Chapter 17
Probability Models
What I will know and be able to do
Know the appropriate conditions under which a
Geometric, Binomial, or Normal model is used and
calculate Geometric and Binomial probabilities.
Assignment:
Bernoulli Trials
The basis for the probability models we will examine in this chapter
is the Bernoulli trial.
We have Bernoulli trials if:
there are two possible outcomes (success and failure). the probability of success, p, is constant.
Independence
One of the important requirements for Bernoulli trials
is that the trials be independent.
When we don’t have an infinite population, the trials are
not independent.
But, there is a rule that allows us to
“pretend” we have independent trials:
The 10% condition
: Bernoulli trials must be independent.
If that assumption is violated, it is still okay to proceed as
long as the sample is smaller than 10% of the
Example 1: Bernoulli or not…..
1. You are rolling 5 die and need to get at least two 6’s to win the
game.
2. We record the distribution of eye colors found in a group of 500
people.
3. A city council of 11 Republicans and 8 Democrats picks a
committee of 4 at random. What’s the probability they choose all Democrats?
4. A 2002 Rutgers University study found that 74% of high school
students have cheated on a test at least once. Mr. McDonough
Example 2: Simulations
20% of cereal boxes contain pictures of LeBron James, 30% a
picture of Danica Patrick and the rest a picture of Serena Williams. You are opening boxes of cereal one at a time looking for LeBron James’ picture. You want to know how many boxes you might have to open in order to find LeBron.
a. Describe how you would simulate the search using a random digit table.
b. Run 30 trials.
c. Based on the simulation, estimate the probabilities that you
The Geometric Model
A single Bernoulli trial is usually not all that interesting.
A
Geometric probability model
tells us the probability for a
random variable that counts the number of Bernoulli trials
until the first success.
Geometric models are completely specified by one
The Geometric Model (cont.)
Geometric probability model for Bernoulli trials: Geom(
p
)
p
= probability of success
q
= 1 –
p
= probability of failure
X
= number of trials until the first success occurs
P(X = x) = q
x-1
p
E
(
X
)
=
m
=
1
Example 3: Speckled Skittles
A new gimmick has 30% of Skittles candies covered with
speckles. These “groovy” candies are mixed randomly with the normal candies as they are put into the bags for distribution and sale. You buy a bag and remove candies one at a time looking for speckles.
a. What is the probability that the first speckled one we see is the fourth candy we get?
b. What’s the probability that the first speckled one is the tenth one?
c. Write the general formula.
d. How many do we expect to have to check, on average, to find a speckled one?
Slide 17- 11
Example 3 answers
a. P(first speckled is fourth candy)=(.7)3(.3)=.1029 b. P(first speckled is tenth candy)=(.7)10(.3)=.0085 c. General Formula: P(X=x)=qx-1p
Slide 17- 12
Example 4: Jumper Cables
Suppose that 40% of the students who drive to Wilson HS carry
jumper cables in their car. Your car has a dead battery and you don’t have jumper cables, so you decide to stop students who are headed to the parking lot and ask them whether they have a pair.
a. What is the probability that the third student you ask will have jumper cables?
b. What is the probability that first student you ask will have jumper cables?
c. What is the probability that three or fewer students must be stopped to find cables?
Slide 17- 13
Example 4 answers
a. P(first person with cables is third one asked)=(.6)2(.4)=.144
b. P(first person with cables is first one asked)=(.6)0(.4)=.4 c. P(x≤3)=P(1 or 2 or 3)=P(1) +P(2)+P(3)
= (.6)0(.4)+ (.6)1(.4)+ (.6)2(.4) =.4 + .24 + .144 = .784
Slide 17- 14
Geometric Model with Calculator
geometpdf(p,x)
p is the probability of success
x is the number you are “waiting” for
Based on the last example, what is the probability of the 4th student have the
jumper cables you need. geometpdf(.4,4) = .0864
What is the probability that you will find jumper cables in 4 or fewer students asked?
geometcdf(.4,4) = .8704 …….or………
Slide 17- 15
Example: Skittles again!
Back to the speckled skittles....what is the probability that
we’ll find two speckled ones in a handful of five candies?
This is a little bit different…notice we are not waiting until
something “special” happens…
Slide 17- 16
So…How do you find the answer?
Would this work?
P(2 speckled out of 5) = (.3)2(.7)3
NO
Why not? This method says that the first 2 would be speckled and the last three are not. Can you get 2 speckled in another way? We need to count the number of ways that getting 2 speckled out of 5 can happen.
The Binomial Model
A Binomial model
tells us the probability for a
random variable that counts the number of
successes in a fixed number of Bernoulli trials.
Two parameters define the Binomial model:
n
, the
number of trials; and,
p
, the probability of success.
The Binomial Model (cont.)
