Chapter 17: Geometric models

AP Statistics B

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

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

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 outcomes2. Probability of success is constant (i.e., doesn’t

change over time)3. Trials are independent

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!)

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)

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

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!

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 f2. s f s f f3. s f f s f4. s f f f s5. f s s f f6. f s f s f7. f s f f s8. f f s s f9. f f s f s10. f f f s s

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 q2. p q p q q3. p q q p q4. p q q q p5. q p p q q6. q p q p q7. q p q q p8. q q p p q9. q q p q p10. q q q p p

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

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

Nature of the geometric model:framing the question

• Q: What are the chances of not getting heads until the 4th toss?

Nature of the geometric model:doing the calculations

1. Probability for failure is q, or q3for 3 successive failures (i.e., not getting heads until the 4th toss)

2. Probability for success on 4th try is p3. Total probability is therefore q3p4. Replace with numbers: (0.5)3×(0.5)=(0.125)

(0.5)=0.0625

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!

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!

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!

Nature of the geometric model:summary

1. P(X=x) = qx-1p, where x= number of trials before first success

2

3

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!

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

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?• Book says 0.0819, or about 8.2% of the time will

he not miss until the fifth shot

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)

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(misses 3 baskets before “success”)=

(.2)3 × 0.8=0.0064• Very straightforward

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 pppppq qpppqp qqp

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 0.128

ppp 0.512

qpq 0.128

qpp 0.128

qqp 0.032

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.

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?

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.

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

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

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.

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

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

Homework for tomorrow

• Ch 17, problems 8, 10, 12, 13, and 14.