#### Transcript Chapter 17: Geometric models

Chapter 17: Geometric models AP Statistics B 1 Overview of Chapter 17 • Two new models: Geometric model, and the Binomial model – Yes, the binomial model involves Pascal’s triangles that (I hope) you learned about in Algebra 2 • Use 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 2 Today’s coverage • Introduction to the vocabulary: – Bernoulli trials – Geometric probability model – Binomial probability model • Examples of both geometric and binomial probability models • Nature of the geometric model (and review of series from Algebra 2) • When to use the geometric model/practice on problems/solutions • Finally, how to use the TI calculators to calculate probabilities by the geometric model 3 Vocabulary: Bernoulli trials • Bernoulli trials – The only kind we do in Chapter 17 – Need to have definition firmly in mind – 3 requirements: 1. There are only two possible outcomes 2. Probability of success is constant (i.e., doesn’t change over time) 3. Trials are independent 4 Vocabulary: 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!) 5 Vocabulary: geometric and binomial models of probability • Geometric probability model: – Counts the number of Bernoulli trials before the first success • Binomial probability model: – Counts the number of successes in the first n trials (doesn’t have to be just one, as in the geometric model) 6 Examples: the geometric models • Example: tossing a coin • Success=heads; failure=tails • COMPETELY ARBITRARY—in the examples we will reverse success and failure without any problems, so don’t get hung up on it • Better way of thinking about it—binary, either/or 7 Examples: 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 p We are talking sequences here! 8 Example: 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: 1. s s f f f 2. s f s f f 3. s f f s f 4. s f f f s 5. f s s f f 6. f s f s f 7. f s f f s 8. f f s s f 9. f f s f s 10. f f f s s 9 Example: binomial model using p and q instead of s and f • An identical model to that of the last slide appears at the left • This one, however, uses the p (success) and q (failure) that the textbook uses • The patterns, however, are identical 1. p p q q q 2. p q p q q 3. p q q p q 4. p q q q p 5. q p p q q 6. q p q p q 7. q p q q p 8. q q p p q 9. q q p q p 10. q q q p p 10 Examples: geometric v. binomial • Today Geometric models, tomorrow, Binomial • The Geometric model is somewhat easier to follow • The Binomial Model requires quite a bit more math • Tomorrow, 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 corny – Irksome mannerisms, but great content 11 Nature of the geometric model: first example (tossing a coin) • Let’s start with flipping coins • What is success? • Let’s define it as getting heads as a result (p) • So tails is q • Probabilities? • p=0.5 • q=0.5 12 Nature of the geometric model: framing the question • Q: What are the chances of not getting heads until the 4th toss? 13 Nature of the geometric model: doing the calculations 1. Probability for failure is q, or q3 for 3 successive failures (i.e., not getting heads until the 4th toss) 2. Probability for success on 4th try is p 3. Total probability is therefore q3p 4. Replace with numbers: (0.5)3×(0.5)=(0.125)(0.5)=0.0625 14 Nature of the geometric model: the formulas (formulae for you pedants) • Unfortunately, to derive most of the formulas we use, you have to use calculus • This 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! 15 Nature 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-1p • Remember our coin-tossing model: 4 times until we got heads (fill in the equation) • You will use this a lot to calculate probabilities! 16 Nature of the geometric model: the mean and standard deviation • Aka expected value, which equals E(X) • μ=1/p, where p=probability of success • Standard deviation • Sadly, you just gotta memorize these! 17 Nature of the geometric model: summary 1. P(X=x) = qx-1p, where x= number of trials before first success 2 3 18 Practice: 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! 19 Practice: Exercise 7(a) • Misses for the first time on his 5th attempt • Use 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 20 Practice, Exercise 7(a): setting up the calculation • P(X=x) = qx-1p is the formula. • Here, this translates as (.8)4 × 0.2 – Yes, I **know** it’s 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 21 Practice, Exercise 7(a): lessons • You can interchange failure for success in the probability model without problems • You 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) 22 Practice, 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=0.0064 • Very straightforward 23 Practice, 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 ppp ppq qpp pqp qqp 24 Practice, Exercise 7(c): the long way • On the right is a chart of Configuration Probability all 7 possibilities pqq 0.032 • With each possibility is ppq 0.128 the percent of the time it happens pqp 0.128 • It al adds up to 0.992 ppp 0.512 • All these had to be assembled by hand qpq 0.128 applying the formulae qpp 0.128 in (a) and (b) qqp 0.032 25 Practice, Exercise 7(c): the easy way • If you have 2 possible outcomes and 3 trials, you will have 23 possible combinations • We could get the 7 of 8 that we did in the previous slide • Or, 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. 26 Practice, Exercise 7(c): the easy way/calculations • So P(X)=1-failure to get any baskets in first three shots • This equals 1-(0.2)3=1-0.008=0.992 • Which 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 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. 28 Practice, 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 29 Practice, 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 30 Practice, 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=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 “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 = 0.7828 • Complement is 1 − 0.7828 = .2172 32 Practice, 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 people • That’s (0.96)9=0.693 33 Homework for tomorrow • Ch 17, problems 8, 10, 12, 13, and 14. 34