Chapter 17: Geometric models

Chapter 17: Geometric models

Chapter 17: Geometric modelsAP Statistics B1Overview of Chapter 17Two new models: Geometric model, and the Binomial modelYes, the binomial model involves Pascals triangles that (I hope) you learned about in Algebra 2Use the geometric model whenever you want to find how many events you have to have before a successUse the binomial model to find out how many successes occur within a specific number of trials

2Todays coverageIntroduction to the vocabulary:Bernoulli trialsGeometric probability modelBinomial probability modelExamples of both geometric and binomial probability modelsNature of the geometric model (and review of series from Algebra 2)When to use the geometric model/practice on problems/solutionsFinally, how to use the TI calculators to calculate probabilities by the geometric model

3Vocabulary: Bernoulli trialsBernoulli trialsThe only kind we do in Chapter 17Need to have definition firmly in mind3 requirements:There are only two possible outcomesProbability of success is constant (i.e., doesnt change over time)Trials are independent

4Vocabulary: nomenclature for Bernoulli trialsWere going to start using s for success and f for failure (duh)Soon, however, we will switch to p for success and q for failure (dont ask why.)Remember, remember, remember!p+q=1(s+f, too!)

5Vocabulary: geometric and binomial models of probabilityGeometric probability model:Counts the number of Bernoulli trials before the first successBinomial probability model:Counts the number of successes in the first n trials (doesnt have to be just one, as in the geometric model)

6Examples: the geometric modelsExample: tossing a coinSuccess=heads; failure=tailsCOMPETELY ARBITRARYin the examples we will reverse success and failure without any problems, so dont get hung up on itBetter way of thinking about itbinary, either/or

7Examples: asking the geometry model questionTypical question: What is the probability of not getting heads until the 5th toss of the coin?Many geometric model questions are going to look like this:f f f f s (no success until 5th toss)

In terms of p and q, it looks like q q q q p

We are talking sequences here!

8Example: contrast geometric with the binomial modelIn the binomial model, we ask questions like how many ways can we have exactly two successes in 5 Bernoulli trials?You would get a distribution like that on the right:

s s f f fs f s f fs f f s fs f f f sf s s f ff s f s ff s f f sf f s s ff f s f sf f f s s

9Example: binomial model using p and q instead of s and fAn identical model to that of the last slide appears at the leftThis one, however, uses the p (success) and q (failure) that the textbook usesThe patterns, however, are identicalp p q q qp q p q qp q q p qp q q q pq p p q qq p q p qq p q q pq q p p qq q p q pq q q p p

10Examples: geometric v. binomialToday Geometric models, tomorrow, BinomialThe Geometric model is somewhat easier to followThe Binomial Model requires quite a bit more mathTomorrow, Im going to show you a lecture by Arthur Benjamin on binomial math ( hour)Professor of Math, Harvey Mudd College (Claremont Colleges)Good instructor, makes my jokes look less cornyIrksome mannerisms, but great content

11Nature of the geometric model: first example (tossing a coin)Lets start with flipping coinsWhat is success?Lets define it as getting heads as a result (p)So tails is qProbabilities?p=0.5q=0.5

12Nature of the geometric model:framing the questionQ: What are the chances of not getting heads until the 4th toss?

13Nature of the geometric model:doing the calculationsProbability for failure is q, or q3 for 3 successive failures (i.e., not getting heads until the 4th toss)Probability for success on 4th try is pTotal probability is therefore q3pReplace with numbers: (0.5)3(0.5)=(0.125)(0.5)=0.0625

14Nature of the geometric model:the formulas (formulae for you pedants)Unfortunately, to derive most of the formulas we use, you have to use calculusThis will be one of the few times where youre simply going to have to memorize the equations (at least until you get to college and take calculus!)Sorry, sorry, sorry!

15Nature of the geometric model:are we there yet?In other words, how many trials do we need until we succeed?Using p and q nomenclature, where x=number of trials until the first success occurs:P(X=x) = qx-1pRemember our coin-tossing model: 4 times until we got heads (fill in the equation)You will use this a lot to calculate probabilities!

16Nature of the geometric model:the mean and standard deviationAka expected value, which equals E(X)=1/p, where p=probability of success

Standard deviation

Sadly, you just gotta memorize these!

