Download Binomial vs. Geometric

Binomial vs. Geometric
Chapter 8
Binomial and Geometric
Distributions
Binomial vs. Geometric
The Binomial Setting The Geometric Setting
1. Each observation falls into 1. Each observation falls into
one of two categories.
one of two categories.
2. The probability of success 2. The probability of success
is the same for each
is the same for each
observation.
observation.
3. The observations are all
3. The observations are all
independent.
independent.
4. There is a fixed number n
of observations.
4. The variable of interest is the
number of trials required to
obtain the 1st success.
Are Random Variables and Binomial
Distributions Linked?
RECALL: Suppose that each of three randomly selected
customers purchasing a hot tub at a certain store chooses
either an electric (E) or a gas (G) model. Assume that
these customers make their choices independently of one
another and that 40% of all customers select an electric
model. The number among the three customers who
purchase an electric hot tub is a random variable.
What is the probability distribution?
Are Random Variables and Binomial
Distributions Linked?
X = number of people who purchase electric hot tub
X 0
1
2
3
P(X) .216 .432 .288 .064
GGG
(.6)(.6)(.6)
EGG
GEG
GGE
(.4)(.6)(.6)
(.6)(.4)(.6)
(.6)(.6)(.4)
EEG
GEE
EGE
(.4)(.4)(.6)
(.6)(.4)(.4)
(.4)(.6)(.4)
EEE
(.4)(.4)(.4)
Combinations
Formula:
nI
n!
F

G
J
Hk K k !an  k f!
Practice:
6
6!
6  5 4  3 2 1 6  5
 15

1.


