Counting Principles;

1
Chapter 8

• Counting Principles
• Further Probability Topics

2
The Multiplication Principle Permutations/Combina
tions

• 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

3
The Multiplication Principle Permutations/Combina
tions

• 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

4
The Multiplication Principle Permutations/Combina
tions

• 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
• 2222 16
• ltTREE DIAGRAMgt

5
The Multiplication Principle Permutations/Combina
tions

• 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 k3 6 k4 5
• k5 4 k6 3 k7 2 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).

6
The Multiplication Principle Permutations/Combina
tions

• 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)!

7
The Multiplication Principle Permutations/Combina
tions

• 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).

8
The Multiplication Principle Permutations/Combina
tions

• 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

9
The Multiplication Principle Permutations/Combina
tions

• 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

10
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

11
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
12
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

13
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 1 2 3 4
• P(X) .4 .3 .2 .1
• E(x) ? X ? P(X)
• 1(.4) 2(.3) 3(.2) 4(.1)
• 2

14
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.

15
Binomial Probability

• When a random variable can take on a large
  number of values with particular characteris-tics
  it is convenient to express the probability
  distribution in terms of a formula.

16
Binomial Probability

• Binomial Probability Formula
• P(X) can be written as b(xn,p)
• is the same as nCx
• Mean (average) ? np
• Variance ?2 np(1-p)

17
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) q1-p
• n number of trials
• x number of successes
• 0 lt x lt n

18
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.

19
Binomial Probability

• Example Given the following Binomial Experiment
  characteristics, use the Binomial Table to find
  the corresponding probabilities
• (a) n 2 p .3 X1
• b(12,.3) .420
• (b) n 12 p .90 X 2
• b(212,.9) 0
• (c) n 20 p .50 X 10
• b(10 20, .5) .176

20
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

21
Binomial Probability

• Given n 10 and p 0.20
• (a) P(Xlt3) P(x0) P(x1) P(x2)
  P(x3)
• 0.107 0.268 0.302 0.201
• 0.878
• (b) P(X3) 0.201
• (c) P(X gt 5) P(x5) P(x6) P(x7)
  P(x8)P(x9)P(x10)
• .026 .006 .001 .000 .000 .000
• .033
• Rework (b) using binomial formula
• b(310,0.2)
• 10! 0.23 (1-.2)10-3
• (10-3)!3!
• 120(.2)
  3 (.8) 7
• 0.201

22
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?

23
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.