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Thinking Mathematically by Robert Blitzer

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Thinking Mathematically by Robert Blitzer Powered By Docstoc
					  Thinking
Mathematically
      Chapter 11:
  Counting Methods and
   Probability Theory
  Thinking
Mathematically
         Section 1:
  The Fundamental Counting
          Principle
The Fundamental Counting Principle

If you can choose one item from a group of
M items and a second item from a group of
N items, then the total number of two-item
choices is M  N.

You MULTIPLY the numbers!
The Fundamental Counting Principle
At breakfast, you can have eggs, pancakes or cereal.
You get a free juice with your meal: either OJ or apple
juice. How many different breakfasts are possible?

      eggs               pancakes           cereal




OJ    apple    OJ         apple     OJ     apple

 1      2       3          4        5         6
 Example: Applying the Fundamental
        Counting Principle
• The Greasy Spoon Restaurant offers 6
  appetizers and 14 main courses. How many
  different meals can be created by selecting
  one appetizer and one main course?
• Using the fundamental counting principle,
  there are 14  6 = 84 different ways a
  person can order a two-course meal.
 Example: Applying the Fundamental
        Counting Principle
• This is the semester that you decide to take your
  required psychology and social science courses.
• Because you decide to register early, there are 15
  sections of psychology from which you can
  choose. Furthermore, there are 9 sections of social
  science that are available at times that do not
  conflict with those for psychology. In how many
  ways can you create two-course schedules that
  satisfy the psychology-social science requirement?
               Solution
The number of ways that you can satisfy the
requirement is found by multiplying the
number of choices for each course.
You can choose your psychology course
from 15 sections and your social science
course from 9 sections. For both courses
you have:
15  9, or 135 choices.
  The Fundamental Counting
          Principle
The number of ways a series of successive
things can occur is found by multiplying the
number of ways in which each thing can
occur.
  Example: Options in Planning a
        Course Schedule
  Next semester you are planning to take three
  courses - math, English, and humanities. Based
  on time blocks and highly recommended
  professors, there are 8 sections of math, 5 of
  English, and 4 of humanities that you find
  suitable. Assuming no scheduling conflicts, there
  are:
8  5  4 = 160 different three course schedules.
                 Example
Car manufacturers are now experimenting with
lightweight three-wheeled cars, designed for a
driver and one passenger, and considered ideal for
city driving. Suppose you could order such a car
with a choice of 9 possible colors, with or without
air-conditioning, with or without a removable
roof, and with or without an onboard computer. In
how many ways can this car be ordered in terms of
options?
                    Solution

This situation involves making choices with
 four groups of items.
color - air-conditioning - removable roof - computer
              9  2  2  2 = 72
Thus the car can be ordered in 72 different
 ways.
Example: A Multiple Choice Test

You are taking a multiple-choice test that
has ten questions. Each of the questions has
four choices, with one correct choice per
question. If you select one of these options
per question and leave nothing blank, in
how many ways can you answer the
questions?
                    Solution
We DON’T blindly multiply the first two numbers
we see. The answer is not 10  4 = 40.
We use the Fundamental Counting Principle to
determine the number of ways you can answer the
test. Multiply the number of choices, 4, for each of
the ten questions
      4444444444
             =1,048,576
Example: Telephone Numbers in
      the United States
Telephone numbers in the United States
begin with three-digit area codes followed
by seven-digit local telephone numbers.
Area codes and local telephone numbers
cannot begin with 0 or 1. How many
different telephone numbers are possible?
                 Solution

  We use the Fundamental Counting Principle
  to determine the number of different
  telephone numbers that are possible.
8  10  10  8  10  10  10  10  10  10
                =6,400,000,000
  Thinking
Mathematically
     Section 2:
    Permutations
             Permutations
• A permutation is an arrangement of
  objects.
  – No item is used more than once.
  – The order of arrangement makes a difference.
      Example: Counting
        Permutations
Based on their long-standing contribution to
rock music, you decide that the Rolling
Stones should be the last group to perform
at the four-group Offspring, Pink Floyd,
Sublime, Rolling Stones concert. Given
this decision, in how many ways can you
put together the concert?
                         Solution
  We use the Fundamental Counting Principle to
  find the number of ways you can put together the
  concert. Multiply the choices:
                 3211=6
         3 choices   2 choices    1 choice    1 choice

