Transcript Chapter 8 - jonesmth110

Chapter 8
Counting Principles;
Further Probability Topics
The Multiplication Principle;
Permutations/Combinations

Counting Rules

When we wish to know the number of all possible
outcomes for a sequence of events.



Fundamental Counting Rule (Multiplication Principle)
Permutation Rule
Combination Rule
The Multiplication Principle;
Permutations/Combinations

Fundamental Counting Rule (Multiplication Principle)

In a sequence of n events in which the first one has k
possibilities and the second event has k and the third
has k, and so forth, the total number of possibilities of
the sequence will be
k1 · k2 · k3 · · · kn
where n is the number of events
and
k is the number of possible outcomes of each event
The Multiplication Principle;
Permutations/Combinations

Example

A quiz with four T/F questions. How many
possible answer keys?
If n = 4 then
k1 = 2 k2 = 2

k3 = 2
k4 = 2
2·2·2·2 = 16

<TREE DIAGRAM>
The Multiplication Principle;
Permutations/Combinations

A store manager wishes to display 8 different
brands of shampoo in a row. How many ways
can this be done?
If n = 8 then
k1 = 8 k2 = 7
k5 = 4 k6 = 3

k3 = 6
k7 = 2
k4 = 5
k8 = 1
8 ·7 ·6 ·5 ·4 ·3 ·2 ·1 = 40,320
or 8! (factorial)
Factorials determine the number of ways in which objects or persons
can be arranged in a line (Recall that 0! = 1).
The Multiplication Principle;
Permutations/Combinations

Permutations

The ordered arrangement of objects where “r” objects
are selected from a set of n distinct objects, i.e., 1st ,
2nd , 3rd place out of 5 contestants.

nPr
=
n! .
(n - r)!
The Multiplication Principle;
Permutations/Combinations

Combinations

The arrangement of objects without regard to
order where “r” objects are selected from a set of
n distinct objects (i.e., any 3 out of 5 contestants).
The Multiplication Principle;
Permutations/Combinations

In a board of directors composed of 8 people, how
many ways can 1 chief executive officer, 1 director,
and 1 treasurer be selected?


n=8
r=3
Need CEO, DIR., TRES.
 Perm or Comb?

8P3
= 8!
= 8!
(8-3)!
5!
= 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1 = 336
5·4·3·2·1
or
40320 = 336
120
The Multiplication Principle;
Permutations/Combinations

How many ways can a committee of 4 people
be selected from a group of 10 people?

n = 10 r = 4

Perm or Comb?

10C4
= 10!
= 10!
4!(10-4)! 4!6!
= 10 · 9 · 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1
4·3·2·1·6·5·4·3·2·1
or
3628800 = 210
17280
Probability Distributions;
Expected Value

Probability Distribution – any device (table,
graph) used to specify all possible values of a
variable along with its probabilities.

Two types of probability distributions:


discrete random variables (r.v.) – only certain values
e.g. whole numbers such as counts – people, cars,
etc.
continuous random variables – continuum of values
e.g. whole numbers and the numbers in between,
such as measurements like height, weight, etc.

random variable - a function that assigns a real number to
each outcome of an experiment
Probability Distributions;
Expected Value

The probabilities that a tutor sees 1, 2, 3, 4, or 5 students in any
one day are 0.10, 0.25, 0.25, 0.20, and 0.20 respectively
X
1
2
3
4
5
P(X)
.10
.25
.25
.20
.20
P(x)
.3
.2
.1
1
2
3
4
5
Probability Distributions;
Expected Value

If a player rolls two dice and gets a sum of 2 or 12, she wins $20. If the person
gets a 7, she wins $5. The cost to play the game is $3. Find the expectation of
the game.
Win
Lose
Gain (X)
$17
$2
-$3
P(X)
.0556 .1667 .7777
P(sum of 2) or P(sum of 12)=
1/36 + 1/36 = 2/36 = 1/18 = .0556
P(sum of 7)=
6/36 = 1/6 = .1667
P(o/w)=
1 - .0556 - .1667 = .7777
E(X) =
17(.0556) + 2(.1667) + -3(.7777) = -$1.05
Means that theoretically there will be an average loss of about a dollar
Probability Distributions;
Expected Value

A recent survey by an insurance company showed the
following probabilities for the number of automobiles each
policyholder owned. Find the expected value.
# of autos, X
P(X)
1
.4
2
.3
3
.2
E(x) =  X  P(X)
= 1(.4) + 2(.3) + 3(.2) + 4(.1)
= 2
4
.1
Binomial Probability

Binomial Experiment


The same experiment is repeated several times (a
fixed number of times).
There are only two possible outcomes:



Success
Failure
The repeated trials are independent, so that the
probability of success remains the same for each
trial.
Binomial Probability
When a random variable can take on a large
number of values with particular characteristics it is convenient to express the probability
distribution in terms of a formula.
Binomial Probability

Binomial Probability Formula
n!
x
n x
P( X ) 
p (1  p)
(n  x)!
x!





P(X) can be written as b(x;n,p)
n!
(n  x)!x! is the same as nCx
Mean (average) =  = np
Variance = 2 = np(1-p)
Binomial Probability

Binomial Distribution Notation









P(S)
Probability of Success
P(F)
Probability of Failure
p
numerical probability of a success
q= 1-p numerical probability of a failure
P(S) = p
P(F) = q=1-p
n
number of trials
x
number of successes
0< x < n
Binomial Probability

Using the Binomial Table




step 1: find page with sample size under consideration.
step 2: find relevant value of p in column headings
step 3: find desired value of x in second column from left.
step 4: find probability at intersection of row x and column p.
Binomial Probability

Example: Given the following Binomial
Experiment characteristics, use the Binomial
Table to find the corresponding probabilities:
(a) n = 2 p= .3 X=1
b(1;2,.3) = .420
(b) n = 12 p = .90 X = 2
b(2;12,.9) = 0
(c) n = 20 p = .50 X = 10
b(10; 20, .5) = .176
Binomial Probability
EXAMPLE
If 20% of the people in a community use the emergency room at a hospital in
one year, find these probabilities for a sample of 10 people:
(a) At most three used the emergency room
(b) Exactly three used the emergency room
(c) At least five used the emergency room
Binomial Probability

Given n = 10 and p = 0.20

(a)



P(X<3) = P(x=0) + P(x=1) + P(x=2) + P(x=3)
= 0.107 + 0.268 + 0.302 + 0.201
= 0.878
(b) P(X=3) = 0.201
(c) P(X > 5) = P(x=5) + P(x=6) + P(x=7) +P(x=8)+P(x=9)+P(x=10)
= .026 + .006 + .001 + .000 + .000 + .000
= .033
Rework (b) using binomial formula
b(3:10,0.2)
=
10!
∙ 0.23 (1-.2)10-3
(10-3)!3!
= 120(.2) 3 (.8) 7
= 0.201
Binomial Probability
Example
In a restaurant, a study found that 42% of all
patrons smoked. If the seating capacity of the
restaurant is 80 people, how many seats should be
available for smoking customers?
Binomial Probability

Given: P(smoker) = .42

n = 80
 (average) = np = 80(.42) = 33.6

Therefore using mean as a “good” estimate, the
restaurant should have about 34 seats available for
smokers.