Probability by fjzhangxiaoquan

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									Math I
                       Probability Experiments
   Probability experiment
   • An action, or trial, through which specific results (counts,
     measurements, or responses) are obtained.
   Outcome
   • The result of a single trial in a probability experiment.
   Sample Space
   • The set of all possible outcomes of a probability
     experiment.
   Event
   • Consists of one or more outcomes and is a subset of the
     sample space.
Larson/Farber 4th ed                                                3
                       Probability Experiments

   • Probability experiment: Roll a die

   • Outcome: {3}

   • Sample space: {1, 2, 3, 4, 5, 6}

   • Event: {Die is even}={2, 4, 6}



Larson/Farber 4th ed                             4
       Example: Identifying the Sample Space

   A probability experiment consists of tossing a coin and
   then rolling a six-sided die. Describe the sample space.

     Solution:
     There are two possible outcomes when tossing a coin:
     a head (H) or a tail (T). For each of these, there are six
     possible outcomes when rolling a die: 1, 2, 3, 4, 5, or
     6. One way to list outcomes for actions occurring in a
     sequence is to use a tree diagram.


Larson/Farber 4th ed                                              5
       Solution: Identifying the Sample Space

     Tree diagram:




           H1 H2 H3 H4 H5 H6    T1 T2 T3 T4 T5 T6

      The sample space has 12 outcomes:
      {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}

Larson/Farber 4th ed                                     6
                       Simple Events

   Simple event
   • An event that consists of a single outcome.
       e.g. “Tossing heads and rolling a 3” {H3}

   • An event that consists of more than one outcome is
     not a simple event.
       e.g. “Tossing heads and rolling an even number”
        {H2, H4, H6}


Larson/Farber 4th ed                                      7
            Example: Identifying Simple Events

   Determine whether the event is simple or not.
   • You roll a six-sided die. Event B is rolling at least a 4.


    Solution:
    Not simple (event B has three outcomes: rolling a 4, a 5,
    or a 6)




Larson/Farber 4th ed                                              8
   Probability is the chance that something will happen.

   Probability is most often expressed as a fraction, a
    decimal, a percent, or can also be written out in words.

   To determine the probability…

        P(event) =         number of true outcomes
                     total number of equally likely outcomes
   Independent: Two events are independent if
    the occurrence of one has no effect on the
    occurrence of the other…
   Example 1: (probability of two events)
    What is the probability of drawing a king and then
    an ace from a standard 52 card deck with
    replacement?                      4 4        16
                     P(King, Ace) =       
                                      52 52     2704
   Example 2:
    What is the probability of flipping heads on a coin
    three times in a row?                1 1 1 1
                          P(H, H, H) =   
                                         2 2 2      8
   Example 3:
    A die is rolled twice. What’s the probability of rolling a 2
    and then an even number?
              1 1 1
    Solution:   
              6 2 12
   Example 4:
    You spin the spinner 3 times.
    What is the probability of spinning a 4, a 3 and then a 1?

                         Solution:      1 2 3  6   3
                                              
                                        8 8 8 512 256
                       Complementary Events

   Complement of event E
   • The set of all outcomes in a sample space that are not
     included in event E.
   • Denoted E ′ (E prime)
   • P(E ′) + P(E) = 1
   • P(E) = 1 – P(E ′)                               E′
   • P(E ′) = 1 – P(E)                    E




Larson/Farber 4th ed                                          13
         Let’s try a few
         Complement
                           Spinner
• In the spinner
  at the right:
• P’(yellow)
• P’(not blue)
• P’(red or yellow)
• P’(white)
• P’(odd number)
   Dependent: Two events such that the
    occurrence of one affects the occurrence of the


                         
    other.
     P(A and B) = P(A) P(B|A)

       **P(B|A) = means the probability of B
           given that event A has already
           occurred.
   Dependent: WITHOUT REPLACEMENT

   Independent: WITH REPLACEMENT
            Independent and Dependent Events

   Independent events
   • The occurrence of one of the events does not affect
     the probability of the occurrence of the other event
   • P(B | A) = P(B) or P(A | B) = P(A)
   • Events that are not independent are dependent




Larson/Farber 4th ed                                        18
        Example: Independent and Dependent
                      Events
   Decide whether the events are independent or dependent.
   1. Selecting a king from a standard deck (A), not
      replacing it, and then selecting a queen from the deck
      (B).

