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Chapter 12 From Randomness to Probability Copyright © 2014, 2012, 2009 Pearson Education, Inc. 1 Objectives: State the definition of trial, outcome, sample space, event and P(A). 41. Apply the Law of Large Numbers. 42. Recognize when events are disjoint and when events are independent. 43. State the basic definitions and apply the rules of probability for disjoint and independent events. 40. Copyright © 2014, 2012, 2009 Pearson Education, Inc. Slide 1- 2 2 12.1 Random Phenomena Copyright © 2014, 2012, 2009 Pearson Education, Inc. 3 Red Light, Green Light Each day you drive through an intersection and check if the light is red, green, or yellow. • Day 1: green • Day 2: red • Day 3: green Before you begin, you know: • The possible outcomes • An outcome will occur. After you finish, you know: • The outcomes that occurred Copyright © 2014, 2012, 2009 Pearson Education, Inc. 4 Random Phenomena Vocabulary Trial • Each occasion which we observe a random phenomena Outcome • The value of the trial for the random phenomena Event • The combination of the trial’s outcomes Sample Space • The collection of all possible outcomes Copyright © 2014, 2012, 2009 Pearson Education, Inc. 5 Flipping Two Coins Trial • The flipping of the two coins Outcome • Heads or tails for each flip Event • HT, for example Sample Space • S = {HH, HT, TH, TT} Copyright © 2014, 2012, 2009 Pearson Education, Inc. 6 The Law of Large Numbers • If you flip a coin once, you will either get 100% heads or 0% heads. • If you flip a coin 1000 times, you will probably get close to 50% heads. The Law of Large Numbers states that for many trials, the proportion of times an event occurs settles down to one number. •This number is called the empirical probability. Copyright © 2014, 2012, 2009 Pearson Education, Inc. 7 The Law of Large Numbers Requirements Identical Probabilities • The probabilities for each event must remain the same for each trial. Independence • The outcome of a trial is not influenced by the outcomes of the previous trials. Empirical probability • # times A occurs P(A) # of trials (in the long run) Copyright © 2014, 2012, 2009 Pearson Education, Inc. 8 Red Light, Green Light: The Law of Large Numbers • After many days, the proportion of green lights encountered is approximately 0.35. • P(green) = 0.35. • If we recorded more days, the probability would still be about 0.35. Copyright © 2014, 2012, 2009 Pearson Education, Inc. 9 The Nonexistent Law of Averages Wrong • If you flip a coin 5 times and get five tails, then you are due for a head on the next flip. • You put 10 quarters in the slot machine and lose each time. You are just a bad luck person, so you have a smaller chance of winning on the 11th try. • Example: suppose a couple has 3 children all of whom are boys, is the couple more likely to have a girl for the next child? • There is no such thing as the Law of Averages for short runs. Copyright © 2014, 2012, 2009 Pearson Education, Inc. 10 12.2 Modeling Probability Copyright © 2014, 2012, 2009 Pearson Education, Inc. 11 Theoretical Probability American Roulette • 18 Red, 18 Black, 2 Green • If you bet on Red, what is the probability of winning? Theoretical Probability • • # of outcomes in A P(A) = # of possible outcomes 18 P(red) 38 Copyright © 2014, 2012, 2009 Pearson Education, Inc. 12 Heads or Tails Flip 2 coins. Find P(HH) • List the sample space: • S = {HH, HT, TH, TT} • P(HH) = ¼ Flip 100 coins. Find the probability of all heads. • The sample space would involve 1,267,650,600,228,229,401,496,703,205,376 different outcomes. • Later, we will see a better way. Copyright © 2014, 2012, 2009 Pearson Education, Inc. 13 Equally Likely? What’s wrong with this logic? • Randomly pick two people. • Find the probability that both are left-handed. • Sample Space S = {LL, LR, RL, RR} • P(LL) = ¼ Since