1. X of Z: MAJOR LEAGUE BASEBALL ATTENDANCE Rather than solving for z score first, we may be given a percentage, then we find the z score, then we find the specific x value Mean MLB attendance in 2009 was 2.43 million with a standard deviation of.69. First, see if percentage is above or below 50%. For example at least 90% verse at least 44%

2. X of Z continued… Since 90% is above 50%, you know z score must be positive and your final x value must be larger than the mean. X= µ+z(ó) 90% = 1.000-90%=.1000 which is associated with a Z score of 1.28 using ‘C’ area; X= 2.43+1.28(.69)= 3.31. For major league team to draw at least 90% of league attendance they needed to draw at least 3.31 million in 2009; What about at least 44% in league attendance? Since percentage is below 50%, Z must be negative, and your final answer must be below the mean;

3. Solving for 44% –Z(ó) +µ When z score has to be subtracted, write like above. 44% is associated with a Z score of.15. –.15(.69) +2.43= 2.23 A MLB team would have to draw at least 2.23 million to be in top 44% of league attendance;

4. Review of Simple Probability The probability of a simple event is a ratio of the number of favorable outcomes for the event to the total number of possible outcomes of the event. The probability of an event a can be expressed as:

5. Find Outcomes of simple events For Simple Events – count the outcomes Examples: One Die- 6 outcomes One coin- 2 outcomes One deck of cards- 52 outcomes; 4 Aces, 12 face cards, and 36 non-face cards(ie 2-10); U.S. Roulette Wheel- 38 total slots, 18 red, 18 black, and 2 green;

6. Basic Probabilities What is the probability of getting a red on a U.S. roulette wheel? U.S. wheel has 18 red slots, 18 black slots, and 2 green slots. Solution: Desired outcome divided by # of possible outcomes=18/38,.474 or a 47.4% chance of landing on red; What is the probability of drawing a black ace from a deck of cards; 2/52=.0385 or 3.85%

7. Compound Events Events that cannot occur at the same time are called mutually exclusive. Suppose you want to find the probability of rolling a 2 or a 4 on a die. P(2 or 4) Since a die cannot show both a 2 and a 4 at the same time, the events are mutually exclusive.

8. Compound Mutually Exclusive

9. Addition Rule of Probability: Two or More Events P(A or B)= P(A) + P(B) – P( A and B) Example: What is probability of drawing a Ace of Spades or a 2 from a deck of shuffled cards? Solution: 1/52 + 4/52=5/52=.0962=9.62% What is the probability of drawing an Ace OR a Spade from a deck? Solution: Ace=4/52, spades equal 13/52 minus 1/52= 16/52=.3076 or 30.76%.

10. Additive Rule… more challenging problem

11. First: Create Your Sample Space 1 2 3 4 5 6 1 (1,1) (1,2) (1,3)(1,4)(1,5)(1,6) 2 (2,1) (2,2) (2,3)(2,4)(2,5)(2,6) 3 (3,1) (3,2) (3,3)(3,4)(3,5)(3,6) 4 (4,1) (4,2) (4,3)(4,4)(4,5)(4,6) 5 (5,1) (5,2) (5,3)(5,4)(5,5)(5,6) 6 (6,1) (6,2) (6,3)(6,4)(5,6)(6,6)

12. Now solve the problem….

13. Probability of Compound events P(jack, tails)

14. Compound Event Notations

15. Compound Events When the outcome of one event does not affect the outcome of a second event, these are called independent events. The probability of two independent events is found by multiplying the probability of the first event by the probability of the second event, minus probability of B given that A has already occurred.

16. Multiplication Rule:( ie. conditional probability] What’s the probability of tossing a head and rolling a 6? Is tossing a head independent of rolling a 6? Since they are independent events we can use P (A ∩ B) = P(A) * P(B)|A. Prob of tossing a head =1/2 and prob of tossing a 6 =1/6. Solution: Prob of tossing both a head AND a 6 =1/2*1/6= 1/12=.083 =8.3 %.

17. Birthday continued… Event A= friend has B-day on June 15 th Event B= friend has B-Day on June 15 th. P (A ∩ B) = P(A) * P(B)|A. Are they independent events? Yes. Solution: 1/365 * 1/365=.0000075 very small chance;

18. Multiplication Rule: Dependent Events P (A ∩ B) = P(A) * P(B)|A. Dependent Events- There are 10 African children and 8 Latin American children ready to be adopted. What’s the probability of choosing an African child and then another African Child? Prob choosing first African child =10/18 but the prob of choosing second African child drops to 9/17 th. Then we simply multiply 10/18 by 9/17 th = 90/306=.294 or 29.4% probability of choosing two consecutive African children without replacement;