        offspring    whichever    only one
        pink floyd   of the two   remaining
                                              stones
        sublime      remaining
Thus, there are six different ways to arrange the
  concert if the Rolling Stones are the final group to
  perform.
      Example: Counting
        Permutations
You need to arrange seven of your favorite
books along a small shelf. How many
different ways can you arrange the books,
assuming that the order of the books makes
a difference to you?
                  Solution
You may choose any of the seven books for the
first position on the shelf. This leaves six choices
for second position. After the first two positions
are filled, there are five books to choose from for
the third position, four choices left for the fourth
position, three choices left for the fifth position,
then two choices for the sixth position, and only
one choice left for the last position.
         7  6  5  4  3  2  1 = 5040
 There are 5040 different possible permutations.
        Factorial Notation
If n is a positive integer, the notation n! is
the product of all positive integers from n
down through 1.
         n! = n(n-1)(n-2)…(3)(2)(1)

note that 0!, by definition, is 1.
                    0!=1
Permutations of n Things Taken r at a
                Time
The number of permutations possible if r
items are taken from n items:

         n!
 nPr = (n – r)!            = n(n – 1) (n – 2) (n – 3) . . . (n – r + 1)


  n! = n(n – 1) (n – 2) (n – 3) . . . (n – r + 1) (n - r) (n - r - 1) . . . (2)(1)

 (n – r)! =                                       (n - r) (n - r - 1) . . . (2)(1)
Permutations of n Things Taken r at a
                Time
The number of permutations possible if
r items are taken from n items:
nPr: starting at n, write down r numbers
going down by one:
  nPr   = n(n – 1) (n – 2) (n – 3) . . . (n – r + 1)
         1 2         3       4                r
                     Problem
A math club has eight members, and it must choose 5
officers --- president, vice-president, secretary, treasurer
and student government representative. Assuming that
each office is to be held by one person and no person can
hold more than one office, in how many ways can those
five positions be filled?
We are arranging 5 out of 8 people into the five distinct
offices. Any of the eight can be president. Once selected,
any of the remaining seven can be vice-president.
Clearly this is an arrangement, or permutation, problem.

  8P5   = 8!/(8-5)! = 8!/3! = 8 · 7 · 6 · 5 · 4 = 6720
 Permutations with duplicates.
• In how many ways can you arrange the
  letters of the word minty?
• That's 5 letters that have to be arranged, so
  the answer is 5P5 = 5! = 120
• But how many ways can you arrange the
  letters of the word messes?
• You would think 6!, but you'd be wrong!
                 messes
                 here are six permutations of messes
me s s e s 1
                 well, all 3! arrangements of the s's
me s s e s 2     look the same to me!!!!

me s s e s       This is true for any arrangement
             3
                 of the six letters in messes, so
me s s e s   4   every six permutations should
                 count only once.
me s s e s   5   The same applies for the 2!
                 arrangement of the e's
             6
 Permutations with duplicates.
• How many ways can you arrange the letters
  of the word messes?
• The problem is that there are three s's and 2
  e's. It doesn't matter in which order the s's
  are placed, because they all look the same!
• This is called permutations with duplicates.
 Permutations with duplicates.
• Since there are 3! = 6 ways to arrange the
  s's, there are 6 permutations that should
  count as one. Same with the e's. There are
  2! = 2 permutations of them that should
  count as 1.
• So we divide 6! by 3! and also by 2!
• There are 6!/3!2! = 720/12 = 60 ways to
  arrange the word messes.
 Permutations with duplicates.
• In general if we want to arrange n items, of which
  m1, m2, .... are identical, the number of
  permutations is


                    n!
                m1!m2!m3!
                 Problem
A signal can be formed by running different
colored flags up a pole, one above the other.
Find the number of different signals
consisting of 6 flags that can be made if 3
of the flags are white, 2 are red, and 1 is
blue


6!/3!2!1! = 720/(6)(2)(1) = 720/12 = 60
  Thinking
Mathematically
     Section 3:
    Combinations
      Combination: definition

A combination of items occurs when:
• The item are selected from the same
  group.
• No item is used more than once.
• The order of the items makes no
  difference.
 How to know when the problem is a
    permutation problem or a
       combination problem
• Permutation:
  – arrangement, arrange
  – order matters
• Combination
  – selection, select
  – order does not matter.
   Example: Distinguishing between
   Permutations and Combinations
• For each of the following problems, explain if the
  problem is one involving permutations or
  combinations.
• Six students are running for student government
  president, vice-president, and treasurer. The
  student with the greatest number of votes becomes
  the president, the second biggest vote-getter
  becomes vice-president, and the student who gets
  the third largest number of votes will be student
  government treasurer. How many different
  outcomes are possible for these three positions?
                  Solution
• Students are choosing three student
  government officers from six candidates.
  The order in which the officers are chosen
  makes a difference because each of the
  offices (president, vice-president, treasurer)
  is different. Order matters. This is a
  problem involving permutations.
   Example: Distinguishing between
   Permutations and Combinations
• Six people are on the volunteer board of
  supervisors for your neighborhood park. A
  three-person committee is needed to study
  the possibility of expanding the park. How
  many different committees could be formed
  from the six people on the board of
  supervisors?
                 Solution
• A three-person committee is to be formed
  from the six-person board of supervisors.
  The order in which the three people are
  selected does not matter because they are
  not filling different roles on the committee.
  Because order makes no difference, this is a
  problem involving combinations.
   Example: Distinguishing between
   Permutations and Combinations

