Transcript of Chapter 17: Geometric models AP Statistics B 1 Overview of Chapter 17 Two new models: Geometric...
- Slide 1
- Slide 2
- Chapter 17: Geometric models AP Statistics B 1
- Slide 3
- Overview of Chapter 17 Two new models: Geometric model, and the
Binomial model Yes, the binomial model involves Pascals 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
- Slide 4
- Todays 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
- Slide 5
- 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., doesnt change over time) 3.Trials are
independent 4
- Slide 6
- Vocabulary: nomenclature for Bernoulli trials Were 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 (dont ask why.)
Remember, remember, remember! p+q=1 (s+f, too!) 5
- Slide 7
- 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 (doesnt have to be just
one, as in the geometric model) 6
- Slide 8
- Examples: the geometric models Example: tossing a coin
Success=heads; failure=tails COMPETELY ARBITRARYin the examples we
will reverse success and failure without any problems, so dont get
hung up on it Better way of thinking about itbinary, either/or
7
- Slide 9
- Examples: asking the geometry model question Typical question:
What is the probability of not getting heads until the 5 th toss of
the coin? Many geometric model questions are going to look like
this: f f f f s (no success until 5 th toss) In terms of p and q,
it looks like q q q q p We are talking sequences here! 8
- Slide 10
- 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
- Slide 11
- 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
- Slide 12
- 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, Im
going to show you a lecture by Arthur Benjamin on binomial math (
hour) Professor of Math, Harvey Mudd College (Claremont Colleges)
Good instructor, makes my jokes look less corny Irksome mannerisms,
but great content 11
- Slide 13
- Nature of the geometric model: first example (tossing a coin)
Lets start with flipping coins What is success? Lets define it as
getting heads as a result (p) So tails is q Probabilities? p=0.5
q=0.5 12
- Slide 14
- Nature of the geometric model: framing the question Q: What are
the chances of not getting heads until the 4 th toss? 13
- Slide 15
- Nature of the geometric model: doing the calculations
1.Probability for failure is q, or q 3 for 3 successive failures
(i.e., not getting heads until the 4 th toss) 2.Probability for
success on 4 th try is p 3.Total probability is therefore q 3 p
4.Replace with numbers: (0.5) 3 (0.5)=(0.125)(0.5)=0.0625 14
- Slide 16
- 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 youre
simply going to have to memorize the equations (at least until you
get to college and take calculus!) Sorry, sorry, sorry! 15
- Slide 17
- 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) = q x-1 p 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
- Slide 18
- Nature of the geometric model: the mean and standard deviation
Aka expected value, which equals E(X) =1/p, where p=probability of
success Standard deviation Sadly, you just gotta memorize these!
17
- Slide 19
- Nature of the geometric model: summary 1.P(X=x) = q x-1 p,
where x= number of trials before first success 2 3 18
- Slide 20
- Practice: Exercise 7 Basketball player makes 80% of his shots.
Lets set things up before we start. p=0.8, so q=0.2 (he makes 80%
of his shots and misses 20%) Dont calculate the mean just yet,
because Im going to show you that the definition of success often
changes in the middle of the question! 19
- Slide 21
- Practice: Exercise 7(a) Misses for the first time on his 5 th
attempt Use the probability model, except notice something really,
really bizarre: the 5 th attempt appears to be a failure! Thats
right, a failure!!! But its considered to be the success, so we
have to reverse things 20
- Slide 22
- Practice, Exercise 7(a): setting up the calculation P(X=x) = q
x-1 p is the formula. Here, this translates as (.8) 4 0.2 Yes, I
**know** its 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
- Slide 23
- 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 4 th 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
- Slide 24
- 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, its
the same as (a): P(X=x) = q x-1 p is the formula. P(miss 3 baskets
before success)= (.2) 3 0.8=0.0064 Very straightforward 23
- Slide 25
- 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): pqqqpqppp
ppqqpp pqpqqp 24
- Slide 26
- 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)
ConfigurationProbability pqq0.032 ppq0.128 pqp0.128 ppp0.512
qpq0.128 qpp0.128 qqp0.032 25
- Slide 27
- Practice, Exercise 7(c): the easy way If you have 2 possible
outcomes and 3 trials, you will have 2 3 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
- Slide 28
- 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
- Slide 29
- 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, whats
p? Here, the success is missing. So the mean is 1/0.2 = 5. 28
- Slide 30
- 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 (were a rare breed) This problem will
involve finding the mean as well as doing the probability
calculations 29
- Slide 31
- 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
- Slide 32
- Practice, Exercise 11(b): the easy way Q: Whats the probability
that there is a Type AB donor among the first 5 people checked?
Problem: there are 32 possible outcomes! (2 5 ) So lets be clever
(again) This is the same as asking whats the probability of getting
NO AB donors in the first 5? Thats equal to (0.96) 5 =0.8154.
Subtract that answer from one for 0.1846, which is the answer to
the question. 31
- Slide 33
- Practice, Exercise 11(c): similar to (b) Asking whats 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
- Slide 34
- Practice, Exercise 11(d): Q: whats the probability that we wont
find an AB donor before the 10 th person? Similar to saying we wont
find any AB donors in the first NINE people Thats (0.96) 9 =0.693
33
- Slide 35
- Homework for tomorrow Ch 17, problems 8, 10, 12, 13, and 14.
34