Probability Models

Chapter 16

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)

Probability

successNumber

observations

Binomial

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 deck 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. Surveying people by asking them what

they think of the current president.

2. 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?”

3. Rolling a fair die 50 times.

4. Rolling a fair die 50 times and finding the

number of times that a 5 occurs.

5. Recording the gender of 250 newborn

babies.

6. Spinning a roulette wheel 12 times.

7. Spinning a roulette wheel 12 times and

finding the number of times that the

outcome is an odd number.

1. NO

2. YES

3. NO

4. YES

5. YES

6. NO

7. YES

Combinations - nCk

• In n trials, there are

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.”

!

! !n k

nC

k n k

Binomial Formula

• Binomial Coefficient

– The number of ways of arranging k successes among nobservations (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 nobservations 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

( ) 1n kk

nP X k p p

k

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

1n kk

np p

k

57

12.75 .25

7

Binomial Distributions on the calculator

• Binomial Probabilities

• B(n,p) with k successes

• binomcdf(n,p,k)

• Corinne makes 75% of

her free throws.

• What is the probability of

making at most 7 of 12

free throws.

• binomcdf(12,.75,7)=.1576

57 6 6

5 7 4 8

3 9 2 10

1 11 0 12

12 12.75 .25 .75 .25

7 6

12 12.75 .25 .75 .25

5 4

12 12.75 .25 .75 .25

3 2

12 12.75 .25 .75 .25

1 0

Binomial Distributions on the calculator

• Binomial Probabilities

• B(n,p) with k successes

• binomcdf(n,p,k)

• Corinne makes 75% of her

free throws.

• What is the probability of

making at least 7 of 12 free

throws.

• 1-binomcdf(12,.75,6) = .9456

57 8 4

9 3 10 2

11 1 12 0

12 12.75 .25 .75 .25

7 8

12 12.75 .25 .75 .25

9 10

12 12.75 .25 .75 .25

11 12

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

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

1

np

np 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 Distributionnp = (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.

• Check that

conditions

of the

model are

met.

• Let X = the number of defective cones in

the shipment (successes).

– 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.

• np ≥ 10 and n(1-p) ≥10

– 1,000,000(.15)≥10 &

1,000,000(.85)≥10.

• Find the

mean &

standard

deviation

• Calculate the

probability of

151,000 or

more

successes

• State your

conclusion

• μ = np and σ = √ np(1-p)

– μ = 1,000,000(.15) = 150,000

– σ = 1,000,000(.15)(.85) = 357.07

• P(X ≥ 151,000)

–

–

• In a shipment of

1 million cones,

the probability of

getting more than

151,000 defective cones is .0026.

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-p)k-1p = qk-1p

E(X) 1

p

2 2

1 p q

p p

Mean 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

2

2 2

1

1 1,

X

X X

p

p p

p p

Useful Geometric Probability Formulas

1( ) , geometpdf(p,n)

( ) 1 , 1-geometcdf(p,n)

( ) 1 1 , geometcdf(p,n)

n

n

n

P X n pq

P X n p

P X n p

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