   Solution:
                                                                    4
        P( B | A)  P(2nd card is a Queen |1st card is a King ) 
                                                                    51
                              4
        P( B)  P(Queen) 
                             52
   Dependent (the occurrence of A changes the probability
   of the occurrence of B)
Larson/Farber 4th ed                                                     19
        Example: Independent and Dependent
                      Events
   Decide whether the events are independent or dependent.
   2. Tossing a coin and getting a head (A), and then
      rolling a six-sided die and obtaining a 6 (B).
   Solution:
                                                      1
        P( B | A)  P(rolling a 6 | head on coin) 
                                                      6
                                   1
        P( B)  P(rolling a 6) 
                                   6

   Independent (the occurrence of A does not change the
   probability of the occurrence of B)
Larson/Farber 4th ed                                      20
   You Spin A Spinner Twice
    ◦ Independent
   You choose a pair of shoes and a pair of
    socks
    ◦ Independent
   You buy two different CD’s
    ◦ Dependent
   You roll a number cube twice so that6 the
    sum is 8
    ◦ Dependent
   You spin a spinner and flip a coin
    ◦ Independent
   You pick a marble and pick again without
    replacing.
    ◦ Dependent
       Example: Using the Multiplication Rule

   Two cards are selected, without replacing the first card,
   from a standard deck. Find the probability of selecting a
   king and then selecting a queen.
    Solution:
    Because the first card is not replaced, the events are
    dependent.
                P( K and Q)  P ( K )  P (Q | K )
                                 4 4
                               
                                52 51
                                 16
                                     0.006
                                2652
Larson/Farber 4th ed                                           23
       Example: Using the Multiplication Rule

   A coin is tossed and a die is rolled. Find the probability
   of getting a head and then rolling a 6.
   Solution:
   The outcome of the coin does not affect the probability
   of rolling a 6 on the die. These two events are
   independent.
                       P ( H and 6)  P ( H )  P (6)
                                       1 1
                                         
                                       2 6
                                        1
                                            0.083
                                       12
Larson/Farber 4th ed                                            24
       Example: Using the Multiplication Rule

   The probability that a particular knee surgery is
   successful is 0.85. Find the probability that three knee
   surgeries are successful.

   Solution:
   The probability that each knee surgery is successful is
   0.85. The chance for success for one surgery is
   independent of the chances for the other surgeries.
     P(3 surgeries are successful) = (0.85)(0.85)(0.85)
                                   ≈ 0.614
Larson/Farber 4th ed                                          25
       Example: Using the Multiplication Rule

   Find the probability that none of the three knee
   surgeries is successful.

   Solution:
   Because the probability of success for one surgery is
   0.85. The probability of failure for one surgery is
   1 – 0.85 = 0.15
  P(none of the 3 surgeries is successful) = (0.15)(0.15)(0.15)
                                           ≈ 0.003

Larson/Farber 4th ed                                              26
       Example: Using the Multiplication Rule

   Find the probability that at least one of the three knee
   surgeries is successful.

   Solution:
   “At least one” means one or more. The complement to
   the event “at least one successful” is the event “none are
   successful.” Using the complement rule
     P(at least 1 is successful) = 1 – P(none are successful)
                                 ≈ 1 – 0.003
                                 = 0.997
Larson/Farber 4th ed                                            27
   Example: Using the Multiplication Rule to
             Find Probabilities
   More than 15,000 U.S. medical school seniors applied to
   residency programs in 2007. Of those, 93% were matched to
   a residency position. Seventy-four percent of the seniors
   matched to a residency position were matched to one of their
   top two choices. Medical students electronically rank the
   residency programs in their order of preference and program
   directors across the United States do the same. The term
   “match” refers to the process where a student’s preference
   list and a program director’s preference list overlap,
   resulting in the placement of the student for a residency
   position. (Source: National Resident Matching Program)
                                                  (continued)
Larson/Farber 4th ed                                              28
   Example: Using the Multiplication Rule to
             Find Probabilities
   1. Find the probability that a randomly selected senior was
      matched a residency position and it was one of the
      senior’s top two choices.
    Solution:
    A = {matched to residency position}
    B = {matched to one of two top choices}
    P(A) = 0.93 and P(B | A) = 0.74
    P(A and B) = P(A)∙P(B | A) = (0.93)(0.74) ≈ 0.688
                       dependent events

