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Chapter 17: Geometric models

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1 Chapter 17: Geometric models
AP Statistics B

2 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

3 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

4 Vocabulary: Bernoulli trials
The only kind we do in Chapter 17 Need to have definition firmly in mind 3 requirements: There are only two possible outcomes Probability of success is constant (i.e., doesn’t change over time) Trials are independent

5 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!)

6 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)

7 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

8 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!

9 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: s s f f f s f s f f s f f s f s f f f s f s s f f f s f s f f s f f s f f s s f f f s f s f f f s s

10 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 p p q q q p q p q q p q q p q p q q q p q p p q q q p q p q q p q q p q q p p q q q p q p q q q p p

11 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

12 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

13 Nature of the geometric model: framing the question
Q: What are the chances of not getting heads until the 4th toss?

14 Nature 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 p Total probability is therefore q3p Replace with numbers: (0.5)3×(0.5)=(0.125)(0.5)=0.0625

15 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!

16 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!

17 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!

18 Nature of the geometric model: summary
P(X=x) = qx-1p, where x= number of trials before first success 2 3

19 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!

20 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

21 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… ? Everybody get that? Books says , or about 8.2% of the time will he not miss until the fifth shot

22 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)

23 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

24 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

25 Practice, Exercise 7(c): the long way
On the right is a chart of all 7 possibilities With each possibility is the percent of the time it happens It al adds up to 0.992 All these had to be assembled by hand applying the formulae in (a) and (b) Configuration Probability pqq 0.032 ppq 0.128 pqp ppp 0.512 qpq qpp qqp

26 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.

27 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= =0.992 Which would you rather have in YOUR wallet? (oops….sorry, wrong commercial)…which would you rather spend your time on?

28 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.

29 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

30 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

31 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= Subtract that answer from one for , which is the answer to the question.

32 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 = Complement is 1 − = .2172

33 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

34 Homework for tomorrow Ch 17, problems 8, 10, 12, 13, and 14.

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