 # Chapters 14 & 15 Probability math2200. Random phenomenon In a random phenomenon we know what could happen, but we don’t know which particular outcome.

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Chapters 14 & 15 Probability math2200

Random phenomenon In a random phenomenon we know what could happen, but we don’t know which particular outcome did or will happen. Example: the color of the traffic light at a particular intersection.

Accumulated percentage Day Light fraction [1,] G 1.0000000 [2,] G 1.0000000 [3,] G 1.0000000 [4,] R 0.7500000 [5,] G 0.8000000 [6,] G 0.8333333 [7,] G 0.8571429 [8,] R 0.7500000 [9,] R 0.6666667 [10,] G 0.7000000

Probability Each occasion upon which we observe a random phenomenon is called a trial. We observe the value of the random phenomenon and call it an outcome. The collection of all possible outcomes forms the sample space. When we combine outcomes, the resulting combination is an event. An event occurs if the outcome of the trial is in this event. Trials are independent if the outcome of one trial does not influence the outcome of the others. The long-run relative frequency of an event is called the probability of the event.

Example: Two coins Trial: flip two coins at the same time Outcome: one of the four combinations of heads (H) or tails (T) Sample space: all possible outcomes –S = {HH,TT,HT,TH} Events: –Event A : two coins give the same results –{ HH,TT } –Event B : two coins give different results –{ HT,TH }

The Law of Large Numbers (LLN) The Law of Large Numbers (LLN) says that the long-run relative frequency of repeated independent events gets closer and closer to a single value. We call the value the probability of the event. this definition of probability is often called empirical probability because it is based on repeatedly observing the event’s outcome,. Jacob Bernoulli

CAUTION: The Nonexistent Law of Averages The LLN says nothing about short-run behavior. Relative frequencies even out only in the long run (infinitely long, in fact). The so called Law of Averages (that an outcome of a random event that hasn’t occurred in many trials is “due” to occur) doesn’t exist at all.

Modeling probability: Example When probability was first studied, a group of French mathematicians looked at games of chance in which all the possible outcomes were equally likely. –It’s equally likely to get any one of six outcomes from the roll of a fair die. –It’s equally likely to get heads or tails from the toss of a fair coin.

If Event A is made up of several equally likely outcomes. P(A) = Example: the probability of drawing a non-face card (A-10) from a deck? Probability (cont.) # of outcomes in A # of possible outcomes

Personal Probability In everyday speech, when we express a degree of uncertainty without basing it on long-run relative frequencies, we are stating subjective or personal probabilities. Personal probabilities don’t display the kind of consistency that we will need probabilities to have.

Formal Probability 1.A probability must be a number between 0 and 1. For any event A, 0 ≤ P(A) ≤ 1.

Formal Probability (cont.) 2.Probability Assignment Rule: –The probability of the set of all possible outcomes of a trial must be 1. P(S) = 1 (S denotes the set of all possible outcomes.)

Formal Probability (cont.) rules of computations 3.Complement Rule:  The set of outcomes that are not in the event A is called the complement of A, denoted A C.The probability of an event occurring is 1 minus the probability that it doesn’t occur: P(A) = 1 – P(A C )

Formal Probability - Notation Notation alert: In this class we use the notation P(A or B) and P(A and B). In other situations, you might see the following: –P(A  B) instead of P(A or B) –P(A  B) instead of P(A and B)

Formal Probability (cont.) 4.Addition Rule: –Events that have no outcomes in common (and, thus, cannot occur together) are called disjoint (or mutually exclusive).