Larson/Farber 4th ed                                             29
   Example: Using the Multiplication Rule to
             Find Probabilities
   2. Find the probability that a randomly selected senior that
      was matched to a residency position did not get matched
      with one of the senior’s top two choices.
    Solution:
    Use the complement:
     P(B′ | A) = 1 – P(B | A)
                       = 1 – 0.74 = 0.26




Larson/Farber 4th ed                                              30
   Example 1:
    What is the probability of drawing a King
    and then an Ace without replacement?
                   4 4        16
    P(King, Ace) = 52  51  2652
   Example 2:
    You randomly select two marbles from a bag that
    contains 14 green, 7 blue, and 9 red marbles. What
    is the probability that the first marble is blue and
    the second marble is not blue if you do not replace
    the first marble?

    Solution:
                 7 23 161
                     
                 30 29 870
   Example 3:
    Your teacher passes around a basket with 6 red
    erasers, 9 blue erasers, and 7 green erasers. If you
    and your two neighbors are the first to randomly
    select an eraser, what is the probability that all
    three of you select green erasers?

Solution:
P(A) and P(B|A) and P(C|A and B)           7 6 5     210
                                                 
                                           22 21 20 9240
   The table shows the number of males and
    females with certain hair colors. Find …

 A) the probability that a listed person has red hair
 B) the probability that a female has red hair

           Brown    Blonde Red           Black     Other
           hair     hair   hair          hair
Male       42       11     3             17        27
Female 47           16         13        9         15
   P(red hair) = # of people with red hair 16   2
                                               
                   total # of people        200 25


    P(red hair | female) = # of red hair females      13

                                                   
                            total # of females       100
 When  you consider the outcomes for
 either of two events A and B, you form
 the union of A and B.
   When you consider only the outcomes shared
    by both A and B, you form the intersection of
    A and B.
   When the sets of A and B have nothing in
    common (no intersection) then they are
    considered mutually exclusive events.
Mutually exclusive
 Two events A and B cannot occur at the same time




      A
                  B               A              B


   A and B are               A and B are not
   mutually                  mutually exclusive
   exclusive
                                       40   Larson/Farber 4th ed
Decide if the events are mutually exclusive.
     Event A: Roll a 3 on a die.
     Event B: Roll a 4 on a die.

Solution:
Mutually exclusive (The first event has one
outcome, a 3. The second event also has one
outcome, a 4. These outcomes cannot occur
at the same time.)


                                    41   Larson/Farber 4th ed
Decide if the events are mutually exclusive.
     Event A: Randomly select a male student.
     Event B: Randomly select a nursing major.


Solution:
Not mutually exclusive (The student can be a
male nursing major.)



                                  42   Larson/Farber 4th ed
   If A and B are mutually exclusive events (one
    event does not have anything in common with
    the other), then…

       P(A or B) = P(A) + P(B)
   A die is rolled one time. What is the
    probability of rolling a 2 or a 6?


Solution:                                     1
                                            
                          1 1 2
P(A or B) = P(A) + P(B) =   
                             6   6     6      3
   A card is randomly selected out of a
    standard deck of 52 cards. What is the
    probability that it is a 2 or a king?