19. Multiplication Rule: Implied AND Sometimes AND is implied. Ex: What is the probability of getting four heads in a row? First, ask yourself what is the probability of getting one head. Which is 50%. multiply this value by the # of events; What is the probability of a woman giving birth to four consecutive female children? First, ask yourself what is the probability of a woman giving birth to a single female child, we will assume it is 50%; multiply this value by the # of events;

20. A= Jason arrives first B= Scott arrives last Are events independent? No. P (A ∩ B) = P(A) * P(B)|A. There is only one Jason and one Scott and they arrive at different times;

21. You TRY!! What is the probability of drawing an Ace and a King from a deck of cards? What is the probability of drawing an Ace OR a King from a deck of cards? What is the probability of drawing a 2 OR a Heart from a deck of cards? Find probability of rolling 3 consecutive 3s with a fair die?

22. Fundamental Counting Principle

23. The total outcomes of each event are found by using a tree diagram or by using the fundamental counting principle. Example: At football games, a student concession stand sells sandwiches on either wheat or rye bread. The sandwiches come with salami, turkey, or ham, and either chips, a brownie, or fruit. Use a tree diagram to determine the number of possible sandwich combinations.

24. Tree diagram with sample space

25. Answer Using the fundamental counting principle bread x meat x side 2 x 3 x 3 = 18 possibilities

26. Fundamental Counting Principle Professor wants to assign a 3-digit code to each of his students. How many total codes are possible? First step: Decide how many independent events there are. There are three events( ie. one event for each ) digit with no restrictions.

27. 10

28. More on the fundamental counting principle Sometimes the number of outcomes changes after each event depending upon the situation Example: There are 8 students in a student action group at SJSU. The students want to stand in a line for their end of the term pictures. How many different ways could the 8 students stand for their picture?

29. Counting principle cont’ The number of ways to arrange the students can be found by multiplying the number of choices for each position. There are eight people from which to choose for the first position. After choosing a person for the first position, there are seven people left from which to choose for the second position.

30. Counting Principle There are now six choices for the third position. This process continues until there is only one choice left for the last position. Let n represent the number of arrangements. Answer:??

31. Factorials and Permutations Symbol that represents factorial is ! 4! = 4*3*2*1= 24 7!= 7*6*5*4*3*2*1=5,040 10!=10*9*8*7*6*5*4*3*2*1=3,628,800 8! Divided by 4!= 8*7*6*5*4*3*2*1 4*3*2*1 =8*7*6*5 =1680

32. Counting Principle Problem

33. Permutations We have four cards each representing one suit. What is probability of drawing a heart and a spade without replacement (ie order matters)? On first draw there is a one in four chance but in second there is only a one in three chance. 1/4 * 1/3= 1/12 or 8.33% chance Permutation formula= n! = 4! (n-r)! (4-2)! = 4*3*2*1 2*1= 12 different ways(4*3) that two out of four cards can occur in a specific order. We are only interested in one of the outcomes(ie a heart then a spade) divide 1 by 12 = 8.33% chance.

34. Permutation problem Using Formula

35. We’re only concerned with placing the top three positions. Since order matters, it is a permutation problem. 10 possible ways a person can come in first, 9 possible ways a person can come in second, 8 possible ways a person can come in third. 10*9*8=720 possible ways a person can come in first, second, and third place assuming no ties Using the formula: 10! = 10*9*8=720 (10-3) !

36. Combinations: Order Doesn’t Matter!!

37. Combination Problems First question we need to ask: Does order matter?

38. More combination problems…. First question: Does order matter?

39. More difficult Combination problem….. Does order matter? Event 1 is picking 2 motors from 8, and Event 2 is picking 2 switches from 5;

40. More Binomial Problems

41. Binomial Distr Requirements

42. Binomial Distribution P(X) = p x q n–x * nCx Where nCx = n!= Combination formula (n-x)! x! p=prob of success in one trial q=prob of failure in one trial n=number of trials x= number of successes in ‘n’ trials

43. More Binomial Problems

44. Binomial Distribution Problems The probability of getting a getting a question correct, a success, is 1/5 or.20. The probability of not getting question correct. A failure, is 4/5 or.80

45. Binomial… There are 6 trials, so n=6. We are concerned with only 2 of them being color blind, so x will be 2. A ‘success’ =.09 and a failure, or those who aren’t color blind =1–.09=.91 * 6 men taken 2 at a time = all different ways men in study can be ordered; p=.09 q=.91 n=6 x=2

46. One more Binomial cause you love this stuff!! p=.20 q=.80 n=18 x=5

47. Probability Questions 1. What is the probability of having four female children (births) in a row? 2. What is the probability of randomly selecting the top three finishers of a race, in the exact order of them finishing, out of a race comprised of six total racers? 3. What is the probability of getting three heads out of seven consecutive flips of a coin? 4. What is the probability of having three or fewer female children out of six consecutive births?