Probability Models
Chapter 17
Objectives:
• Binomial Distribution
–
–
–
–
–
Conditions
Calculate binomial probabilities
Cumulative distribution function
Calculate means and standard deviations
Use normal approximation to the
binomial distribution
• Geometric Distribution
– Conditions
– Cumulative distribution function
– Calculate means and standard
deviations
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.
– the trials are independent.
• Examples of Bernoulli Trials
– Flipping a coin
– Looking for defective products
– Shooting free throws
The Binomial Distribution
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. We denote this
Binom(n, p).
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 population.
Binomial Setting
• Conditions required for a binomial distribution
1. Each observation falls into one of just two categories,
“success” or “failure” (only two possible outcomes).
2. There are a fixed number n of observations.
3. The n observations are all
independent.
4. The probability of success,
called p, is the same for each
observation.
Definition:
• Binomial Random Variable – The random
variable X = number of successes produced in a
binomial setting.
– Examples:
• The number of doubles in 4 rolls of a pair
of dice (on each roll you either get
doubles or you don’t).
• The number of patients with type A
blood in a random sample of 10
patients (either each person has
type A blood or they don’t).
Definition:
• Binomial Distribution (model)– A class of
discrete random variable distribution, where the
count X = the number of successes, in the
binomial setting with parameters n and p.
– n is the number of observations
– p is the probability of a success
on any one observation.
– Notation: B(n,p)
Binomial
Number
observations
Probability
success
Example
• Blood type is inherited. If both parents carry genes
for the O and A blood types, each child has
probability 0.25 of getting two O genes and so of
having blood type O. Different children inherit
independently of each other.
The number of O blood types
among 5 children of these
parents is the count X off
successes in 5 independent
observations.
• How would you describe this
with “B” notation?
• X=B(5,.25)
Example
• Deal 10 cards from a shuffled disk and count the
number X of red cards. These 10 observations,
and each gives either a red or a black card. A
“success” is a red card.
• How would you describe this
using “B” notation?
• This is not a Binomial distribution
because once you pull one card
out, the probabilities change.
Determining if a situation is (yes) or is not
(no) binomial?
1.
2.
3.
4.
5.
6.
7.
Surveying people by asking them what
they think of the current president.
Surveying 1012 people and recording
whether there is a “should not” response
to this question: “do you think the cloning
of humans should or should not be
allowed?”
Rolling a fair die 50 times.
Rolling a fair die 50 times and finding the
number of times that a 5 occurs.
Recording the gender of 250 newborn
babies.
Spinning a roulette wheel 12 times.
Spinning a roulette wheel 12 times and
finding the number of times that the
outcome is an odd number.
1.
NO
2.
YES
3.
4.
NO
YES
5.
YES
6.
7.
NO
YES
Combinations - nCk
• In n trials, there are