left-handed and right-handed are not equally likely, this method does not work. Copyright © 2014, 2012, 2009 Pearson Education, Inc. 14 Personal Probability What’s your chance of getting an A in statistics? • You cannot base this on your long-run experience. • There is no sample space of events with equal probabilities to list. • You can only base your answer on personal experience and guesswork. • Probabilities based on personal experience rather than long-run relative frequencies or equally likely events are called personal probabilities. Copyright © 2014, 2012, 2009 Pearson Education, Inc. 15 12.3 Formal Probability Copyright © 2014, 2012, 2009 Pearson Education, Inc. 16 Rules 1 and 2 Rule 1: 0 ≤ P(A) ≤ 1 • You can’t have a −25% chance of winning. • A 120% chance also makes no sense. • Note: Probabilities are written in decimals. • 45% chance → P(A) = 0.45 Rule 2: P(S) = 1 • The set of all possible outcomes has probability 1. • There is a 100% chance that you will get a head or a tail. Copyright © 2014, 2012, 2009 Pearson Education, Inc. 17 Rule 3: The Complement Rule Complements • Define AC as the complement of A. • AC is the event of A not happening. • If A is the event of rolling a 5 on a six sided die, then AC is the event of not rolling a 5: {1, 2, 3, 4, 6} • P(A) = 1/6. P(AC) = ? • P(AC) = 5/6 = 1 – 1/6 The Rule of Complements: P(AC) = 1 – P(A) Copyright © 2014, 2012, 2009 Pearson Education, Inc. 18 Red Light Green Light and Complements We know that P(green) = 0.35. Find P(not green) • Not green is the complement of green. • Use the rule of complements: • P(not green) = P(greenC) = 1 – P(green) = 1 – 0.35 = 0.65 • The probability of the light not being green is 0.65. Copyright © 2014, 2012, 2009 Pearson Education, Inc. 19 Rule 4: The Addition Rule Suppose P(sophomore) = 0.2 and P(junior) = 0.3 • Find P(sophomore OR junior) • Solution: 0.2 + 0.3 = 0.5 • This works because sophomore and junior are disjoint events. They have no outcomes in common. The Addition Rule • If A and B are disjoint events, then P(A OR B) = P(A) + P(B) Copyright © 2014, 2012, 2009 Pearson Education, Inc. 20 Examples: suppose you roll a die. What is the probability that you roll a 5 or 6? Are these events disjoint? What about the probability that you roll a even number or a number less than 3? Are these events disjoint? Copyright © 2014, 2012, 2009 Pearson Education, Inc. 21 Red Light, Green Light, Yellow Light Given that P(green) = 0.35 and P(yellow) = 0.04 • Find P(red). • Solution: Use the Rule of Complements and the Addition Rule. • P(red) = 1 – P(redC) = 1 – P(green OR yellow) = 1 – [P(green) + P(yellow)] = 1 – [0.35 + 0.04] = 1 – 0.39 = 0.61 Copyright © 2014, 2012, 2009 Pearson Education, Inc. 22 The Sum of Probabilities The sum of all the probabilities of every disjoint event must equal 1. • What’s wrong with the following statement? • Probabilities for freshmen, sophomore, junior, senior are: 0.25, 0.23, 0.22, 0.20. • 0.25 + 0.23 + 0.22 + 0.20 = 0.90 • Since they do not add to 1, something is wrong. • How about the following? • P(owning a smartphone) = 0.5 and P(owning a computer) = 0.9 • This is fine, since they are not disjoint. Copyright © 2014, 2012, 2009 Pearson Education, Inc. 23 Rule 5: The Multiplication Rule The probability that an Atlanta to Houston flight is on time is 0.85. • If you have to fly every Monday, find the probability that your first two Monday flights will be on time. Multiplication Rule: For independent events A and B: P(A AND B) = P(A) × P(B) • P(1st on time AND 2nd on time) = P(1st on time) × P(2nd on time) = 0.85 × 0.85 = 0.7225 Copyright © 2014, 2012, 2009 Pearson Education, Inc. 24 Example: the outcomes of rolling 2 distinct dice are independent. What is the probability of rolling a 1 then a 2? What is the probability of rolling an even number on the first die and then a number greater than 4 on the