Binomial probability model for Bernoulli trials: Binom(n,
p
)
n
= number of trials
p
= probability of success
q
= 1 –
p
= probability of failure
X
= # of successes in
n
trials
P(X = x) =
nC
xp
xq
n–xnp
The Binomial Model (cont.)
In
n
trials, there are
ways to have
k
successes.
Read
nC
kas “
n
choose
k.
”
Note: , and
n
! is read as “
n
factorial.”
(
!
)
!
!
n kn
C
k n k
=
Slide 17- 20
Back to the Skittles….
Now lets solve it:
P(2 speckled out of 5) =
5C
2(.3)
2(.7)
3Slide 17- 21
Calculator for Binomial Models
binompdf(n,p,x)
n is the number in the group p is the probability of success
x is the number of successes you want
Based on the last example, what is the probability of 2 out 5 being speckled? P(X=2) = binompdf(5,.3,2) =
What is the probability that you will get 2 or fewer speckled Skittles out of 5? P(X<=2) = binomcdf(5,.3,2) =
Slide 17- 22
Before you do a problem:
Check to see if there are Bernoulli trials.
1. 2. 3.
If you can’t say “independent”, you must
show
the 10% conditionto be true.
State the model that you are using, with correct symbolic
notation.(Geom(p) or Binom(n,p))
Example: Blood Donors
People with O-negative blood are called “universal” donors because O-negative blood can be given to anyone else, regardless of the recipient’s blood type. Only about 6% of people have O-negative blood.
a. If donors line up at random for a blood drive, how many do you expect to examine before you find someone who has O-negative blood?
b. What’s the probability that the first O-negative donor found is the one of the first four people in line?
c. What is the probability that there are 4 O-negative donors amount in a group of 50 donors standing in line?
d. What is the probability that 7 out of the 100 donors that donated blood last Wednesday were O-negative?
e. What is the probability that 10 or fewer donors out of a 87 donor pool are O-negative? f. What is the probability that 4 or 5 of the donors out of a group 63 donors are O-negative? g. What is the probability that at least 1 out of 20 donors is type O-negative?
h. What is the probability that at least 6 of the donors out of a 200 donor pool is type O-negative? i. Suppose 20 donors come to the blood drive. What are the mean and standard deviation of the
Example – More Blood
Suppose the Red Cross anticipates the need for at least 1850 units of O-negative blood this year. It estimates that it will collect blood from 32,000 donors. How great is the risk that they will fall short of
meeting its need?
We need to calculate the probability of getting exactly 1850 units of O-neg blood from 32,000 donors. What would that be?
That’s just the beginning – they need at least 1850 units – so we will
The Normal Model to the Rescue!*
When dealing with a large number of trials in a Binomial situation,
making direct calculations of the probabilities becomes tedious (or outright impossible).
The Normal Model to the Rescue (cont.)
As long as the
Success/Failure Condition
holds, we can
use the Normal model to approximate Binomial
probabilities.
Success/failure condition
: A Binomial model is
approximately Normal if we expect at least
10 successes and 10 failures:
Slide 17- 29
Blood and more blood….
Check Bernoulli and success failure condition:
np ≥ 10 and nq ≥ 10
32000(.06) ≥ 10 and 32000(.94) ≥ 10
≥ 10 ≥ 10
So, we can use a Normal model, using the mean and standard deviation from the Binomial model.
μ = np = 32000(.06) = 1920 and SD=
N(1920,42.48)
So we can find the probability P(X≥1850)=P(z≥-1.65)=.95
Continuous Random Variables
When we use the Normal model to approximate the Binomial
model, we are using a continuous random variable to approximate a discrete random variable.
So, when we use the Normal model, we no longer calculate the
What Can Go Wrong?
Be sure you have Bernoulli trials.
You need two outcomes per trial, a constant probability of
success, and independence.
Remember that the 10% Condition provides a reasonable
substitute for independence.
Don’t confuse Geometric and Binomial models. Don’t use the Normal approximation with small n.
You need at least 10 successes and 10 failures to use the
What have we learned?
Bernoulli trials show up in lots of places.
Depending on the random variable of interest, we might be dealing
with a
What have we learned? (cont.)
Geometric model
When we’re interested in the number of Bernoulli trials until the next
success.
Binomial model
When we’re interested in the number of successes in a certain number
of Bernoulli trials.
Normal model
To approximate a Binomial model when we expect at least 10 successes
AP Tips
The AP rubrics usually have three requirements for probability
problems:
Name of distribution
AP Tips
The Name can be identified with
words (binomial) or with
the proper formula (P(X = x) = nCx px qn–x)
(Your teacher will probably ask you to write both and that’s a
good thing!)
The parameters should use standard notation:
n = 10, p = 0.7 or
