1. Probability Models
Chapter 17

2. Objectives: Binomial Distribution Geometric Distribution Conditions
Calculate binomial probabilities
Cumulative distribution function
Calculate means and standard deviations
Use normal approximation to the
binomial distribution
Geometric Distribution
Calculate means and standard
deviations

3. 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

4. The Binomial Distribution

5. 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).

6. 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.

7. Binomial Setting Conditions required for a binomial distribution
Each observation falls into one of just two categories, “success” or “failure” (only two possible outcomes).
There are a fixed number n of observations.
The n observations are all
independent.
The probability of success,
called p, is the same for each
observation.

8. 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).

9. 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
success
Number
observations
Binomial

10. 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)

11. 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.

12. Determining if a situation is (yes) or is not (no) binomial?
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.

13. 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.”

14. 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”

15. 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

16. 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,

17. 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
Mean
Standard Deviation

18. Binomial Probabilities

19. Binomial Mean & Standard Deviation

20. 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.
What is the probability that exactly
8 seeds will germinate?
What is the probability that
at least 8 seeds will germinate?

21. Solution: Binomial Setting? Yes
Each trial has only two outcomes, success or failure?
Fixed number of trials?
Trials are independent?
Probability of success is the same for each trial?

22. Solution:
We wish to find P(X = 8), the probability of exactly eight success.
n=10 p=.7 (1-p)=.3 k=8

23. Solution:
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) =
P(X ≥ 8) = .382

24. 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.

25. 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.

26. 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

27. 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

28. 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

29. 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.
Let X be the number of bad switches
in the sample, what is the
distribution of X?
B(10,.1)
What is the probability of exactly
two bad switches?
P(X = 2)
Binompdf(10,.1,2)
.1937

30. Continued
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

31. 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

32. 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…

33. 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 failures:
np ≥ 10 and nq ≥ 10

34. 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.

35. 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.

36. 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.

37. Normal Approximation of Binomial Distribution
Remember

38. Example 1: Binomial Distribution
np = (3)(.25) = .75
n(1-p) = (3)(.75) = 2.25

39. Example 2: Binomial Distribution
np = (10)(.25) = 2.5
n(1-p) = (10)(.75) = 7.5

40. Example 3: Binomial Distribution
np = (25)(.25) = 6.25
n(1-p) = (25)(.75) = 18.75

41. Example 4: Binomial Distribution
np = (50)(.25) = 12.5
n(1-p) = (50)(.75) = 37.5

42. 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?

43. 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.

44. 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) =
P(X ≥ 151,000)
In a shipment of
1 million cones,
the probability of
getting more than
151,000 defective cones is

45. 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”.

46. TI-83 calculator B(2500,.6) and P(X>1520) 1-binomcdf(2500,.6,1519)

47. Review: Binomial Probability

48. Review: Binomial Probability

49. Geometric Distributions

50. 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).

51. 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.

52. Geometric Setting (Conditions)
Each observation falls into one of just two categories, success (p) or failure (1-p).
The probability of a success (p) is the same for each observation.
The observations are all
independent.
The variable of interest is the
number of trials required to
obtain the first success.

53. 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
Mean
Standard Deviation

54. 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

55. 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 … n
P(X) p (1-p)p (1-p)2p (1-p)n-1p
Geometric probability distributions are skewed right.

56. Example: Geometric Probability Distributions

57. 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.

58. Using the TI-83/84 Single values Cumulative 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.

59. 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.

60. 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 on the second roll or before:
P(X<2)=(1/6) +(5/6)∙(1/6) = 11/36.
geometcdf(1/6,2) =
= 11/36.

61. 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) =
= 25/36.

62. 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

63. Geometric Distribution Mean & Standard Deviation

64. Useful Geometric Probability Formulas

65. Review:

66. Review:

67. 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.

68. 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

69. What have we learned? Geometric model Binomial model Normal 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.

70. Assignment
Pg. 401 – 404:# odd, 25, 29, 35