second? What is the probability of rolling a 5 four times in Copyright © 2014, 2012, 2009 Pearson Education, Inc. 25 Putting the Rules to Work In most situations where we want to find a probability, we’ll often use the rules in combination. A good thing to remember is that sometimes it can be easier to work with the complement of the event we’re really interested in. Copyright © 2014, 2012, 2009 Pearson Education, Inc. Slide 1- 26 26 11) the plastic arrow on a spinner for a child’s game stops rotating to point at a color that determines what will happen next. Which of the following probability assignments are possible? Red Yellow Green Blue a) 0.25 0.25 0.25 0.25 b) 0.10 0.20 0.30 0.40 c) 0.20 0.30 0.40 0.50 d) 0 0 1.00 0 e) 0.10 0.20 1.20 -1.50 Copyright © 2014, 2012, 2009 Pearson Education, Inc. Slide 1- 27 27 Red Light AND Green Light AND Yellow Light Find the probability that the light will be red on Monday, green on Tuesday, and yellow on Wednesday. • The multiplication rule works for more than 2 events. • P(red Mon. AND green Tues. AND yellow Wed.) = P(red Mon.) × P(green Tues.) × P(yellow Wed.) = 0.61 × 0.35 × 0.04 = 0.00854 Copyright © 2014, 2012, 2009 Pearson Education, Inc. 28 At Least One Red Light Find the probability that the light will be red at least one time during the week. • Use the Complement Rule. • P(at least 1 red) = 1 – P(no reds) = 1 – (0.39 × 0.39 × 0.39 × 0.39 × 0.39 × 0.39 × 0.39) ≈ 0.9986 Copyright © 2014, 2012, 2009 Pearson Education, Inc. 29 In a large introductory statistics class, the professor reports that 55% of the students have never taken Calculus, 32% have taken one semester of Calculus, and the rest have taken two or more semesters of Calculus. The professor randomly assigns students to work in groups of three. • What is the probability that the first groupmate • 2+ semesters of Calculus? • Some Calculus? • No more than one semester of Calculus? you meet has studied • What is the probability that, of your two groupmates • Neither has studied Calculus? • Both have studied at least one semester of Calculus? • At least one has had more than one semester of Calculus? Copyright © 2014, 2012, 2009 Pearson Education, Inc. Slide 1- 36 36 32) The American Red Cross says that about 45% of the U.S. population has Type O blood, 40% Type A, 11% Type B, and the rest Type AB. • Someone volunteers to give blood, what is the probability that this donor • Has type AB blood? • Has type A or Type B blood? • Is not Type O? • Among four potential donors, what • All are type O? • No one is Type AB? • They are not all Type A? • At least one person is Type B? is the probability that Copyright © 2014, 2012, 2009 Pearson Education, Inc. Slide 1- 37 37 34) The American Red Cross says that about 45% of the U.S. population has Type O blood, 40% Type A, 11% Type B, and the rest Type AB. • If you examine one person, are the events that the person is Type A and that the person is Type B disjoint, independent, or neither? • If you examine two people, are the events that the first is Type A and the second Type B disjoint, independent, or neither? • Can disjoint events ever be independent? Copyright © 2014, 2012, 2009 Pearson Education, Inc. Slide 1- 38 38 What Can Go Wrong? • Beware of probabilities that don’t add up to 1. • If they add to less than 1, look for another category. • If they add to more than 1, maybe they are not disjoint. • Don’t add probabilities of events if they are not disjoint. • Events must be disjoint to use the Addition Rule. Copyright © 2014, 2012, 2009 Pearson Education, Inc. 39 What Can Go Wrong? • Don’t multiply probabilities of events if they are not independent. • P(over 6’ and on basketball team) is not equal to P(over 6’) × P(on basketball team) • Don’t confuse disjoint and independent • Disjoint events are never independent. Copyright © 2014, 2012, 2009 Pearson Education, Inc. 40