    P(A or B) = P(A) + P(B) =
                                    2
               4
                 
              52 52
                   4
                     
                       8
                       52
                                 
                                   13
 If A and B are not mutually exclusive, then
  there are some outcomes in common.
 Therefore, the intersection of A and B are
  counted twice when P(A) and P(B) are added.
 So, P(A and B) must be subtracted once from
  the sum…
    P(A or B) = P(A) + P(B) – P(A and B)
 A die is rolled one time. What is the
  probability of rolling an odd number or
  a prime number?
                           3
 Odd = 1, 3, 5     P(A) = 6
  Prime = 2, 3, 5 P(B) = 3
                         6
                                    2
 Odd and prime = 3, 5 P(A and B) =
                                    6
 A card is randomly selected from standard
  deck of 52 cards. What is the probability
  that it is a red card or a king?
 Red cards = 26 =     26
                       52
  Kings = 4 =   4
               52
                       2
   Red Kings = 2 =   52
   The probability that it will rain today is 40%. The
    probability that is will rain tomorrow is 30%. The
    probability that it will rain both days is 20%. Find
    the probability that it will rain either today OR
    tomorrow.
   Solution: P(A) + P(B) – P(A and B)

    P(today) + P(tomorrow) – P(today and tomorrow)
    =
           40%  30%  20%  50%
                       The Addition Rule

   Addition rule for the probability of A or B
   • The probability that events A or B will occur is
       P(A or B) = P(A) + P(B) – P(A and B)
   • For mutually exclusive events A and B, the rule can
     be simplified to
       P(A or B) = P(A) + P(B)
       Can be extended to any number of mutually
        exclusive events


Larson/Farber 4th ed                                       50
               Example: Using the Addition Rule

   You select a card from a standard deck. Find the
   probability that the card is a 4 or an ace.

   Solution:
   The events are mutually exclusive (if the card is a 4, it
   cannot be an ace)                     Deck of 52 Cards
        P (4 or ace)  P (4)  P( ace)
                                               4♣
                          4   4                        4♥
                                           4♠               A♣
                         52 52                    4♦        A♠ A♥
                          8                                  A♦
                            0.154      44 other cards
                         52
Larson/Farber 4th ed                                                51
               Example: Using the Addition Rule

   You roll a die. Find the probability of rolling a number
   less than 3 or rolling an odd number.

   Solution:
   The events are not mutually exclusive (1 is an
   outcome of both events)             Roll a Die
                                         4     6
                                        Odd    Less than
                                    3         1 three
                                         5         2


Larson/Farber 4th ed                                          52
               Solution: Using the Addition Rule
                                     Roll a Die
                                 4      6
                                Odd     Less than
                            3          1 three
                                 5           2



               P(less than 3 or odd )
           P(less than 3)  P(odd )  P(less than 3 and odd )
            2 3 1 4
               0.667
            6 6 6 6
Larson/Farber 4th ed                                             53
               Example: Using the Addition Rule

   The frequency distribution shows     Sales volume ($)   Months
   the volume of sales (in dollars)            0–24,999      3
   and the number of months a sales       25,000–49,999      5
   representative reached each sales      50,000–74,999      6
   level during the past three years.     75,000–99,999      7
   If this sales pattern continues,     100,000–124,999      9
   what is the probability that the     125,000–149,999      2
   sales representative will sell       150,000–174,999      3
   between $75,000 and $124,999         175,000–199,999      1

   next month?

Larson/Farber 4th ed                                                54
               Solution: Using the Addition Rule

   • A = monthly sales between                Sales volume ($)   Months
     $75,000 and $99,999                             0–24,999      3
   • B = monthly sales between                  25,000–49,999      5
     $100,000 and $124,999                      50,000–74,999      6

   • A and B are mutually exclusive             75,000–99,999      7
                                              100,000–124,999      9
            P ( A or B )  P ( A)  P ( B )   125,000–149,999      2
                            7   9             150,000–174,999      3
                             
                           36 36              175,000–199,999      1
                           16
                              0.444
                           36
Larson/Farber 4th ed                                                      55
               Example: Using the Addition Rule

   A blood bank catalogs the types of blood given by
   donors during the last five days. A donor is selected at
   random. Find the probability the donor has type O or
   type A blood.

