1. Chapter 17 Probability Models
Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

2. Bernoulli Trials
The basis for the probability models we will examine in this chapter is the
We have Bernoulli trials if:
there are possible outcomes ( and ).
the probability of success, p, is
the trials are
Bernoulli trial.
two
success
failure
constant.
independent.

3. 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
Geometric models are completely specified by one parameter, , and are denoted Geom( ).
first success.
p, the probability of success p

4. The Geometric Model (cont.)
Geometric probability model for Bernoulli trials: Geom(p)
p = probability of success
q = 1 – p = probability of failure
X = # of trials until the first success occurs
P(X = x) =
qx-1p

5. Independence
One of the important requirements for Bernoulli trials is that the trials be
When we don’t have an infinite population, the trials are But, there is a rule that allows us to pretend we have trials:
The % 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 % of the population.
independent.
not independent.
independent 10 10

6. TI Tips Under Distributions you will find both geometpdf and geometcdf
Use geometpdf(p,x) to find the probability of any outcome. p is the individual probability of success and x is the number of trials until you get a success. Ex. geometpdf(.2,5) ≈
Use geometcdf(p, x) to find the probability of success the xth trial.
Ex. Geometcdf(.2, 4) ≈
individual
0.082
on or before
0.590

7. Ex. Approximately 3.6% of all (untreated) Jonathan apples had bitter pit in a study conducted by the botanists Ratkowsky and Martin (Australian Journal of Agricultural Research, Vol. 23, pp ). Bitter pit is a disease of apples resulting in a soggy core, which can be caused either by over watering the apple tree or by a calcium deficiency in the soil. Let n be a random variable that represents the first Jonathan apple chosen at random that has bitter pit.
a) Find the probability that n = 3, n = 5, n= 12.
b) Find the probability that n < 5.
c) Find the probability that n > 5.
d) State the mean and standard deviation.
geometpdf(.036,3) ≈ 0.033
geometpdf(.036,5) ≈ 0.031
geometpdf(.036,12) ≈ 0.024
geometcdf(.036,5) ≈ 0.167
P(5 - ∞) =
1 – P(1-4) =
1 - geometcdf(.036,4) ≈ 0.864
μ =
≈ apples
σ =
≈ apples

8. Assignment
P. 398 #7-12

9. Chapter 17 Probability Models (2)
Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

10. The Binomial Model
A Binomial model tells us the probability for a random variable that counts the number of in a fixed number of
Two parameters define the Binomial model: ; and, We denote this Binom(n, p).
successes
Bernoulli trials. n,
the number of trials
p, the probability of
success

11. The Binomial Model (cont.)
In n trials, there are
ways to have k successes.
Read nCk as
Note: n! = n x (n-1) x … x 2 x 1, and n! is read as
Can also be shown as
“n choose k.”
“n factorial.”
Find the following.
a) b)
10C4
= 210
= 10

12. The Binomial Model (cont.)
Binomial probability model for Bernoulli trials: Binom(n,p)
n =
p =
q =
X =
P(X = x) =
number of trials
probability of success
1 – p = probability of failure
# of successes in n trials
nCx px qn-x

13. TI tips Under Distributions you will find both binompdf and binomcdf
Use binompdf(n,p,x) to find the probability of any outcome. n is the number of trials, p is the individual probability of success and x is the number of successes. Ex. binompdf(5,.2,2)
Use binomcdf(n,p, x) to find the probability of getting successes in n trials.
Ex. binomcdf(5,.2, 2)
individual
≈ 0.205
x or fewer
≈ 0.942

14. Ex. A basketball player makes 70% of his freethrows
Ex. A basketball player makes 70% of his freethrows. Assuming the shots are independent, what is the probability of each of the following?
a) He makes exactly 3 out of the next 5 attempts.
b) He makes at most 3 out of the next 5.
c) He makes at least 3 out of the next 5.
binompdf(5,.7,3)
≈ 0.309
P(0-3) =
binomcdf(5,.7,3)
≈ 0.472
P(3-5) =
1 – P(0-2) =
1 – binomcdf(5,.7,2)
≈ 0.837

15. The Normal Model to the Rescue
As long as the 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 successes and failures:
np ≥ 10 and nq ≥ 10
Success/Failure 10 10

16. Continuous Random Variables
When we use the Normal model to approximate the Binomial model, we are using a random variable to approximate a random variable.
So, when we use the Normal model, we no longer calculate the probability that the random variable equals a value, but only that it lies two values.
continuous
discrete
particular
between

17. Ex. An orchard owner knows that he’ll have to use about 6% of the apples he harvests for cider because they will have bruises or blemishes. He expects a tree to produce about 300 apples.
a) Verify that you can use the normal model to approximate the distribution.
b) Find the probability there will be no more than a dozen cider apples.
c) Is it likely there will be more than 50 cider apples? Explain.
np = 18 > 10 and nq = 282 > 10, therefore N(18, 4.113)
is appropriate.
.5 – N(12,18,18,4.113) ≈ 0.072
P(x < 12) = P(0-12) = binomcdf(300,.06,12) ≈ .085
No, 50 is ≈ 7.8 SD above the mean.

18. What Can Go Wrong? Be sure you have Bernoulli trials.
You need outcomes per trial, a constant , and
Remember that the provides a reasonable substitute for
Don’t confuse models.
Don’t use the Normal approximation with
You need at least successes and failures to use the Normal approximation.
two
probability of success
independence.
10% Condition
independence.
Geometric and Binomial
small n. 10 10

19. Assignment
P. 399 #13-15, 17, 19-21, 26, 29