#### 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
• 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
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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
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)
•
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
pqq
qpq
ppp
ppq
qpp
pqp
qqp
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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
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
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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
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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.
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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
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Homework for tomorrow
• Ch 17, problems 8, 10, 12, 13, and 14.
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