Chapter 16
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).
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
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 conducts a survey in homerooms and
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
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-1p
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?
Example continued….
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)=q
x-1p
Example 4: Jumper Cables
Suppose that 40% of the students who drive to BHS 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?
d. What is the average number of students you would have
Example 4 continued….
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
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………
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? We are
How do you find the answer?
Would this work?
P(2 speckled out of 5) = (.3)
2(.7)
3NO
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
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
n
C
kas “
n
choose
k.
”
Note: , and
n
! is read
as “
n
factorial.”
(
!
)
!
!
n k
n
C
k n k
=
Back to the Skittles….
Now lets solve it:
P(2 speckled out of 5) =
5C
2(.3)
2(.7)
3Calculator 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) =
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% condition to 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. Suppose 20 donors come to the blood drive. What are the mean and
standard deviation of the number of universal donors among them?
b. What is the probability that there are 4 universal donors in a group of 100
donors?
c. What is the probability that there are 2 or 3 universal donors?
d. What is the probability that there are at most 3 universal donors in 55
donations?
e. What is the probability that there are at least 4 universal donors in a 150
donations?
Don’t forget……
f. 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?
Blood Example
What do you have to do first? Lot’s of stuff.
E(X)=16.7
P(X<=4)=.2193
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?
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).
Fortunately, the Normal model comes to the
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:
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 probability that the random variable
equals a
particular
value, but only that it lies
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
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
Geometric model
Binomial model
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
AP Tips
The AP rubrics usually have three requirements
for probability problems:
Name of distribution
Identification of correct parameters
AP Tips
The Name can be identified with
words (binomial)
or
with
the proper formula (
P(X = x) =
n
C
xp
x
q
n–x)
(Your teacher will probably ask you to write both
and that’s a good thing!)
The parameters should use standard notation:
