The Basic Practice of Statistics

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The Basic Practice of Statistics
Chapter 6 Introducing Probability
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A random phenomenon is a situation in which we
know what outcomes could happen, but we dont
know which particular outcome did or will happen.

• The probability of an event is its long-run
  relative frequency.
• While we may not be able to predict a particular
  individual outcome, we can talk about what will
  happen in the long run.

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• For any random phenomenon, each attempt, or
  trial, generates an outcome.
• Something happens on each trial, and we call
  whatever happens the outcome.
• These outcomes are individual possibilities, like
  the number we see on top when we roll a die.

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This is the sample space of event A. Event A is
rolling a total of five on a pair of dice.
Always express sample space in brackets.

• An event is any set or collection of outcomes.

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Sample Spaces

• Collection of all possible outcomes
• e.g. All six faces of a die
• e.g. All 52 cards in a deck

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All the possible outcomes of rolling two dice
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Events

• Simple event
• Outcome from a sample space with one
  characteristic
• e.g. A red card from a deck of cards
• Joint event
• Involves two outcomes simultaneously
• e.g. An ace that is also red from a deck of cards

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Simple Events
The Event of a Triangle
There are 5 triangles in this collection of 18
objects
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Joint Events
The event of a happy face and purple in color
One purple happy face
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Visualizing Events

• Contingency tables
• Tree diagrams

Ace Not Ace
Total
Black 2 24 26
Red 2 24 26

Total 4 48 52

Ace
Red Cards
Not an Ace
Full Deck of Cards
Ace
Black Cards
Not an Ace
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Contingency Table
A Deck of 52 Cards
Red Ace
Not an Ace
Total
Ace
Red
2
24
26
Black
2
24
26
Total
4
48
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Sample Space
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Joint Probability Using Contingency Table
Event
Total
B1
B2
Event
P(A1 and B2)
P(A1)
P(A1 and B1)
A1
P(A2 and B1)
A2
P(A2 and B2)
P(A2)
Total
1
P(B1)
P(B2)
Marginal (Simple) Probability
Joint Probability
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Compound Probability (Addition Rule)
P(A1 or B1 ) P(A1) P(B1) - P(A1 and B1)
Event
Total
B1
B2
Event
P(A1)
P(A1 and B1)
P(A1 and B2)
A1
P(A2 and B1)
A2
P(A2 and B2)
P(A2)
Total
1
P(B2)
P(B1)
For Mutually Exclusive Events P(A or B) P(A)
P(B)
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Conditional Probability Using Contingency Table
Color
Type
Total
Black
Red
2
2
4
Ace
24
24
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Non-Ace
26
26
52
Total
Revised Sample Space
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Tree Diagram
Event Possibilities
Ace
Red Cards
Not an Ace
Full Deck of Cards
Ace
Black Cards
Not an Ace
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• Tree Diagrams
• A tree diagram helps us think through conditional
  probabilities by showing sequences of events as
  paths that look like branches of a tree.
• Making a tree diagram for situations with
  conditional probabilities is consistent with our
  make a picture mantra.

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Probability Trees

• A method for solving probability problems
• Given probabilities for some events (perhaps
  union, intersection, or conditional)
• Find probabilities for other events
• Record the basic information on the tree
• Usually three probability numbers are given
• Perhaps two probability numbers if events are
  independent
• The tree helps guide your calculations
• Each column of circled probabilities adds up to 1
• Circled prob times conditional prob gives next
  probability
• For each group of branches
• Conditional probabilities add up to 1
• Circled probabilities at end add up to
  probability at start

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Probability Tree (continued)

• Shows probabilities and conditional probabilities

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Example Appliance Purchases

• First, record the basic information
• Prob(Washer) 0.20, Prob(Dryer) 0.25
• Prob(Washer and Dryer) 0.15

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Example (continued)

• Next, subtract 10.20 0.80, 0.250.15 0.10

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Example (continued)

• Now subtract 0.200.15 0.05, 0.800.10 0.70

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Example (completed tree)

• Now divide to find conditional probabilities
• 0.15/0.20 0.75, 0.05/0.20 0.25
• 0.10/0.80 0.125, 0.70/0.80 0.875

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Example (finding probabilities)

• Finding probabilities from the completed tree
• P(Washer) 0.20
• P(Dryer) 0.150.10 0.25
• P(Washer and Dryer) 0.15
• P(Washer or Dryer)
• 0.150.050.10 0.30
• P(Washer and not Dryer) 0.05
• P(Dryer given Washer) 0.75
• P(Dryer given not Washer) 0.125
• P(Washer given Dryer) 0.15/0.25 0.60
• (using the conditional probability formula)

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Tree Diagrams Figure is a nice example of a tree
diagram and shows how we multiply the
probabilities of the branches together
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Probability tree
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Conditional Probability Example

• Of the cars on a used car lot, 70 have air
  conditioning (AC) and 40 have a CD player (CD).
  20 of the cars have both.

• What is the probability that a car has a CD
  player, given that it has AC ?
• i.e., we want to find P(CD AC)

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Using Decision Trees
(continued)
P(CD and AC) 0.2
Given CD or no CD
Has AC
P(CD) 0.4
Does not have AC
P(CD and AC) 0.2
Has CD
All Cars
Does not have CD
P(CD and AC) 0.5
Has AC
P(CD) 0.6
Does not have AC
P(CD and AC) 0.1
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Using Decision Trees
P(AC and CD) 0.2
Given AC or no AC
Has CD
P(AC) 0.7
Does not have CD
P(AC and CD) 0.5
Has AC
All Cars
Does not have AC
P(AC and CD) 0.2
Has CD
P(AC) 0.3
Does not have CD
P(AC and CD) 0.1
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Probability trees

• Conditional probabilities can get complex and it
  is often a good strategy to build a probability
  tree that represents all possible outcomes
  graphically and assigns conditional probabilities
  to subsets of events.

