2. PROBABILITY! Let’s learn about probability and chance!

3. What is probability? Probability is the measure of how likely an event or outcome is. Different events have different probabilities!

4. The outcomes of an experiment are the ways it can happen. The event is the particular outcome you are looking for. 6101252

5. Mutually exclusive events: If one and only one of them can take place at a time. BA

6. Equally likely Event: Each event has an equal chance of occurrence. Complementary Event : The complement of an event is the set of all the outcomes in a sample space that are not included in The event. P(A’) = 1- P(A)

7. Probability is expressed in numbers between 0 and 1. Probability = 0 means the event never happens; probability = 1 means it always happens. The total probability of all possible event always sums to 1.

8. How do we express probabilities? Usually, we express probabilities as fractions. –The numerator is the number of ways the event can occur. –The denominator is the number of possible events that could occur. Let’s look at an example!

9. Theoretical Probability Theoretical Probability is based upon the number of favorable outcomes divided by the total number of outcomes

10. Theoretical Probability Formula Theoretical Probability : P= Number of favorable outcomes Total number of outcomes Example: In the roll of a die, the probability of getting an even number is 3/6 or ½.

11. What is the probability the spinner will land on the number 3? 1 4 3 2

12. 1. How many 3’s are on the spinner? 2. How many possible numbers could the spinner land on? 1 4 3 2 1 4

13. What is the probability the die will land on an even number? Remember, a die has six sides. Numbers 1, 2, 3, 4, 5, and 6 are each depicted once on the die.

14. Ask yourself the following questions: 1. How many even numbers are on the die? 2. How many possible numbers could the die land on? 3 6

15. What is the probability that I will choose a red marble? In this bag of marbles, there are: –3 red marbles –2 white marbles –1 purple marble –4 green marbles

16. Ask yourself the following questions: 1. How many red marbles are in the bag? 2. How many marbles are in the bag IN ALL? 3 10

17. B A B A A B 1414 1212 3535 2525 1414 a) b) P(red) P(blue) P(yellow or blue) 1515

18. What is the probability of getting 4 fives? P(4 fives) = CARDS 1 13

19. What is the probability of getting an even number? 6 12 DICE P(even) =

20. What is the probability of rolling two coins and getting H first and then T? P(H & then T) = COINS 1414 H T H TT H

21. Experimental Probability As the name suggests, Experimental Probability is based upon repetitions of an actual experiment. Example: If you toss a coin 10 times and record that the number of times the result was 8 heads, then the experimental probability was 8/10 or 4/5

22. Experimental Probability: P = Number of favorable outcomes Total number trials

23. In an experiment a coin is tossed 15 times. The recorded outcomes were: 6 heads and 9 tails. What was the experimental probability of the coin being heads? P (heads) = # Heads = 6 Total # Tosses 15

24. Odds

25. Another way to describe the chance of an event occurring is with odds. The odds in favor of an event is the ratio that compares the number of ways the event can occur to the number of ways the event cannot occur. We can determine odds using the following ratios: Odds in Favor = number of successes number of failures Odds against = number of failures number of successes

26. Suppose we play a game with 2 die. If the sum of the numbers rolled is 6 or less – you win! If the sum of the numbers rolled is not 6 or less – you lose In this situation we can express odds as follows: Odds in favor =numbers rolled is 6 or less numbers rolled is not 6 or less Odds against = numbers rolled is not 6 or less `numbers rolled is 6 or less

27. A bag contains 5 yellow marbles, 3 white marbles, and 1 black marble. What are the odds drawing a white marble from the bag? Odds in favor = number of white marbles3 number of non-white marbles = 6 Odds against = number of non-white marbles 6 number of white marbles3 Therefore, the odds for are 1:2 and the odds against are 2:1

28. Your Turn - Probability Find the probability of randomly choosing a red or white marble from the given bag of red and white marbles. 1.Number of red marbles 16 Total number of marbles 64 2.Number of red marbles8 Total number of marbles 40 3.Number of white marbles7 Total number of marbles 20 4.Number of white marbles 24 Total number of marbles32

29. Find the favorable odds of choosing the indicated letter from a bag that contains the letters in the name of the given state. 5.S; Mississippi 6.N; Pennsylvania 7.A; Nebraska 8.G; Virginia

30. You toss a six-sided number cube 20 times. For twelve of the tosses the number tossed was 3 or more. 9.What is the experimental probability that the number tossed was 3 or more? 10.What are the favorable odds that the number tossed was 3 or more?

