1. + Probability

2. + THE BASIC LAW OF PROBABILITY ALL OUTCOMES BEING LIKELY, AN EVENT’S PROBABILITY = THE WAYS OF THE EVENT YOU’RE INTERESTED IN OCCURING/TOTAL POSSIBLE NUMBER OF OUTCOMES (n) Written as a formula, this would be: P(A)=number of events in A / total number of trials n

3. + Some basic probabilities What is the probability that you roll a three on a six sided die? What is the probability of drawing the 10 of hearts from a deck of cards? What is the probability that you flip a heads? These are easy, what about when we have to deal with events that occur in several stages?

4. + We make trees

5. + Example You flip a coin 3 times, what is the probability of 3 heads? Heads, Tails, Heads At least one Tails 2 or more heads?

6. + Make your tree Each new section represents a trial

7. + A handy formula for finding n TO FIND n (THE TOTAL POSSIBLE OUTCOMES) HANDY FORMULA: Number of possible outcomes per object ^ number of objects 4 sides of each COIN ^ 3 individual COINS This is your n=2 ^ 3 = 8

8. + EXAMPLE YOU ROLL A PAIR OF 4-SIDED DICE. EACH OUTCOME HAS A PROBABILITY OF ______________

9. + HOW TO SOLVE IT: FIRST, FIND N (THE TOTAL POSSIBLE OUTCOMES) HANDY FORMULA: Number of possible outcomes per object ^ number of objects 4 sides of each die ^ 2 individual dice This is your n THE PROBABILITY FOR EACH OUTCOME IS 1/N OR 1/16 THIS COULD ALSO BE SHOWN BY A TREE

10. + Probability tree

11. + EXAMPLE YOU ROLL A PAIR OF 4-SIDED DICE. WHAT IS THE PROBABILITY THAT THE SUM IS EVEN?

12. + HOW TO SOLVE IT: We already made the tree: Just go through it, find the even sums, and divide by 16

13. + EXAMPLE YOU ROLL A PAIR OF 4-SIDED DICE. WHAT IS THE PROBABILITY THAT THE FIRST ROLL IS BIGGER THAN YOUR SECOND ROLL?

14. + HOW TO SOLVE IT: USE THE TREE

15. + Rules of Probability Unions, Intersections

16. + A trick to remember the difference Intersection: MIT DUSP is located at the intersection of Mass Ave AND Vasser. An intersection contains the elements in A AND B Example: You have two sets A={2,4,6,8,10} B={1,2,3,4,5} What is A B? Intersections:

17. + Unions Union: Think of a union as a marriage between two sets: When people get married they bring their belongings into one house. Items which either he OR she owned are now in the new house. A Union contains elements in A OR in B Example: A={2,4,6,8,10} B={1,2,3,4,5} What it A B? The number is in A or B A trick to remember the difference

18. + The General Rule for addition Used for unions of probabilities What is the probability that either A or B happens? The formula is P(A U B) = P(A) + P(B) – P(A∩B )

19. + The probability of 3 events occurring (A and B and C) P( A U B U C)= P(A) + P(B) + P(C) – P(A B) –P(A C)-P(B C) ∩∩∩

20. + Example You qualify to be in the English civil service if you have a degree, you are a member of Pi Alpha, or you pass an exam. What is the probability a person is on the list because they passed an exam, had a degree, or was a member of Pi Alpha? Exam result DegreeNo Degree Subtotal Passed261440 Failed41620 Subtotal30 Total 60 Pi alpha member Exam Result DegreeNo degree subtotal Passed641680 Failed3674110 Subtota l 10090190 Total 190 Non Member Total Pass 120 Fail 130 Total 250

21. + Add the probabilities up… This question is asking you to calculate an EITHER probability, so you use the addition rule It is asking for three events, so you need to add all three, subtract shared, and then re-add the overlap of all three Find probability of Passing: 120/250 Find the probability of Having a degree: 130/250 Being a member: 60/250 Exam result DegreeNo Degree Subtotal Passed261440 Failed41620 Subtotal30 Total 60 Exam Result DegreeNo degree subtotal Passed641680 Failed3674110 Subtota l 10090190 Total 190 Total Pass 120 Fail 130 Total 250 Pi alpha memberNon Member

22. + Last steps: The probability of passing, having a degree, or being a member is: 0.48+0.52+0.24 The answer is 1.24 (which we know can’t be right) We now need to subtract P(A and B) P(A and C) and P(B and C) Go back to the tables to get these numbers This is P(member and Pass) =(40/250)=0.16 P(member and degree) =(30/250)=0.12 P(Pass and degree)= (90/250)=0.36

23. + Last steps: Then re-add P(A and B and C) The table says it is 26/250=0.104 So: 1.24-(0.16+0.36+0.12) + 0.104= 0.704 is the probability a person is on the list because they passed an exam, had a degree, or was a member of Pi Alpha?

