2Overview of Chapter 17Two new models: Geometric model, and the Binomial modelYes, the binomial model involves Pascal’s 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 “success”Use the binomial model to find out how many successes occur within a specific number of trials
3Today’s coverage Introduction 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
4Vocabulary: Bernoulli trials The 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., doesn’t change over time)Trials are independent
5Vocabulary: nomenclature for Bernoulli trials We’re 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 (don’t ask why….)Remember, remember, remember!—p+q=1(s+f, too!)
6Vocabulary: geometric and binomial models of probability Geometric probability model:Counts the number of Bernoulli trials before the first successBinomial probability model:Counts the number of successes in the first n trials (doesn’t have to be just one, as in the geometric model)
7Examples: the geometric models Example: tossing a coinSuccess=heads; failure=tailsCOMPETELY ARBITRARY—in the examples we will reverse success and failure without any problems, so don’t get hung up on itBetter way of thinking about it—binary, either/or
8Examples: asking the geometry model question Typical 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 pWe are talking sequences here!
9Example: contrast geometric with the binomial model In 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
10Example: binomial model using p and q instead of s and f An 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
11Examples: geometric v. binomial Today Geometric models, tomorrow, BinomialThe Geometric model is somewhat easier to followThe Binomial Model requires quite a bit more mathTomorrow, I’m going to show you a lecture by Arthur Benjamin on binomial math (½ hour)Professor of Math, Harvey Mudd College (Claremont Colleges)Good instructor, makes my jokes look less cornyIrksome mannerisms, but great content
12Nature of the geometric model: first example (tossing a coin) Let’s start with flipping coinsWhat is success?Let’s define it as getting heads as a result (p)So tails is qProbabilities?p=0.5q=0.5
13Nature of the geometric model: framing the question Q: What are the chances of not getting heads until the 4th toss?
14Nature of the geometric model: doing the calculations Probability 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
15Nature 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 you’re simply going to have to memorize the equations (at least until you get to college and take calculus!)Sorry, sorry, sorry!
16Nature 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!
17Nature of the geometric model: the mean and standard deviation Aka expected value, which equals E(X)μ=1/p, where p=probability of successStandard deviationSadly, you just gotta memorize these!
18Nature of the geometric model: summary P(X=x) = qx-1p, where x= number of trials before first success23
19Practice: Exercise 7 Basketball player makes 80% of his shots. Let’s set things up before we start.p=0.8, so q=0.2 (he makes 80% of his shots and misses 20%)Don’t calculate the mean just yet, because I’m going to show you that the definition of “success” often changes in the middle of the question!
20Practice: 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!That’s right, a failure!!!But it’s considered to be the “success”, so we have to reverse things
21Practice, Exercise 7(a): setting up the calculation P(X=x) = qx-1p is the formula.Here, this translates as (.8)4 × 0.2Yes, I **know** it’s bizarre looking at the success as a failure, but hey…..Multiply this out on your calculators, and you should get… ?Everybody get that?Books says , or about 8.2% of the time will he not miss until the fifth shot
22Practice, Exercise 7(a): lessons You 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 answer“Failure” 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)
23Practice, Exercise 7(b): a more normal set-up Q: “he makes his first basket on his fourth shot.”Except for reversing p and q, it’s 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
24Practice, Exercise 7(c): a trick you need to learn Question (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):pqq qpq pppppq qpppqp qqp
25Practice, Exercise 7(c): the long way On 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)ConfigurationProbabilitypqq0.032ppq0.128pqpppp0.512qpqqppqqp
26Practice, Exercise 7(c): the easy way If 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.
27Practice, Exercise 7(c): the easy way/calculations So P(X)=1-failure to get any baskets in first three shotsThis equals 1-(0.2)3= =0.992Which would you rather have in YOUR wallet? (oops….sorry, wrong commercial)…which would you rather spend your time on?
28Practice, Exercise 9: “expected number of shots until miss” This 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, what’s p?Here, the “success” is missing. So the mean is 1/0.2 = 5.
29Practice, Exercise 11: the AB blood problem NB: your instructor has AB+ blood. The Red Cross is always VERY glad to see me.0.04 of all people have AB blood (we’re a rare breed)This problem will involve finding the mean as well as doing the probability calculations
30Practice, Exercise 11(a): using the mean Q: 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 = 25
31Practice, Exercise 11(b): the easy way Q: What’s the probability that there is a Type AB donor among the first 5 people checked?Problem: there are 32 possible outcomes! (25)So let’s be clever (again)This is the same as asking what’s the probability of getting NO AB donors in the first 5?That’s equal to (0.96)5=Subtract that answer from one for , which is the answer to the question.
32Practice, Exercise 11(c): similar to (b) Asking “what’s 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 =Complement is 1 − = .2172
33Practice, Exercise 11(d): Q: what’s the probability that we won’t find an AB donor before the 10th person?Similar to saying we won’t find any AB donors in the first NINE peopleThat’s (0.96)9=0.693
34Homework for tomorrowCh 17, problems 8, 10, 12, 13, and 14.