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Stats chapter 6
 

Stats chapter 6

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Stats chapter 6 Stats chapter 6 Presentation Transcript

  • Chapter 6 Probability and Simulation
  • 6.1 SIMULATION
  • What is Simulation (stats) • Define a scenario whose probabilistic outcomes are know • Use a mathematical model to carry out a number of likely outcomes for the scenario • Compare the distribution of the outcomes with “alternative models”
  • Steps to Simulation 1. State the problem and the random phenomenon (What are we trying to determine?) 2. State assumptions (what are the probabilities involved?) 3. Create a mathematical model (use your calculator or table B) 4. Carry out many repetitions of trial (don‟t forget to record outcomes!) 5. State your conclusions
  • Creating a mathematical model • Using a calculator, most probabilities can be computed with the outcomes 1-100 – RandInt(1,100) • Using the table, probabilities can be computed with digits 00-99 – Like with exp design, you should use ID‟s with the same number of digits – You may treat 00 as 100 • You may want to simplify your model into two outcomes – „Success‟ or „Failure‟
  • Penultimate Last thoughts • Make sure you follow through all 5 steps when you perform a simulation • It may be helpful to list all possible outcomes Coin 1 Coin 2 Outcome Heads Heads HH Heads Tails HT Tails Heads TH Tails Tails TT
  • Last thoughts • You will not receive credit for calculator notation – Record your mathematical model (i.e. #1-32 is a success, #33-99 and 00 are failures) – Record your method of producing random integers “I will use my calculator” “I will use line 131 from table B” • Record your observations (when possible) – Create a table with „trial #‟ „observation‟ and „outcome‟ trial# observation outcome 16 42 failure 17 12 success
  • Assignment 6.1 Pg. 397 #1, 3, 5, 9, 13, 19
  • 6.2 PROBABILITY MODELS
  • The idea of probability • Random behavior does not mean “haphazard” • Random behavior is: – Unpredictable in the short term – Has a regular and predictable pattern in the long run • Observation of random behavior to determine probability model is “empirical probability” • Remember- the regular predictable pattern only appears after many repetitions
  • Probability Models • A list of all possible outcomes of a random phenomenon is called the “Sample Space” or „S‟ • An event is a one or more set of outcomes for the random phenomenon, it is a subset of the sample space – It‟s helpful to think of events in terms of “success” or “failure” • A Probability Model describes the random phenomenon. Consists of two parts: – Sample Space (S) – Probability for each event (P)
  • Probability Model- Coins S = {H, T} P(H) = 0.5, P(T) = 0.5 • Notice that the sum of probabilities for the sample space is 1.00 • This model also assumes that the coin is „fair‟
  • Tree Diagrams • Using a tree is helpful when there is more than one „event‟ or mechanism for each event • These outcomes must be independent – The outcome of one event does not effect the outcome of the next event
  • Tree Diagram
  • Tree Diagram Event #1 „Heads‟ or „Tails‟
  • Tree Diagram Event #2 #1 - #6 Notice that each of the outcomes from Event #1 branches to all outcomes of Event #2
  • Tree Diagram Sample Space It does not matter what order the events are placed in the tree!
  • Probability Model • Provided that the outcomes of each event are equally likely, then the probability of each outcome is the same: 1/n • Multiplication Principle – If the Sample Space consists two events, event 1 has n1 outcomes, event 2 has n2 outcomes, then the Sample Space has n1 x n2 outcomes – I like to call this the menu principle
  • Confusions and Clarification • The concepts of “events” and “outcomes” easily confused. • Outcome – The product of some mechanism- dice, cards, etc. • Event – An outcome or set of outcomes with significance • Many times, the “event of interest” is the result of many outcomes
  • Assignment 6.2A • Pg 411 #23, 24, 27-29, 33, 35, 36
  • With or Without Replacement Many events involve some kind of repeated sampling- think of drawing cards from a deck • Sampling with Replacement – After a sample, the card is put back into the deck – The probability model for the second card is the same as the probability model for the first card • Sampling without Replacement – The card is not put back in the deck – The probability model for the second card is not the same as the probability model for the first card
  • Probability RULZ Suppose a sample space „S‟ has events „A‟ and „B‟ 1. 0 < P(A) < 1 -The probability of an event is between 0 and 1 -P(A) = 0 means the event never happens -P(A) = 1 means the event always happens
  • Probability RULZ Suppose a sample space „S‟ has events „A‟ and „B‟ 2. P(A) + P(B) + … +P(n) = 1 -or- P(S) = 1 -The sum of all possible outcomes is 1. -One of the outcomes in the S must happen!