24. + General multiplication rule Used when you want to find the joint probability of two events This is an “and” probability The probability of A and B, or P(A B) Equals P(A) * P(B | A) If A and B are independent, you can just take P(A) * P(B) This is super simple and is best illustrated with an example ∩

25. + Independence When the probability of an event is not influenced by the event before it

26. + Independence A jar has 3 red marbles, 3 blue marbles, and 2 yellow marbles. You pick out one marble. What is the probability it is red? It was red. What is the probability that your next one is red, if you don’t put the red back in the jar? What if you put it back in and then pick?

27. + Example : conditional probability and independence You have 4 females and 2 males in a group. You need to select two people to be on a committee. You want to choose at random, and you can’t choose the same person twice. What is the probability of…. (F F) M F) ∩ ∩

28. + Make a tree

29. + The formula for what you just did P(M F) = P(M) * P(F | M) P(F F)=P(F) * P(F | F) This is where the conditional is important, if you choose a female first, the total number of females that can be selected from decreased ∩ ∩

30. + Are these events independent? Use this formula for independence: If A and B are independent then P(B)= P(B|A) and P(A)=P(A|B) In this example, that means that the probability of choosing a male is equal to the probability that you choose a male given you chose a female first. It also means the probability of choosing a female equals the probability that you chose a female given you chose a male first

31. + Example General Multiplication Rule You are taking pizza orders. A customer can order a small, medium, or large. They can choose thin or thick crust. They can choose up to two toppings, peperoni or mushrooms. Are these likelihoods independent? This is a real life example of conditional probability (several conditions across several stages)

32. + Sample Problem: General Rule of multiplication: What is P(Small Thin No peperoni mushrooms)? What is the probability of small ∩ thick crust? What is the probability of small ∩ thick crust ∩ peperoni? ∩ ∩ ∩

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35. + Examples where you’d use both Multiplication and Addition rules Often, in probability you don’t use one rule on its own Trees help you determine when and where to use each rule When you make a tree and move left to right, it is the multiplication rule When you are going up and down, it is the addition rule. Let’s show this with an example.

36. + Sample problem 3 There are three restaurants in town. They get 50%, 30%, and 20% of the business. You know that 70% of the customers that leave Restaurant 1 are satisfied. 60% at Restaurant 2 are satisfied and 50% that leave restaurant 3 are satisfied. What is the probability that someone eating in this town leaves satisfied?

37. + Make your tree

38. + Multiply through each branch to get the conditional probabilities

39. + Add down the line to get total p for satisfaction You’re first multiplying across to get the conditionals Then you’re adding up and down, to get the probability of being satisfied at 1, 2, OR 3.35+.18+.1=.63 You have a 63% change of being satisfied when eating out in town The formula for what you just did: What is the P[(eat at R1 satisfies) (eat at R2 satisfied) eat at R 3 satisfied] ∩∩ ∩ ∪∪

40. + One last example You’re playing in a chess tournament. Your probability of winning against half of the players (type 1) is 0.3. Your probability of winning against a quarter of the players (type 2) is 0.4, and it is.5 when playing against the other quarter of the players (type 3). What’s the probability of winning when a random opponent is chosen?

41. + Make your table

42. + Multiply out your conditionals P(win|type 1), P(win|type 2), and P(win|type3) They are independent so you can just multiply through L to R

43. + Sum according to addition rule 0.15+0.1+0.125=0.375 Your chance of winning given a randomly chosen opponent is 0.375 or just over 37% The formula shows why you’d conclude with the addition rule: P(win|type 1) (win|type2) (win|type 3) ∪∪

44. + Conditional Probability “Given X, what is the probability of Y” Example: You’re picking one person at random from the class. Given the person in the class is a female, what the the probability he or she is blonde? What statisticians would write: P(Blonde | Female) Tips: your total (n, or the number you divide by is only the girls! Not the whole class) (# of blondes/#of girls)

45. + From your pset! Given your cloth is from a hand loom, what is the probability that the quality is poor? Locate Handloom cloth How many total pieces are there made by a hand loom? How many of those are of poor quality?

46. + Hints for solving Probability word problems When there is already a table, diagramming a tree is unnecessary Be careful to take the right total (n, denominator) Especially in conditional probabilities! The simplest example of a conditional probability is the blonde | woman example we did above, store that in your head for easy reference, and so you’re not intimidated by the “ | “

47. + The Normal Distribution

48. + Probability Distributions A probability distribution is a table or an equation that links each outcome of a statistical experiment with its probability of occurrence.

