so we have a.298 probability of selecting a car with two occupants.

22. PROBABILITY AND STATISTICS CHAPTER 3 NOTES SECTION 3-4 WARM-UP The percent distribution of the number of occupants in vehicles crossing the Tacoma Narrows Bridge in Washington is as follows: One --55.5% Two --29.8% Three -7.6% Four--4.7% Five - 1.4% Six or More --1.0% Find each probability: B)Randomly selecting a car with two or more occupants. This is the COMPLEMENT of “One” 1 -.555 =.445 Could also add.298 +.076 +.047 +.014 +.01

23. PROBABILITY AND STATISTICS CHAPTER 3 NOTES SECTION 3-4 WARM-UP The percent distribution of the number of occupants in vehicles crossing the Tacoma Narrows Bridge in Washington is as follows: One --55.5% Two --29.8% Three -7.6% Four--4.7% Five - 1.4% Six or More --1.0% Find each probability: C)Randomly selecting a car with between two and five occupants, inclusive. ADD.298 +.076 +.047 +..014 =.435

28. PROBABILITY AND STATISTICS SECTION 3-4 EXAMPLES Calculate the number of ways each of the following can be done. Then answer the corresponding probability question. A.Your company routes all materials from NY to Chicago, then from Chicago to Denver and then from Denver to L.A. There are 4 routes from NY to Chicago, 6 routes from Chicago to Denver and 3 routes from Denver to L.A. How many possible routes are there from NY to L.A.? What is the probability that one specified route is taken? Use the multiplication principle. Since there are 4 ways to perform the first action, and 6 ways to perform the second, and 3 ways to perform the third, there are (4)(6)(3) = 72 possible routes. The probability that any one specified route is taken is 1/72 =.0139

29. PROBABILITY AND STATISTICS SECTION 3-4 EXAMPLES Calculate the number of ways each of the following can be done. Then answer the corresponding probability question. B.You are supervisor of nurses and have 12 nurses working for you. In how many ways can you choose 4 to have the weekend off? What is the probability that 4 specific individuals will have the weekend off? In this problem, the order does not matter, so we will use a combination. n = 12 and r = 4. 12 C 4 = 12!/4!(12-4)! = 12!/4!8! = 495 The probability that 4 specific individuals will all have the same weekend off is 1/495 =.0020

30. PROBABILITY AND STATISTICS SECTION 3-4 EXAMPLES Calculate the number of ways each of the following can be done. Then answer the corresponding probability question. C.You are still in charge of 12 nurses (congratulations!) and must assign one to be head of the first floor ward, one to be head of the second floor ward, one for the third floor, and one for the fourth. In how many ways can this be done? What is the probability that the assignments are made in one specified order? This is a permutation, since it matters which floor they are on. n= 12, r = 4 12 P 4 = 12!/(12-4)! = 12!/8! = 11,880 The probability that the assignments are made in one specified order is 1/11,880 =.00008.

31. PROBABILITY AND STATISTICS SECTION 3-4 ASSIGNMENTS Classwork:Page 178, #1-18 All Homework:Pages 178-181, #19, 20, 22, 25-28