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 k
1
· k
2
· k
3
· · · k n 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 k
1
= 2 k
2
= 2 k
3
= 2 k
4
= 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 k
1 k
5
= 8 k
2
= 4 k
6
= 7 k
3
= 3 k
7
= 6
= 2 k
4 k
8
= 5
= 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., 1 st ,
2 nd , 3 rd place out of 5 contestants.
 n
P r
= 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?
8
P
3
= 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?
10
C
4
= 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
P(X) .0556 .1667
-$3
.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
P ( X )

( n
 n !
x
 
)!
x !
p x
( 1
 p ) n
 x




P(X) can be written as b(x;n,p) n !
( n
 x )!
x !
is the same as
Mean (average) =

= np n
C x
Variance =

2 = np(1-p)

Binomial Probability
 Binomial Distribution Notation









P(S) Probability of Success
P(F) p
Probability of Failure numerical probability of a success q= 1-p numerical probability of a failure
P(S) = p
P(F) = q=1-p n x number of trials 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! ∙ 0.2
3 (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.