Formal Probability (cont.) 4.Addition Rule (cont.): –For two disjoint events A and B, the probability that one or the other occurs is the sum of the probabilities of the two events. P(A or B) = P(A) + P(B), provided that A and B are disjoint. For more events: P(A or B or C) = P(A) + P(B) + P(C), where A, B and C are mutually exclusive

Be careful about natural language! –Often, “or” in our natural language has an exclusive meaning as in “Would you like the steak or the vegetarian entrée?”. –In this class, when we ask for the probability that A or B occurs, we mean A or B or both. –P(A or B but not both) = P(A or B) – P(A and B) = P(A) + P(B) – 2 * P(A and B)

The General Addition Rule (cont.) General Addition Rule: –For any two events A and B, P(A or B) = P(A) + P(B) – P(A and B) –P(A or B but not both) = P(A or B) – P(A and B) = P(A) + P(B) – 2 * P(A and B)

General addition rule For three events: P(A or B or C) = P(A) + P(B) + P(C)- P(A and B) – P(A and C) – P(B and C) + P(A and B and C)

Formal Probability 5.Multiplication Rule (cont.): –For two independent events A and B, the probability that both A and B occur is the product of the probabilities of the two events. P(A and B) = P(A) x P(B), provided that A and B are independent. P(A and B and C) = P(A) x P(B) x P(C), if A and B and C are mutually independent

Formal Probability (cont.) 5.Multiplication Rule (cont.): –Two independent events A and B are not disjoint, provided the two events have probabilities greater than zero:

Multiplication rule For two independent events A and B, the probability that both A and B occur is the product of the probabilities of the two events. P(A and B) = P(A) * P(B), if A and B are independent For more events, P(A and B and C) = P(A) * P(B) * P(C), if A and B and C are mutually independent

Example: M&M In 2001, Masterfoods decided to add another color to the standard color lineup of brown, yellow, red, orange, blue and green. To decide which color, they surveyed kids in nearly every country and asked them to vote for purple, pink and teal.

Example: M&M In Japan, the result is somehow different. –38% for pink –36% for teal –16% purple What is the probability that a Japanese M&M’s survey respondent selected at random preferred either pink or teal?

Example (Cont’) If we pick two respondents at random, what’s the probability that they both said pink or teal?

Example (cont’) If we pick three respondents at random, what’s the probability that at least one preferred purple? Solution : Let A be the event that at least one preferred purple. Therefore A C is the event that non of the three prefer purple. P(A)=1-P(A C ) = 1- (0.84)^3=40.73% Solution 2 : P(A) =P (Exactly one respondent likes purple) + P (Exactly two like purple)+ P (Exactly three like purple)= 3*0.16*0.84*0.84 +3*0.16*0.16*0.84 + 0.16^3 = 40.73%

What can go wrong? Be aware of probabilities that don’t add up to 1. Don’t add probabilities of events if they are NOT disjoint. Don’t multiply probabilities of events if they are NOT independent. Independent ≠ Disjoint

Disjoint events cannot be independent! Well, why not? –Since we know that disjoint events have no outcomes in common, knowing that one occurred means the other didn’t. –Thus, the probability of the second occurring changed based on our knowledge that the first occurred. –It follows that the two events are not independent. A common error is to treat disjoint events as if they were independent, and apply the Multiplication Rule for independent events— don’t make that mistake.

Example Police report that 78% of drivers stopped on suspicion of drunk driving are given a breath test, 36% a blood test, and 22% both tests. What is the probability that a randomly selected suspect is given –a blood test or a breath test? –neither test?

A = {the suspect is given a breath test} B = {the suspect is given a blood test} We know that –P(A) = 0.78 –P(B) = 0.36 –P(A and B) = 0.22

P(A or B) = P(A) + P(B) – P(A and B) =0.78 + 0.36 – 0.22 = 0.92 P(A or B but NOT both) –P(A or B) – P(A and B) = 0.92 – 0.22 = 0.70 –P(A or B c ) + P(B or A c ) P(A c and B c ) = 1 – P(A or B)

Checking disjoint Are giving a suspect a blood test and a breath test mutually exclusive? –This is to see whether P(A and B) = 0 –In this case, P(A and B) = 0.22 –So, not mutually exclusive

Checking independence Are giving a suspect a blood test and a breath test independent? –P(A and B) = P(A) * P(B)?