  • Probability RULZ Suppose a sample space „S‟ has events „A‟ and „B‟ -Two events are „disjoint‟ or „mutually exclusive‟ if they have no common outcomes 3. If events A and B are disjoint, then P(A or B) = P(A) + P(B) (more on disjoint events later)
  • Probability RULZ Suppose a sample space „S‟ has events „A‟ and „B‟ 4. P(A) + P(AC) = 1 -The probability of an event occurring plus the probability that an event does not occur is „1‟ -“Either an event happens, or it doesn‟t” -Also P(A) = 1 - P(AC)
  • Terminology • “Union,” “OR,” “U” – “A U B” – “Either A or B or both” • “Intersect,” “AND,” “∩” – “A∩B” – Both A and B occurred simultaneously • Empty set – Event has no outcomes! – i.e. “A∩B = ” or “A∩B = { }”
  • THE FOUR RELATIONS • Two events can be related in one of four ways: 1. Complimentary Events 2. Disjoint Events 3. Implied Events 4. Independent Events
  • Terminology Complimentary Events A and AC - Either A occurs or AC occurs B (or AC )
  • Terminology Disjoint Events If A occurs, then B does not occur. If B occurs, then A does not occur.
  • Terminology Implied Events • All the outcomes for one event are contained in another event • If B happens, then A also happens S A B
  • Terminology Independent Events • Events share outcomes • It is possible that an outcome qualifies event A and event B
  • Terminology Independent Events • Events share outcomes • It is possible that an outcome qualifies event A and event B • I like this diagram: A BA and B Ac and Bc P(A) P(AC) P(B) P(BC)
  • Equally likely outcomes • If a random phenomenon has k equally likely outcomes, then the probability of each outcome is 1/k. • For an event A: # of outcomes in A A # of outcomes in S # of outcomes in A P k
  • Equally likely outcomes • What is the probability of drawing a „King‟ from a standard deck of cards • Let A = drawing a king A = {King of Spades, King of Diamonds, King of Clubs, King of Hearts} # of outcomes in A A # of outcomes in S 4 1 52 13 P
  • Multiplication rule • Two events are independent if the outcome of the first trial does not affect the outcome of the second trial. • This is not the same as „disjoint‟ – If two events are disjoint, then by definition the outcome of the first event affects the outcome of the second event
  • Multiplication Rule • If A and B are independent events, then P(A and B) = P(A) x P(B) • Ex. A = 3 on a die, B = Heads P(3 and Heads) = P(A and B) =P(A) x P(B) =(1/6) x (1/2) =1/12 • Always check that the events are independent before using the multiplication rule!
  • Multiplication Rule • If we think in terms of the Venn Diagram, then the probability for independent events is just the area of the appropriate rectangle A BA and B Ac and Bc P(A) P(AC) P(B) P(BC)
  • Using the diagram • In fact, this diagram can be used for all four of the relations! (pay attention to the zeros) • COMPLIMENTS: A B0 0 P(A) P(AC) P(B) P(BC)
  • Using the diagram • In fact, this diagram can be used for all four of the relations! (pay attention to the zeros) • Disjoint: A B0 AC and Bc P(A) P(AC) P(B) P(BC)
  • Using the diagram • In fact, this diagram can be used for all four of the relations! (pay attention to the zeros) • Implied: 0 BA and B AC and Bc P(A) P(AC) P(B) P(BC)
  • Using the diagram • This is the “Normal” diagram. • Multiply the edges to find the small box probabilities A BA and B AC and Bc P(A) P(AC) P(B) P(BC)
  • Using the diagram • This is the “Normal” diagram. • All four boxes add to 1 A and Bc Ac and BA and B AC and Bc P(A) P(AC) P(B) P(BC)
  • Assignment 6.2B • Pg 423 #37, 39, 43, 44
  • 6.3 GENERAL PROBABILITY RULES
  • Rules of Probability 1. 0 < P(A) < 1 2. P(S) = 1 3. If A and B are disjoint, P(A or B) = P(A) + P(B) 4. P(AC) = 1 – P(A) 5. If A and B are independent P(A and B) = P(A) x P(B)
  • General Addition Rule • For any two events: P(A or B) = P(A) + P(B) – P(A and B) Let‟s examine this by looking at the four possible event pairs
  • General Addition Rule P(A or B) = P(A) + P(B) – P(A and B) If the events are disjoint or complimentary, P(A and B) = 0 B (or AC )
  • A BA and B Ac and Bc P(A) P(AC) P(B) P(BC) General Addition Rule If A and B are overlapping sets P(A or B) = P(A) + P(B) – P(A and B)
  • A BA and B Ac and Bc P(A) P(AC) P(B) P(BC) General Addition Rule If A and B are overlapping sets P(A or B) = P(A) + P(B) – P(A and B)
  • A BA and B Ac and Bc P(A) P(AC) P(B) P(BC) General Addition Rule P(A or B) = P(A) + P(B) – P(A and B) If A and B are overlapping sets Added twice! Need to subtract one of these!