49. + Real life normal distribution

50. + Distributions fit with different types of variables: Discrete variables: takes on a countable number of values -the number of job classifications in an agency -the number of employees in a department -the number of training sessions Continuous variables: takes a countless (or super big) range of numerical values -temperature -pressure -height, weight, time -Dollars: budgets, income. (not strictly continuous) but they can take so many values that are so close that you may as well treat them that way

51. + The Normal Distribution Characteristics -continuous variables only -The bell curve shape is familiar -most values cluster around the mean mu -As values fall at a greater distance from the mean, their likelihood of occurring shrinks -Its shape is completely determined by its mean and its standard deviation -The height of the curve is the greatest at the mean (where probability of occurrence is highest)

52. + -68.26% of values fall within one standard deviation of the mean in either direction -95.44% of values fall within 2 standard deviations of the mean in either direction -99.72% of values fall within 3 standard deviations of the mean in either direction

53. + z scores The number of standard deviations a score of interests lies away from the mean in a normal distribution It is used to convert raw data into their associated probability of occurrence with reference to the mean The score we are interested in is X. To find the z score of X, subtract the mean mu from it then divide by the amount of standard deviations (sigma) to determine how many SD’s the score is from the mean

54. + z scores The z score itself equals the number of SD’s (sigma) that a score of interest (X) is from the mean (mu) in a normal distribution A data value X one standard deviation above the mean has a z score of 1 A data value X 2 SD’s above the mean has a z score of 2 The probability associated with a z score of one is 0.3413; see below in the blue oval: (68.26/2)=34.13% of the data values lie between the mean and 1 SD above it

55. + z scores The z score for 1 SD below the mean will be the same in magnitude (0.3413) but with a negative z score of -1.0 Thus, the z score of -1.0 contains 34.13% of the data i.e. just over one third of the data fall between mu and 1 SD below it

56. + Example: What is the likelihood that a value has a z score of 2.0?

57. + Example: What is the likelihood that a value has a z score of 2.0? It is equal to 95.55/2=47.72% (Meaning, just over 47% of the data fall between mu and 2 SD above it)

58. + The normal distribution table Displays the percentage of data values falling between the mean mu and each z score the first 2 digits are in the far left column the third digit is on the top row The associated probability is where they meet

59. + Locating z score for 1.0

60. + example What percent of the data lies between mu and 1.33 SD away?

61. + Locate 1.33 on the z table The answer is that 40.824% of the data fall between the mean and 1.33 SD from the mean

62. + Application Example: The police chief is reviewing the academy’s exam scores. The police department’s entrance exam has a normal distribution with a mean of 100 and SD of 10. Someone scored 119.2 on the exam. Is this a good score?

63. + Solution -another way of asking this is: what is the probability that any random applicant takes the test and scores a 119.2? -If the probability is high, then it is an average or mediocre score, if the probability is low, then it is an exceptional score -Step 1: convert the test score to a z score using the formula: (119.2-100)/10=1.92

64. + Solution - Step 2: Use the z score of 1.92 (how many standard deviations the score is above the mean, since it is a positive z score) Look it up in the z table.

65. + Solution -Step 3: The value here is.4726 -But you’re not done. Here is what you just found: -We also need to add in the part of the curve shaded in green, or all of The scores under the mean. (0.5+0.4726=) 97.26 is the percentile, or in other words, 97.26% of the scores fall below this score -The probability that a randomly selected individual will get this score or better is 1-97.26=.0274.4726.50

66. + Tips Always draw a picture, it helps you reason through your answer The z curve is symmetric, so if a your score was a -1.92, it would still contain ~47.2% of the data.

67. + The Binomial Distribution The last section, I promise.

68. + A gem from the reading

69. + Probability Distributions A probability distribution is a table or an equation that links each outcome of a statistical experiment with its probability of occurrence.

70. + Binomial Distribution Definition The probability an event will occur a specified number of times within a specified number of trials Examples: mail will be delivered before a certain time every day this week equipment in a factory remains operational in a 10 day period This is a DISCRETE distribution that deals with the likelihood of observing a certain number of events in a set number of repeated trials

71. + A Bernoulli process The Binomial distribution can be used when the process is Bernoulli Bernoulli characteristics: The outcome of a trial is either a success or a failure The outcomes are mutually exclusive The probability of a success is constant from trial to trial One trial’s probability of success is not affected by the trial before it (INDEPENDENCE) Examples of independent events could be multiple coin tosses, A fire occurring in a community isn’t affected by if one happened the night before

72. + When looking at Bernoulli Events You can calculate their probability with the binomial distribution Examples of Bernoulli events: coin flip is either heads, or not heads A crime is either solved or not solved

73. + To calculate a probability using the Binomial Distribution you need n=number of trials r=number of successes p=probability that the event will be a success q=(1-p)

74. + Breaking down the formula Is a combination, it is read, “a combination of n THINGS taken r at a time” The formula is :

75. + Example We flip a coin three times, and we want to know the probability of getting three heads

76. + Step 1 Define N, P, R, and Q n (number of trials) =3 r (successes)=3 [number of heads] p (probability of getting a heads on a flip)= 0.5 q (1-p)=0.5 Now fill in the formula

77. + Important when solving 0!=1 Any number raised to the power of 0 = 1