n!
n Ck 
k ! n  k !
ways to have k successes.
– Read nCk as “n choose k.”
• Note: n! = n  (n – 1)  …  2  1,
and n! is read as “n factorial.”
Binomial Formula
• Binomial Coefficient
– The number of ways of arranging k successes among n
observations (combinations).
– The binomial coefficient
counts
the number of ways in which k
successes can be distributed
among n observations.
–
is read “combinations n choose k”
Binomial Coefficient
• Uses factorial notation, for any positive whole
number n, its factorial is
– Also,
– TI-83/84 can find the combination
function under the math menu/prb.
–
Find
= 10
Binomial Probability Model
• If X has the binomial distribution with n
observations and probability p of success on each
observation, the possible values of X are
0, 1, 2,…,n. If k is any one of these
values,
The Binomial Model (cont.)
Binomial probability model for Bernoulli trials:
Binom(n,p)
n = number of trials
p = probability of success
1 – p = probability of failure = q
k = # of successes in n trials
n k
nk
P( X  k )     p  1  p 
k 
  np   np 1  p   npq
Mean
Standard Deviation
Binomial Probabilities
n k
nk
 n  k nk
P( X  k )     p  1  p     p q
k
k 
Binomial Mean & Standard Deviation
  np
  np 1  p 
Example:
A biologist is studying a new hybrid tomato. It is
known that the seeds of this hybrid tomato have
probability 0.70 of germinating. The biologist
plants 10 seeds.
a) What is the probability that exactly
8 seeds will germinate?
b) What is the probability that
at least 8 seeds will germinate?
Solution:
Binomial Setting?
1. Each trial has only two
outcomes, success or
failure?
2. Fixed number of trials?
3. Trials are independent?
4. Probability of success is
the same for each trial?
1. Yes
2. Yes
3. Yes
4. Yes
Solution:
a) We wish to find P(X = 8), the probability of exactly
eight success.
n=10 p=.7 (1-p)=.3 k=8
Solution:
b) In this case, we are interested in the probability
of 8 or more seeds germinating, P(X ≥ 8).
P(X ≥ 8) = P(X = 8) + P(X = 9) + P(X = 10)
P(X ≥ 8) = .233 + .121 + .028
P(X ≥ 8) = .382
Finding Binomial Probabilities
using the TI-83/84
• pdf
– Given a discrete random variable X, the probability
distribution function (pdf) assigns a probability to each
value of X.
– Used to find binomial probabilities
for single values of X (ie. X = 3).
– Found under 2nd (DISTR)/0:binompdf
– Input binompdf(n,p,X) for one value
of X or binompdf(n,p,{X1,X2,…,Xn})
for multiple values of X.
Finding Binomial Probabilities
using the TI-83/84
• cdf
– Given a discrete random variable X, the cumulative
distribution function (cdf) calculates a cumulative
probability from X=0 to and including a specified value
of X.
– Used to find binomial probabilities for
an interval of values of X (ie. 0≤X≤3).
– Found under 2nd (DISTR)/0:binomcdf
– Input binomcdf(n,p,X) for interval X=0
up to and including X.
Binomial Distributions on the calculator
•
•
•
•
Binomial Probabilities
B(n,p) with k successes
binompdf(n,p,k)
Corinne makes 75% of
her free throws.
• What is the probability of
making exactly 7 of 12
free throws.
• binompdf(12,.75,7)=.1032
n k
nk
   p  1  p 
k 
12 
5
7
  .75  .25 
7 
Binomial Distributions on the calculator
•
•
•
•
•
•
12 
12 
5
7
6
6
Binomial Probabilities
.75
.25