Example Two psychologists surveyed 478 children. They asked the students whether their primary goal was –to get good grades –to be popular –or to be good at sports Purpose of the study –Did boys and girls at this age have similar goals?

Conditional distributions gradespopularsportsTotal Girl130 (51.79%)91 (36.25%)30 (11.95%)251 gradespopularsportsTotal Boy117 (51.54%)50 (22.03%)60 (26.43%)227

Conditional probability the probability of an event from a conditional distribution is noted P(B|A) and pronounce it “the probability of B given A.” A probability that takes into account a given condition is called a conditional probability.

Conditional probability of B given A Note: P(A) cannot equal 0, since we know that A has occurred.

The General Multiplication Rule Rearranging the equation in the definition for conditional probability, we get the General Multiplication Rule: –For any two events A and B, P(A and B) = P(A) x P(B|A) or P(A and B) = P(B) x P(A|B)

Independence Definition: Events A and B are independent whenever P(B|A) = P(B) (or P(A|B) = P(A)). Using the general multiplication rule, we can see that it’s equivalent to P(A and B) = P(A) * P(B). P(A and B) = P(A) * P(B|A) = P(A) * P(B) ( or P(A) and B) = P(B)*P(A|B) = P(B) *P(A) )

Trial: select a student at random (equally likely) and asks about his/her goal. Sample space: 478 students What is the probability the selected student is a girl? (251/478 = 0.525) P (girl and popular) = ? P (sports) = 90/478 = 0.188 gradespopularsportsTotal Boy1175060227 Girl1309130251 Total24714190478

Conditional probability Given that the selected student is a girl, what is the probability the selected student’s goal is sports? P (sports | girl) = P (sports and girl) / P (girl) = (30/478) / (251/478) =30/251= 0.12 P (sports | boy) = 60/227 = 0.264 P (sports) ≠ P (sports | girl) + P (sports | boy) P (girl | sports) = ?

Example Police report that 78% of drivers stopped on suspicion of drunk driving are given a breath test, 36% a blood test, and 22% both tests. Are giving a DWI suspect a blood test and a breath test independent? P(B|A) = P(A and B) / P(A) = 0.22 / 0.78 = 0.28 P(B) = 0.36 Therefore P(B|A) ≠ P(B), the event of blood test and the event of breath test are not independent

Drawing Without Replacement Sampling without replacement means that once one object is drawn it doesn’t go back into the pool. (It does not matter to large population). Suppose that 12 rooms left when it is time for you and your friend to draw. What is the probability that both of you get rooms in Gold Hall? (1/22) –Three are in Gold Hall –Four in Silver Hall –Five in Wood Hall

Example: Binge drinking A study by Harvard School of Public Health, for college students –44% engage in binge drinking –37% drink moderately –19% abstain entirely Meanwhile, another study –Among binge drinkers, 17% involved in alcohol- related automobile accident –Among students who drink moderately, only 9% have been involved

Tree Diagrams A tree diagram helps us think through conditional probabilities by showing sequences of events as paths that look like branches of a tree.

Reversing the conditioning If we know a student has had an alcohol- related accident, what is the probability that the student is a binge drinker? –P (binge | accident) = ? –P (binge and accident) / P (accident) –P (accident) = P (binge and accident) + P (moderate and accident) + P (abstain and accident) = 0.075+ 0.033 + 0 = 0.108

Bayes’ rule What is P(B|A), how do we get P(A|B) P(B|A) = P(A and B)/P(A) = P(A|B)*P(B)/P(A) P(A) = P(A and B) + P(A and B c ) = P(A|B)*P(B) + P (A | B c )* P(B c ) P(B|A) = P(A|B)*P(B)/[P(A|B)* P (B) + P(A| B c )* P(B c ). We need P(B), P(A|B) and P(A | B c )

What can go wrong? Be aware of probabilities that don’t add up to 1. Don’t add probabilities of events if they are NOT disjoint. Don’t multiply probabilities of events if they are NOT independent. Independent ≠ Disjoint

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