                       Type O   Type A   Type B   Type AB   Total
       Rh-Positive      156      139      37        12      344
       Rh-Negative       28       25       8         4       65
       Total            184      164      45        16      409




Larson/Farber 4th ed                                                56
               Solution: Using the Addition Rule

   The events are mutually exclusive (a donor cannot have
   type O blood and type A blood)
                         Type O    Type A   Type B   Type AB    Total
         Rh-Positive       156      139       37        12      344
         Rh-Negative        28       25        8        4        65
         Total             184      164       45        16      409

              P (type O or type A)  P (type O )  P (type A)
                                     184 164
                                        
                                     409 409
                                     348
                                         0.851
                                     409
Larson/Farber 4th ed                                                    57
               Example: Using the Addition Rule

   Find the probability the donor has type B or is Rh-
   negative.
                       Type O   Type A   Type B   Type AB   Total
         Rh-Positive    156      139      37        12      344
         Rh-Negative     28       25       8         4       65
         Total          184      164      45        16      409


   Solution:
   The events are not mutually exclusive (a donor can have
   type B blood and be Rh-negative)

Larson/Farber 4th ed                                                58
               Solution: Using the Addition Rule

                       Type O   Type A   Type B   Type AB   Total
         Rh-Positive    156      139      37        12      344
         Rh-Negative     28       25       8         4       65
         Total          184      164      45        16      409


            P(type B or Rh  neg )
         P(type B)  P( Rh  neg )  P(type B and Rh  neg )
           45    65     8    102
                                0.249
          409 409 409 409

Larson/Farber 4th ed                                                59
   Formula




       Means the probability of B given that A has
        already occurred equals
        ◦ The probability of A and B occurring
          Divided by the probability of A

   Example 1:
    ◦ Gus rolled a six-sided cube, with faces numbered 1 to 6,
      twice. If you know that he rolled a 2 on his first roll,
      what is the probability that he rolled two even numbers?
      Find the probability of getting two even numbers given that
       he rolled a two
      Probability of A and B is 1/6 times ½ which equals 1/12
        Divide that by 1/6
        = 1/2
   Example 2:
    ◦ Antoine will toss a fair coin three times. What is the
      probability that all 3 tosses will result in heads if you
      know that his first toss resulted in heads?
      Find the probability of getting all three heads given that the
       first one is heads
      Probability of getting all three heads is ½ times ½ times ½
       which is 1/8
         Divide that by 1/2
         = 1/4
   Example 3:

      A bag contains 4 red marbles, 3 blue marbles, and 2
       green marbles. What is the probability that Nick reached
       into the bag and picked a red marble if you know that he did
       NOT pick a green marble?
      Did not say given that
        Don’t Over Think…..If he did not pick green, that means there
         were only 7 marbles left. His probability was 4/7.
   Example 4:
    ◦ Calculate the probability of rolling a total of 8 if the first
      roll is a 5.
       P(5) = 1/6 P(sum 8) = 3/6 P(A and B)= 3/36
       Divide by 1/6
         = 1/6
   The expected value is often referred to as the
    “long-term” average or mean .

   This means that over the long term of doing an
    experiment over and over, you would expect this
    average.

   To find the expected value or long term average,
    simply multiply each value of the random
    variable by its probability and add the products.
   Suppose that the following game is played. A
    man rolls a die. If he rolls a 1, 3, or 5, he loses
    $3, if he rolls a 4 or 6, he loses $2, and if he
    rolls a 2, he wins $12. What gains or losses
    should he expect on average? (What is his
    expected value?)
   We must find the probabilities of each outcome. We
    can make a chart to help us see this.


                    Possible        Probability
                    Losses/Gains    P(losses/gains)

                         -$3               3
                                           6

                         -$2                2
                                            6

                         $12
                                            1
                                            6
   Now, we multiply the probability for each outcome by
    the amount of money either gained or lost for that
    outcome.
                                                   3    9
                                                $3     $1.50
      Expected value of rolling a 1, 3, or 5: ________________
                                                   6    6
                                                   2    4
                                                $2     $0.67
      Expected value of rolling a 4 or 6:    ________________
                                                   6    6

                                                   1  12
                                              $12        $2.00
      Expected value of rolling a 2:         _______________
                                                  6 6
   To find the expected value for the entire game
    (the answer), simply add up the expected
    value for each outcome.

   Expected Value =
 $1.50  $0.67  $2.00  $0.17
   This means that the man playing this game is
    expected to lose an average of $0.17 each game
    he plays.

								
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