.75
.25







 
 
7 
6 


B(n,p) with k successes
12 
12 
5
7
binomcdf(n,p,k)
   .75 .25     .754 .258 
5 
4 
Corinne makes 75% of
12 
12 
her free throws.
3
9
   .75 .25     .752 .2510 
What is the probability of  3 
2 
making at most 7 of 12
12 
12 
1
11
   .75 .25     .750 .2512 
free throws.
1 
0 
binomcdf(12,.75,7)=.1576
Binomial Distributions on the calculator
•
•
•
•
Binomial Probabilities
12 
12 
5
7
8
4
B(n,p) with k successes
.75
.25

.75
.25







 
 
7 
8 


binomcdf(n,p,k)
Corinne makes 75% of her  12  .759 .253   12  .7510 .252 
 
 
9
free throws.
 
10 
• What is the probability of
12 
12 
11
1
   .75 .25     .7512 .250 
making at least 7 of 12 free 11 
12 
throws.
• 1-binomcdf(12,.75,6) = .9456
Example:
• A quality engineer selects an SRS of 10 switches
from a large shipment for detailed inspection.
Unknown to the engineer, 10% of the switches in
the shipment fail to meet the specifications.
1. Let X be the number of bad switches
in the sample, what is the
distribution of X?
•
B(10,.1)
1. What is the probability of exactly
two bad switches?
•
•
•
P(X = 2)
Binompdf(10,.1,2)
.1937
Continued
3. What is the probability that no more than 1 of the 10
switches in the sample fail inspection?
•
P(X ≤ 1) = P(X = 0) + P(X = 1)
binomcdf (10,.1,1)
P(X1) = .7361
Binomial Simulations
•
•
•
•
Corinne makes 75% of her free throws.
Simulate shooting 12 free throws.
randBin(n,p) will do one simulation
randBin(n,p,t) will do t simulations
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 rescue…
The Normal Model to the Rescue!
• 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:
np ≥ 10 and nq ≥ 10
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
between two values.
Normal Approximation of the
Binomial Distribution
• Before the technology became available, this was the
preferred technique for calculation of binomial probabilities
when n was large or when there were a great number of
cases or successes to consider.
• This approximation method works best for binomial
situations when n is large and when the value
of p is not close to either 0 or 1.
• In this approximation, we use the
mean and standard deviation of the
binomial distribution as the mean
and standard deviation needed for
calculations using the normal distribution.
Normal Approximation for Binomial Distribution
• Given a count X has the binomial distribution with
n trials and success probability p.
• When n is large, the distribution of X is
approximately normal, N(np, √np(1-p)).
• When is the approximation valid?
– np ≥ 10 and n(1-p) ≥ 10
– The accuracy of the normal
distribution improves as the sample
size n increases. It is most
accurate for any fixed n when p is
close to ½ and least accurate when p is near 0 or 1.
Normal Approximation of
Binomial Distribution
• Remember
  np
  np 1  p 
Example 1: Binomial Distribution
np = (3)(.25) = .75
n(1-p) = (3)(.75) = 2.25
Example 2: Binomial Distribution
np = (10)(.25) = 2.5
n(1-p) = (10)(.75) = 7.5
Example 3: Binomial Distribution
np = (25)(.25) = 6.25
n(1-p) = (25)(.75) = 18.75
Example 4: Binomial Distribution
np = (50)(.25) = 12.5
n(1-p) = (50)(.75) = 37.5
Problem:
• A shipment of ice cream cones has the
manufacturer’s claim that no more than 15% of
the shipment will be defective (broken cones).
What is the probability that in a shipment
of 1 million cones, Dairy Heaven
Corporate Distribution Center
will find more than 151,000
broken cones?
• Define the
random
variable
and specify
the model.
• Let X = the number of defective cones in
the shipment (successes).
• Check that
conditions
of the
model are
met.
• np ≥ 10 and n(1-p) ≥10
– X is binomial with n = 1,000,000 and
p = 0.15.
– Most calculators will not handle a problem of
this magnitude.
– Use a normal model with μ = np and
σ = √ np(1-p) as an approximation.
– 1,000,000(.15)≥10 &
1,000,000(.85)≥10.
• Find the
mean &
standard
deviation
• μ = np and σ = √ np(1-p)
• Calculate the
probability of
151,000 or
more
successes
• P(X ≥ 151,000)
• State your
conclusion
• In a shipment of
1 million cones,
the probability of
getting more than
151,000 defective cones is .0026.
– μ = 1,000,000(.15) = 150,000
– σ = 1,000,000(.15)(.85) = 357.07
–
–
Example:
• A recent survey asked a nationwide random
sample of 2500 adults if they agreed or disagreed
that “I like buying new clothes, but shopping is
often frustrating and time-consuming.”
Suppose that in fact 60% of all
adults would “agree”. What is
the probability that 1520 or
more of the sample “agree”.
TI-83 calculator
• B(2500,.6) and P(X>1520)
• 1-binomcdf(2500,.6,1519)
• .2131390887
Review: Binomial Probability
Review: Binomial Probability
Geometric Distributions
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 parameter, p, the
probability of success, and are
denoted Geom(p).
Geometric Distributions
• Definition
– A special case of a binomial random variable where the
random variable X is defined as the number of trials
needed to obtain the first success.
Geometric Setting (Conditions)
1. Each observation falls into one of just two
categories, success (p) or failure (1-p).
2. The probability of a success (p) is the same for
each observation.
3. The observations are all
independent.
4. The variable of interest is the
number of trials required to
obtain the first success.
Geometric Probability Model
Geometric probability model for Bernoulli trials: Geom(p)
p = probability of success
1 – p = probability of failure = q
k = number of trials until the first success occur
P(X = k) =
1
E(X)   
p
Mean
(1-p)k-1p