31. 1.¼ 2.1/5 3.13/20 4.¼ 5.4/11 6.¼ 7.¼ 8.1/8 9.3/5 10.3/2

32. Two events A and B are independent in case P(AB) = P(A)P(B) A set of events {A i } is independent in case

33. P(A) =Only one event A can take place. Additional rule If events are mutually exclusive,the probability of either A or B happening P(A or B) = P(A) + P(B) /U A B P(A or B)=

34. Five equally capable students are waiting for a job interview with a company That has announced that it will hire only one of them. the group consists of John,Bill,Sally,Helen and Walter. What is the Probability that John will be the candidate? What is the probability that either John or Helen will be candidate? Probability of getting either ace or a heart from a deck of cards?

35. If events are not mutually exclusive,the probability of either A or B happening P(A or B) = P(A) + P(B) – P(AB) /U The employees of a certain company have elected 5 of their number to represent them on the employee- management probability council. 1.male—30 years, 2. male– 32,3.female– 45, 4. female—20, 5. male—40 What is the probability the spokesperson will be either female or over 35?

36. Conditional Probability If A and B are events with Pr(A) > 0, the conditional probability of B given A is

37. Bayes’ Theorem Suppose we have estimated prior probabilities for events we are concerned with, and then obtain new information. We would like to a sound method to computed revised or posterior probabilities. Bayes’ theorem gives us a way to do this.

38. Probability Revision using Bayes’ Theorem Prior Probabilities New Information Application of Bayes’ Theorem Posterior Probabilities

39. Application of Bayes’ Theorem Consider a manufacturing firm that receives shipment of parts from two suppliers. Let A 1 denote the event that a part is received from supplier 1; A 2 is the event the part is received from supplier 2

40. We get 65 percent of our parts from supplier 1 and 35 percent from supplier 2. Thus: P(A 1 ) =.65 and P(A 2 ) =.35

41. Quality levels differ between suppliers Percentage Good Parts Percentage Bad Parts Supplier 1982 Supplier 2955 Let G denote that a part is good and B denote the event that a part is bad. Thus we have the following conditional probabilities: P(G | A 1 ) =.98 and P(B | A 2 ) =.02 P(G | A 2 ) =.95 and P(B | A 2 ) =.05

42. Tree Diagram for Two-Supplier Example Step 1 Supplier Step 2 Condition Experimental Outcome A1A1 A2A2 G B G B (A 1, G) (A 1, B) (A 2, G) (A 2, B)

43. Each of the experimental outcomes is the intersection of 2 events. For example, the probability of selecting a part from supplier 1 that is good is given by:

44. Probability Tree for Two-Supplier Example Step 1 Supplier Step 2 Condition Probability of Outcome P(A 1 ) P(A 2 ).65.35 P(G | A 1 ) P(B | A 2 ).98.95.02.05

45. A bad part broke one of our machines—so we’re through for the day. What is the probability the part came from suppler 1? We know from the law of conditional probability that: Observe from the probability tree that:

46. The probability of selecting a bad part is found by adding together the probability of selecting a bad part from supplier 1 and the probability of selecting bad part from supplier 2. That is:

47. Bayes’ Theorem for 2 events By substituting equations (4.15) and (4.16) into (4.14), and writing a similar result for P(B | A 2 ), we obtain Bayes’ theorem for the 2 event case:

49. Bayes’ Theorem

50. Tabular Approach to Bayes’ Theorem— 2-Supplier Problem (1) Events A i (2) Prior Probabilities P(A i ) (3) Conditional Probabilities P(B | A 1 ) (4) Joint Probabilities P(A i ∩ B) (5) Posterior Probabilities P(A i | B) A1A1.65.02.0130. 0130/.0305 =.4262 A2A2.35.05.0175.0175/.0305 =.5738 1.00 P(B)=.0305 1.0000

51. Probability distributions We use probability distributions because they work –they fit lots of data in real world

52. Random variable The mathematical rule (or function) that assigns a given numerical value to each possible outcome of an experiment in the sample space of interest.

53. Random variables Discrete random variables Continuous random variables

54. The Binomial Distribution Bernoulli Random Variables Imagine a simple trial with only two possible outcomes Success (S) Failure (F) Examples Toss o a coin (heads or tails) Survival of an organism in a region (live or die) Jacob Bernoulli (1654-1705)

55. The Binomial Distribution Overview Suppose that the probability of success is p What is the probability of failure? –q = 1 – p Examples –Toss of a coin (S = head): p = 0.5  q = 0.5 –Roll of a die (S = 1): p = 1/6  q = 1-1/6=5/6

56. What is the probability of obtaining x successes in n trials? Example –What is the probability of obtaining 2 heads from a coin that was tossed 5 times? P(HHTTT) = (1/2) 5 = 1/32

57. What is the probability of obtaining x successes in n trials? Example –What is the probability of obtaining 2 heads from a coin that was tossed 5 times? P(HHTTT) = (1/2) 5 = 1/32

58. In general, if trials result in a series of success and failures, FFSFFFFSFSFSSFFFFFSF… Then the probability of x successes in that order is P(x)= q  q  p  q   = p x  q n – x

59. However, if order is not important, then where is the number of ways to obtain x successes in n trials, and i! = i  (i – 1)  (i – 2)  …  2  1 Mean=np, Variance= npq n!n! x!(n – x)! p x  q n – x P(x) = n!n! x!(n – x)!

60. Definition A binomial experiment is a probability experiment that satisfies the following conditions: 1. The experiment is repeated for a fixed number of trials, where each trial is independent of the other trials. 2. There are only two possible outcomes of interest for each trial. The outcomes can be classified as a success (S) or as a failure (F). 3. The probability of a success, P(S), is the same for each trial. 4. The random variable, x, counts the number of successful trials.

61. Notation for Binomial Experiments SymbolDescription nThe number of times a trial is repeated. p = P(S)The probability of success in a single trial. q = P(F)The probability of failure in a single trial (q = 1 – p) xThe random variable represents a count of the number of successes in n trials: x = 0, 1, 2, 3,... n.

62. Ex. 1: Binomial Experiments Decide whether the experiment is a binomial experiment. If it is, specify the values of n, p and q and list the possible values of the random variable, x. If it is not, explain why. 1. A certain surgical procedure has an 85% chance of success. A doctor performs the procedure on eight patients. The random variable represents the number of successful surgeries.

63. Solution: the experiment is a binomial experiment because it satisfies the four conditions of a binomial experiment. In the experiment, each surgery represents one trial. There are eight surgeries, and each surgery is independent of the others. Also, there are only two possible outcomes for each surgery—either the surgery is a success or it is a failure. Finally, the probability of success for each surgery is 0.85. n = 8 p = 0.85 q = 1 – 0.85 = 0.15 x = 0, 1, 2, 3, 4, 5, 6, 7, 8

64. A jar contains five red marbles, nine blue marbles and six green marbles. You randomly select three marbles from the jar, without replacement. The random variable represents the number of red marbles. The experiment is not a binomial experiment because it does not satisfy all four conditions of a binomial experiment. In the experiment, each marble selection represents one trial and selecting a red marble is a success. When selecting the first marble, the probability of success is 5/20. However because the marble is not replaced, the probability is no longer 5/20. So the trials are not independent, and the probability of a success is not the same for each trial.

65. In a survey, American workers and retirees are asked to name their expected sources of retirement income. The results are 36% of working Americans expect to rely on social security for retirement income. Seven workers who participated in the survey are asked whether they expect to rely on social security for retirement income. Create a binomial probability distribution for the number of workers who respond yes. p = 0.36 and q = 0.64. Because n = 7, the possible values for x are 0, 1, 2, 3, 4, 5, 6 and 7.

66. xP(x) 00.044 10.173 20.292 30.274 40.154 50.052 60.010 70.001  P(x) = 1

67. Finding a Binomial Probability Using a Table Fifty percent of working adults spend less than 20 minutes commuting to their jobs. If you randomly select six working adults, what is the probability that exactly three of them spend less than 20 minutes commuting to work? Use a table to find the probability. Solution: A portion of Table 2 is shown here. Using the distribution for n = 6 and p = 0.5, you can find the probability that x = 3, as shown by the highlighted areas in the table.

68. In a city, 57% of the days in a year are cloudy. Find the mean, variance, and standard deviation for the number of cloudy days during the month of June. What can you conclude? Solution: There are 30 days in June. Using n=30, p = 0.57, and q = 0.43, you can find the mean variance and standard deviation as shown. Mean:  = np = 30(0.57) = 17.1 Variance:  2 = npq = 30(0.57)(0.43) = 7.353 Standard Deviation:  = √npq = √7.353 ≈2.71

69. The Poisson Distribution Overview When there is a large number of trials, but a small probability of success, binomial calculation becomes impractical –Example: Number of deaths from horse kicks in the Army in different years The mean number of successes from n trials is µ = np –Example: 64 deaths in 20 years from thousands of soldiers Simeon D. Poisson (1781-1840)

70. If we substitute µ/n for p, and let n tend to infinity, the binomial distribution becomes the Poisson distribution: P(x) = e -µ µ x x!x!

71. Poisson distribution is applied where random events in space or time are expected to occur Deviation from Poisson distribution may indicate some degree of non- randomness in the events under study Investigation of cause may be of interest