4
4! 6  4 ! 4  3  2  1 2  1
2 1
F
I
G
J
HK a f
af
8I 8! 8  7  6  5!
F

2. G

 56
J
H5K 5!3! 5! 3  2 1
Developing the Formula
Ex. 8.1 Blood Type; pg. 440 – Pract. Of Stats 2nd Ed
Blood type is inherited. If both parents carry genes
for the O and A blood types, each child has a
probability 0.25 of getting two O genes, thus having
blood type O. Let X = the number of O blood types
among 3 children. Different children inherit genes
independently of each other.
Since X has the binomial distribution with n = 3 and
p = 0.25. We say that X is B(3, 0.25)
Developing the Formula
X
2
3
0
1
P( X ) .4219 .4219 .1406 .0156
Outcomes
Probability
Rewritten
3
0
3
c
c
c
1
.
25
.
75
(.75)(.75)(.75)
OOO
0
1
2
OOcOc
3
(.
25
)(.
75
)(.
75
)
3 .25 .75
c
c
O OO
1
c
c
OOO
OOOc
OOcO
OcOO
OOO
(.25)(.25)(.75)
(.25)(.25)(.25)
F
I
G
J
a
fa
f
HK
F
I
a
fa
f
G
J
HK
3I
F
.25fa
.75f
3a
G
J
H2K
3I
F
.25f a
.75f
1a
G
J
H3K
2
1
3
0
Developing the Formula
Be able to recognize the formula even though you can set up a Prob.
Distribution without it.
nI
F
P( X  k )  GJp a
1 pf
Hk K
k
F
I
G
J
a
fa
f
HK
F
I
a
fa
f
G
J
HK
3I
F
.25fa
.75f
3a
G
J
H2K
3I
F
.25f a
.75f
1a
G
J
H3K
Rewritten
n = # of observations
p = probablity of success
k = given value of variable
n k
3
0
3
1
.
25
.
75
0
1
2
3
3 .25 .75
1
2
1
3
0
Working with probability
distributions
State the distribution to be used
State important numbers
Binomial: n & p
Geometric: p
Define the variable
Write proper probability notation
Twenty-five percent of the customers entering a
grocery store between 5 p.m. and 7 p.m. use an
express checkout. Consider five randomly selected
customers, and let X denote the number among the
five who use the express checkout.
binomial
n=5
p = .25
X = # of people use express
What is the probability that two used the
express checkout?
Get started with the required components:
1. Binomial, n = 5, p = .25
2. X = # of people use express
3. P(X = 2)
REQUIRED COMPONENTS 1, 2, 3
1. Define the situation: Binomial, with n = 5 and p = 0.25
2. Identify X: let X = # of people use express
3. Write proper probability notation that the
question is asking: P( X = 2)
Calculator option #1:
nCr = n choose r = 5 math prb 3:nCr
2 enter
2
3
5
P  X  2     .25  .75   .2637
 2
Calculator option #2:
2nd vars 0:binompdf(5, 0.25, 2) = .2637
What is the probability that no more than three
used express checkout?
1. Binomial, n = 5, p = .25
2. X = # of people use express
3. P(X ≤3)
P(X ≤3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)
Calculator option:
P(X ≤3) = binomcdf(5, 0.25, 3) = .9844
What is the probability that at least four used
express checkout?
1. Binomial, n = 5, p = .25
2. X = # of people use express
3. P(X ≥ 4)
5
5
5
4
1
P  X  4     .25  .75     .25   .0156
5
 4
Calculator option:
P(X
 4) = 1-P(X<4)
1-binomcdf(5, 0.25, 3) = .0156
“Do you believe your children will have a higher
standard of living than you have?” This question was
asked to a national sample of American adults with
children in a Time/CNN poll (1/29,96). Assume that
the true percentage of all American adults who
believe their children will have a higher standard of
living is .60. Let X represent the number who believe
their children will have a higher standard of living
from a random sample of 8 American adults.
binomial
n=8
p = .60
X = # of people who believe…
Interpret P(X = 3) and find the numerical answer.
binomial
n=8
p = .60
X = # of people who believe
The probability that 3 of the people from the
random sample of 8 believe their children will
have a higher standard of living.
Interpret P(X = 3) and find the numerical answer.
binomial
n=8
p = .60
X = # of people who believe
P(X=3) = binomialpdf(8, .6, 3) = .1239
or
8
3
5
P  X  3    .6  .4 
 3
 .1239
Find the probability that none of the parents
believe their children will have a higher standard.
•binomial, n = 8, p = 0.60
•X = # of people who believe
•P(X = 0 )
Find the probability that none of the parents
believe their children will have a higher standard.
binomial
n=8
p = .60
X = # of people who believe
8
0
8
P  X  0     .6  .4 
0
 .00066
or
P(X=0) = binomialpdf(8, .6, 0) = .00066
Binomial vs. Geometric
The Binomial Setting The Geometric Setting
1. Each observation falls into 1. Each observation falls into
one of two categories.
one of two categories.
2. The probability of success 2. The probability of success
is the same for each
is the same for each
observation.
observation.
3. The observations are all
3. The observations are all
independent.
independent.
4. There is a fixed number n
of observations.
4. The variable of interest is the
number of trials required to
obtain the 1st success.
Developing the Geometric
Formula
X
1
2
3
4
Probability
1
a fa f
P X  n  1 p
6
d6id6i
d56id56id16i
5
5
5
1
d6id6id6id6i
5
1
n 1
p
The Mean and Standard
Deviation of a Geometric Random
Variable
If X is a geometric random variable with
probability of success p on each trial,
the expected value of the random
variable (the expected number of trials
to get the first success) is
1 p
1


2
p
p
X
Suppose we have data that suggest that 3% of a
company’s hard disc drives are defective. You have
been asked to determine the probability that the first
defective hard drive is the fifth unit tested.
•geometric p = .03
•X = # of disc drives till defective
•Find P(X = 5)
P  X  5   .97  .03  .0266
4
A basketball player makes 80% of her free throws.
We put her on the free throw line and ask her to
shoot free throws until she misses one. Let X = the
number of free throws the player takes until she
misses.
•Geometric, p = .20
•X = # of free throws till miss
What is the probability that she will make 5 shots
before she misses? (When is the 1st success?)
geometric
p = .20
X = # of free throws till miss
P  X  6   .80  .20   .0655
5
What is the probability that she will miss 5 shots
before she makes one?
geometric
p = .80
Y = # of free throws till make
P Y  6   .20  .80   .00026
5
What is the probability that she will make at most 5
shots before she misses?
geometric
p = .20
X = # of free throws till miss
P  X  6   .20   .80 .20   .80  .20 
2
 .80  .20   .80  .20   .80  .20 
3
4
 .7379
Or 1 – P(makes all 6 shots) = 1 – (.806) = .7379
5