1  p 
p
2

k-1
=q p
q
p2
Standard Deviation
Calculating Geometric
Probabilities
• The probability that the first success occurs on the
nth trial is
• P(X=n) = (1-p)n-1 p
• Example:
– If p = .25, find P(X=3)?
– P(X=3) means “failure on 1st trial
(X=1) and failure on 2nd trial (X=2)
and success on 3rd trial (X=3). Each
trial is independent so,
P(X=3) = (1-p)(1-p)(p) = (1-p)2(p).
– Therefore, P(X=3) = (.75)2(.25) = .1406
Geometric Probability Distribution
• Geometric random variable never ends.
• The probabilities are the terms of a geometric
sequence, arn-1 (hence the name). The a is
probability of success (p), r is the
probability of failure (1-p), and n
is the value of X.
• X 1
2
3 … n
P(X) p (1-p)p (1-p)2p (1-p)n-1p
• Geometric probability
distributions are skewed right.
Example: Geometric Probability Distributions
Geometric Probability Distribution
• The sum of all the probabilities is still equal to
one, even though geometric probability
distribution is infinite.
• For geometric series S∞ = a/(1-r),
where a = p and r = (1-p),
therefore,
∑P(Xi) = p/(1-(1-p)) = p/p =1.
Using the TI-83/84
• Single values
– Distr/geometpdf(p,x)
– Distr/geometpdf(p,{x1,x2,x3,…,xn})
• Cumulative values
– Distr/geometcdf(p,x)
– Calculates the sum of the
probabilities from X=0 to X.
Calculating Probabilities
• The probability of rolling a 6 = 1/6
• The probability of rolling the first 6 on the first roll:
– P(X=1) = 1/6.
– geometpdf(1/6,1) = 1/6.
• The probability of rolling the first 6
after the first roll:
– P(X>1) = 1-1/6 = 5/6.
– 1-geometpdf(1/6,1) = 5/6.
Calculating Probabilities
• The probability of rolling a 6 = 1/6
• The probability of rolling the first 6 on the second
roll:
– P(X=2) = (5/6)∙(1/6) = 5/36.
– geometpdf(1/6,2) = .1388888889 = 5/36.
• The probability of rolling the first
6 on the second roll or before:
– P(X<2)=(1/6) +(5/6)∙(1/6) = 11/36.
– geometcdf(1/6,2) = .305555556
= 11/36.
Calculating Probabilities
• The probability of rolling a 6 = 1/6
• The probability of rolling the first 6 on the second
roll:
– P(X=2) = (5/6)∙(1/6) = 5/36.
– geometpdf(1/6,2) = 5/36.
• The probability of rolling the first
6 after the second roll:
– P(X>2)=1-((1/6) +(5/6)∙(1/6)) = 25/36
– 1-geometcdf(1/6,2) = .69444444
= 25/36.
Mean and Standard Deviation
• Mean μ = 1/p
• Variance σ2 = 1-p/p2
• for P(X>n), the probability that X
takes more than n trials to see
the first success is
P(X>n) = (1-p)n
Geometric Distribution Mean & Standard Deviation
1
X 
p
1 p
1 p
  2 , X 
2
p
p
2
X
Useful Geometric Probability Formulas
P( X  n)  pq
n 1
, geometpdf(p,n)
P( X  n)  1  p  , 1-geometcdf(p,n)
n
P( X  n)  1  1  p  , geometcdf(p,n)
n
Review:
Review:
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 Normal
approximation.
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
– Normal model
What have we learned?
– 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
and 10 failures.
Assignment
• Pg. 401 – 404:#1 - 19 odd, 25, 29, 35