  • General Addition Rule • Implied events P(A or B) = P(A) + P(B) – P(A and B) S A B
  • General Addition Rule • Implied events P(A or B) = P(A) + P(B) – P(A and B) S A B
  • General Addition Rule • Implied events P(A or B) = P(A) + P(B) – P(A and B) S A B Added Twice! We must subtract it out.
  • Assignment 6.3A • Page 430 #45-49 odd, 61, 66, 67, 69
  • Conditional Probability • When two events are not independent, then their probabilities are known as “conditional” • Notation: P(A | B) reads “the probability of A given B” this is “the probability that event A occurs, if event B has already occurred”
  • Conditional Probability • P(A and B) = P(A) x P(B|A) • This should make sense: “the probability that A and B occurs is the probability of A occurs times the probability that B occurs if A has occurred. • Really, this is just the multiplication principle again!
  • Conditional Probability • After rearranging the previous equation, we arrive at the definition for conditional probability: A and B (B | A) A P P P
  • Conditional Probability • We also surmise a mathematical definition for “independent events” Two events are said to be independent if both of the following are true (1) P(B|A) = P(B) and (2) P(A|B) = P(A).
  • Tree Diagrams and Probability • When multiple events occur, many times a tree diagram is helpful in computing the probabilities for each outcome of the sample space. • Outcomes are written on the “nodes” • Probabilities are written on the “branches” • Probabilities for all branches from the same node must add to „1‟ • Probabilities of each outcome in the sample space is a product of each branch in the pathway
  • Tree Diagram and Probability Of all high school male athletes who attend college, 1.7% will become professional athletes. Of the high school male athlete who does not go to college, .01% will go on to become professionals. 5% of all high school male athletes go to college. What percent of high school male athletes become professional athletes?
  • Tree Diagrams and Probability • Let‟s define events: A = “a high school male athlete goes to college” B = “becomes a professional” • Notice what AC and BC are.
  • Tree Diagrams and Probability
  • Tree Diagrams and ProbabilityBranches from the same node add to „1‟
  • Probability is product of the branches P(A and B) .05 x .017 .00085 P(A and BC) .04915 P(ACand BC) .949905 P(AC and B) .000095
  • Tree Diagrams and Probability • P(B) = P(A and B) + P(AC and B) P(B) = .00085 + .000095 P(B) = .000945 summarize: “.09% of all high school male athletes become professionals”
  • Two way tables and probability • An alternate way to work on these problems is to use a two way table. • Choose a sufficiently large number for your population • Use the multiplication property and the complementary sets to complete the table.
  • Two way tables and probability A new test for disease “rawr” is developed. If a patient has rawr the test will give a positive result (the patient has rawr) 98% of the time. Unfortunately, if a patient does not have rawr, the test will give a positive 1% of the time. Approximately 4% of the population has rawr.
  • Two way tables and probability 1- What percent of the population will get a positive test result? 2- What is the probability that patient has rawr if he gets a positive result?
  • Two way tables and probability • Lets assume our population is 10000! Positive Negative Total Rawr No Rawr Total 10000
  • Two way tables and probability • 4% of the population has Rawr. Positive Negative Total Rawr 400 No Rawr 9600 Total 10000
  • Two way tables and probability • 98% of the “Rawrs” will test positive Positive Negative Total Rawr 392 8 400 No Rawr 9600 Total 10000
  • Two way tables and probability • 1% of the “no Rawrs” will test positive Positive Negative Total Rawr 392 8 400 No Rawr 96 9504 9600 Total 10000
  • Two way tables and probability • Do some quick addition Positive Negative Total Rawr 392 8 400 No Rawr 96 9504 9600 Total 488 9512 10000
  • Two way tables and probability 1- What percent of the population will get a positive test result? 488/10000 = .0488
  • Two way tables and probability 2- What is the probability that patient has rawr if he gets a positive result? • P(rawr | positive result) – Look at the table! – 392/488 = .8033 • Conditional probability becomes conditional distribution problem from chapter 4! • These results of this example should worry you. Why?
  • Assignment 6.3B • P441 #70-73, 80-83, 86(a)-(d), 87, 90, 91