17Nature of the geometric model:summaryP(X=x) = qx-1p, where x= number of trials before first success

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18Practice:Exercise 7Basketball player makes 80% of his shots. Lets set things up before we start.p=0.8, so q=0.2 (he makes 80% of his shots and misses 20%)Dont calculate the mean just yet, because Im going to show you that the definition of success often changes in the middle of the question!

19Practice:Exercise 7(a)Misses for the first time on his 5th attemptUse the probability model, except notice something really, really bizarre: the 5th attempt appears to be a failure!Thats right, a failure!!!But its considered to be the success, so we have to reverse things

20Practice, Exercise 7(a): setting up the calculationP(X=x) = qx-1p is the formula.Here, this translates as (.8)4 0.2Yes, I **know** its bizarre looking at the success as a failure, but hey..Multiply this out on your calculators, and you should get..0.08192? Everybody get that?Books says 0.0819, or about 8.2% of the time will he not miss until the fifth shot

21Practice, Exercise 7(a):lessonsYou can interchange failure for success in the probability model without problemsYou have to read the problem VERY carefully and not simply apply a formula. Had you done so here, and raised the MISSED basket to the 4th power, you would have gotten a completely wrong answerFailure depends on context! What normally seems like failure (i.e., not making a basket) can be defined as success. Binary would probably be a better term than success and failure(you heard it here, first)

22Practice, Exercise 7(b):a more normal set-upQ: he makes his first basket on his fourth shot.Except for reversing p and q, its the same as (a): P(X=x) = qx-1p is the formula. P(miss 3 baskets before success)= (.2)3 0.8=0.0064Very straightforward

23Practice, Exercise 7(c):a trick you need to learnQuestion (c): makes his first basket on one of his first three shots.Here, we need to make a chart of all possibilities that fit the configuration (p=success/made basket, q=failure/missed):pqqqpqpppppqqpppqpqqp

24Practice, Exercise 7(c):the long wayOn the right is a chart of all 7 possibilitiesWith each possibility is the percent of the time it happensIt al adds up to 0.992All these had to be assembled by hand applying the formulae in (a) and (b)ConfigurationProbabilitypqq0.032ppq0.128pqp0.128ppp0.512qpq0.128qpp0.128qqp0.032

25Practice, Exercise 7(c):the easy wayIf you have 2 possible outcomes and 3 trials, you will have 23 possible combinationsWe could get the 7 of 8 that we did in the previous slideOr, we can be clever: getting at least one basket in your first three shots is the complement of getting NO baskets in your first three shots, i.e., having three misses.

26Practice, Exercise 7(c):the easy way/calculationsSo P(X)=1-failure to get any baskets in first three shotsThis equals 1-(0.2)3=1-0.008=0.992Which would you rather have in YOUR wallet? (oops.sorry, wrong commercial)which would you rather spend your time on?27

Practice, Exercise 9:expected number of shots until missThis is really a reading problem.what does expected number of shots until misses mean?It means, if you will excuse an unintentional pun, the mean, which equals 1/p.Now, the only question is, whats p? Here, the success is missing. So the mean is 1/0.2 = 5.28

Practice, Exercise 11:the AB blood problemNB: your instructor has AB+ blood. The Red Cross is always VERY glad to see me.0.04 of all people have AB blood (were a rare breed)This problem will involve finding the mean as well as doing the probability calculations29

Practice, Exercise 11(a):using the meanQ: On average, how many donors must be checked to find someone with Type AB blood?Classic case (on average is a clue!) of using the mean.Mean is 1/p = 1/0.04 = 2530

Practice, Exercise 11(b):the easy wayQ: Whats the probability that there is a Type AB donor among the first 5 people checked?Problem: there are 32 possible outcomes! (25)So lets be clever (again)This is the same as asking whats the probability of getting NO AB donors in the first 5?Thats equal to (0.96)5=0.8154. Subtract that answer from one for 0.1846, which is the answer to the question.31

Practice, Exercise 11(c):similar to (b)Asking whats the probability that the first AB donor will be found among the first 6 people is the same thing as subtraction the probability of NO AB donors from 1.No AB donors is (0.96)6 = 0.7828Complement is 1 0.7828 = .217232

Practice, Exercise 11(d):

Q: whats the probability that we wont find an AB donor before the 10th person? Similar to saying we wont find any AB donors in the first NINE peopleThats (0.96)9=0.69333

Homework for tomorrow

Ch 17, problems 8, 10, 12, 13, and 14.34