• Baskin-Robbins offers 31 different flavors
  of ice cream. One of their items is a bowl
  consisting of three scoops of ice cream,
  each a different flavor. How many such
  bowls are possible?
                    Solution
• A three-scoop bowl of three different flavors is to
  be formed from Baskin-Robbin’s 31 flavors. The
  order in which the three scoops of ice cream are
  put into the bowl is irrelevant. A bowl with
  chocolate, vanilla, and strawberry is exactly the
  same as a bowl with vanilla, strawberry, and
  chocolate. Different orderings do not change
  things, and so this problem is combinations.
Combinations of n Things Taken r at a
               Time

          n                     n!
                 = nCr =
          r                  r!(n – r)!



Note that the sum of the two numbers on the bottom
(denominator) should add up to the number on the
top (numerator).
    Computing Combinations
• Suppose we need to compute 9C3
             9!        9!
   9 C3            
                    
   9
          3!(9  3)! 3!6!
                      3!6!
• r = 3, n – r = 6
• The denominator is the factorial of smaller of
  the two: 3!
    Computing Combinations
• Suppose we need to compute 9C3
             9!        9!
   9 C3            
                    
   9
          3!(9  3)! 3!6!
                      3!6!
• r = 3, n – r = 6
• In the numerator write (the product of) all the
  numbers from 9 down to n - r + 1 = 6 + 1 = 7:
• There should be the same number of terms in
  the numerator and denominator: 9  8  7
    Computing Combinations
• If called upon, there's a fairly easy way to
  compute combinations.
   – Given nCr , decide which is bigger: r or n – r.
   – Take the smaller of the two and write out the
     factorial (of the number you picked) as a
     product.
   – Draw a line over the expression you just wrote.
    Computing Combinations
• If called upon, there's a fairly easy way to
  compute combinations.
  – Now, put n directly above the line and directly
    above the leftmost number below.
  – Eliminate common factors in the numerator and
    denominator.
  – Do the remaining multiplications.
  – You're done!
    Computing Combinations
• Suppose we need to compute 9C3 .
  – n – r = 6, and the smaller of 3 and 6 is 3.
       3 4
        987
                   = 3  4  7 = 84
        321
        1 1
   Finding Probabilities from
             Odds
• If the odds in favor of an event E are a to b,
  then the probability of the event is given by


         P( E )   a
                  ab
   Finding Probabilities from
             Odds
• If the odds against an event E are a to b,
  then the probability of the event is given
  by

        P( E )   b
                 ab
  Finding Probabilities from
            Odds
• Example:
  – Suppose Bluebell is listed as 7:1 in the
    third race at the Meadowlands.
  – The odds listed on a horse are odds
    against that horse winning, that is, losing.
  – The probability of him losing is
    7 / (7+1) = 7/8.
  – The probability of him winning is 1/8.
   Finding Probabilities from
             Odds
• Example:
  – Suppose Bluebell is listed as 7:1 in the third
    race at the Meadowlands. (a:b against)
  – The odds listed on a horse are odds against that
    horse winning, that is, losing.
  – The probability of him losing is
                       a
    7 / (7+1) = 7/8. a  b
                                              b
  – The probability of him winning is 1/8. a  b
                                           ab
  Thinking
Mathematically
         Section 7:
  Events Involving And;
  Conditional Probability
        Independent Events
• Two events are independent events if the
  occurrence of either of them has no effect
  on the probability of the other.
• For example, if you roll a pair of dice two
  times, then the two events are independent.
  What gets rolled on the second throw is not
  affected by what happened on the first
  throw.
       And Probabilities with
        Independent Events
• If A and B are independent events, then
          P(A and B) = P(A)  P(B)

• The example of choosing from four pairs of
  socks and then choosing from three pairs of
  shoes (= 12 possible combinations) is an
  example of two independent events.
          Dependent Events
• Two events are dependent events if the occurrence
  of one of them does have an effect on the
  probability of the other.
• Selecting two Kings from a deck of cards by
  selecting one card, putting it aside, and then
  selecting a second card, is an example of two
  dependent events.
• The probability of picking a King on the second
  selection changes because the deck now contains
  only 51, not 52, cards.
      And Probabilities with
       Dependent Events
• If A and B are dependent events, then
• P(A and B) =
   P(A)  P(B given that A has occurred)
• written as
   P(A)  P(B|A)
      Conditional Probability
• The conditional probability of B, given A,
  written P(B|A), is the probability that event
  B will occur computed on the assumption
  that event A has occurred.

• Notice that when the two events are
  independent, P(B|A) = P(B).
     Conditional Probability
• Example:
  – Suppose you are picking two cards from a deck
    of cards. What is the probability you will pick a
    King and then another face card?
  – The probability of an King is 4 =
                                         1 .
                                   52 13
  – Once the King is selected, there are 11 face cards
    left in a deck holding 51 cards.
  – P(A) = 1 . P(B|A) = 11
              13            51
  – The probability in question is 1  11
                                   13 51
    Applying Conditional
Probability to Real-World Data
                   P(B|A) =

 observed number of times B and A occur together
         observed number of times A occurs
                      Review
P(not E)           1 – P(E)
P(A or B)          P(A) + P(B)            mutually
                             – P(A and B) exclusive:
                                          P(A) + P(B)
P(A and B)         P(A)  P(B|A)          independent:
                                          P(A)  P(B)
 Odds in favor -   P(E) / P(not E)        probability is
      a:b                                 a/(a+b)
 Odds against -    P(not E) / P(E)        probability is
      a:b                                 b/(a+b)
  Thinking
Mathematically
      Section 8:
    Expected Value
               Expected Value
• Expected value is a mathematical way to use
  probabilities to determine what to expect in various
  situations over the long run.
• For example, we can use expected value to find the
  outcomes of the roll of a fair dice.
• The outcomes are 1, 2, 3, 4, 5, and 6, each with a
  probability of 1 . The expected value, E, is computed
                  6
  by multiplying each outcome by its probability and
  then adding these products.
• E = 1 1 + 2 1 + 3 1 + 4 1 + 5 1 + 6 1
         6      6      6      6      6      6

    = (1+2+3+4+5+6)/6 = 21 = 3.5
                            6
             Expected Value
E = 1 1 + 2 1 + 3 1 + 4 1 + 5 1 + 6 1
        6      6      6      6      6       6
= (1 + 2 + 3 + 4 + 5 + 6)/6 = 21 = 3.5
                               6

  Of course, you can't roll a 3½ . But the
  average value of a roll of a die over a long
  period of time will be around 3½.
  Example Expected Value and Roulette
    A roulette wheel has 38 different
    "numbers."
• One way to bet in roulette is to place $1
  on a single number.
• If the ball lands on that number, you are
  awarded $35 and get to keep the $1 that
  you paid to play the game.
• If the ball lands on any one of the other
  37 slots, you are awarded nothing and
  the $1 you bet is collected.
Example Expected Value and Roulette
• 38 different numbers.
• If the ball lands on your number,
  you win awarded $35 and you keep
  the $1 you paid to play the game.
• If the ball lands on any of the other
  37 slots, you are awarded nothing
  and you lose the $1 you bet.
• Find the expected value for playing roulette if
  you bet $1 on number 11 every time. Describe
  what this means.
                        Solution
           Outcome           Gain/Loss   Probability
                                           1
                11           $35           38

                              $1
                                           37
               Not 11                      38

E = $35( 1 ) + (-$1)( 37)
          38            38
     35    37      2
=   $  -  $ =   -$ ≈  -$0.05
     38    38     38
      This means that in the long run, a player can
      expect to lose about 5 cents for each game
      played.
               Expected Value
• A real estate agent is selling a house. She gets a 4-
  month listing. There are 3 possibilities:
   – she sells the house:  (30% chance) earns $25,000
   – another agent sells
     the house:            (20% chance) earns $10,000
   – house not sold: (50% chance) loses $5,000
• What is the expected profit (or loss)?
• If the expected profit is at least $6000 she would
  consider it a good deal.
               Expected Value
 Outcome      Probability   Profit or    product
                              loss
she sells     0.3            +$25,000    +$7,500
other sells   0.2            +$10,000    +$2,000
doesn't sell 0.5              -$5,000     -$2,500
                                          +$7,000
The realtor can expect to make $7,000.
Make the deal!!!!

				
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