P(chatting) 0.136 0.099 0.017
0.252 About 25 of all adult Internet users visit
chat rooms.
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Breast cancer screening
If a woman in her 20s gets screened for breast
cancer and receives a positive test result, what
is the probability that she has breast cancer?
She could either have a positive test and have
breast cancer, or have a positive test but not
have cancer (false positive).
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0.8
0.0004
0.2
0.1
0.9996
Incidence of breast cancer among women ages 2030
0.9
Mammography performance
Possible outcomes given the positive diagnosis
positive test and breast cancer, or positive test
but not cancer (false positive).
This value is called the positive predictive
value, or PV. It is an important piece of
information but, unfortunately, is rarely
communicated to patients.
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• Events
• When outcomes are equally likely, probabilities
  for events are easy to find just by counting.
• When the k possible outcomes are equally likely,
  each has a probability of 1/k.
• For any event A that is made up of equally likely
  outcomes

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Computing Probabilities

• The probability of an event E
• Each of the outcomes in the sample space is
  equally likely to occur

e.g. P( ) 2/36
(There are 2 ways to get one 6 and the other 4)
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We can see this in the following example. If we
flip a coin 500 times and it lands on heads 248
times, then the empirical probability is given by
Remember
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The probability of drawing a king is given by
Remember
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The probability of tossing a die and rolling a 7
is given byThe probability of tossing a die
and rolling a number less than 7 is given by
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The probability of drawing a heart is given by
Remember
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The probability of drawing a king is given by
Remember
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• Formal Probability
• Two requirements for a probability
• A probability is a number between 0 and 1.
• For any event A, 0 P(A) 1.

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Probability

• Probability is the numerical measure of the
  likelihood that an event will occur
• Value is between 0 and 1
• Sum of the probabilities of all mutually
  exclusive and collectively exhaustive
  events is 1

Certain
1
.5
Impossible
0
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• Formal Probability
• Something has to happen rule
• The probability of the set of all possible
  outcomes of a trial must be 1.
• P(S) 1 (S represents the set of all possible
  outcomes.)

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Independent versus disjoint events
Two events are independent if the probability
that one event occurs on any given trial of an
experiment is not influenced in any way by the
occurrence of the other event.
Imagine coins spread out so that half were heads
up, and half were tails up. Pick a coin at
random. The probability that is was head-up is
0.5. But, if you dont put it back, the
probability of picking up another head-up coin is
now less than 0.5. Without replacement,
successive trials are not independent.
In this example, the trials are independent only
when you put the coin back (sampling with
replacement) each time.
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Two events are disjoint if they have no outcomes
in common and can never happen together.
Events A and B are disjoint.
Events A and B are NOT disjoint.
If you flip a coin once, it may turn out head or
tail. However, you cannot obtain both head and
tail on the same flip. Head and tail are disjoint
events. If a couple gets a child, the child
could be a boy or a girl. The child cannot be
both boy and girl. Boy and girl are disjoint
events.
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General addition rule

• General addition rule for any two events A and B
• The probability that A occurs, or B occurs, or
  both events occur is
• P(A or B) P(A) P(B) P(A and B)

What is the probability of randomly drawing
either an ace or a heart from a pack of 52
playing cards? There are 4 aces in the pack and
13 hearts. However, one card is both an ace and a
heart. Thus P(ace or heart) P(ace)
P(heart) P(ace and heart) 4/52 13/52 -
1/52 16/52 0.3
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Mutually Exclusive Events

• Two events are Mutually Exclusive if they cannot
  both happen, that is, if
• Prob(A and B) 0
• No overlap
• in Venn diagram
• Examples
• Profit and Loss (for a selected business
  division)
• Green and Purple (for a manufactured product)
• Country Squire and Urban Poor (marketing
  segments)
• Mutually exclusive events are dependent events

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• Formal Probability
• Complement Rule
• The set of outcomes that are not in the event A
  is called the complement of A, denoted AC.
• The probability of an event occurring is 1 minus
  the probability that it doesnt occur P(A) 1
  P(AC)

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Since any event will either occur or it will not
occur, by rule 4 previously discussed, we get
RememberRule 4 the sum of the probabilities of
all possible outcomes of an experiment is 1.
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can also be stated
as
So the probability of tossing a die and not
rolling a 4 is
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Special Events
Null Event

• Impossible event
• e.g. Club diamond on one card draw
• Complement of event
• For event A, all events not in A
• Denoted as A
• e.g. A queen of diamonds A all cards
  in a deck that are not
  queen of diamonds

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• Formal Probability Addition Rule
• Events that have no outcomes in common (and,
  thus, cannot occur together) are called disjoint
  (or mutually exclusive).

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Special Events

• Mutually exclusive events
• Two events cannot occur together
• e.g. -- A queen of diamonds B queen of clubs
• Events A and B are mutually exclusive
• Collectively exhaustive events
• One of the events must occur
• The set of events covers the whole sample space
• e.g. -- A all the aces B all the black cards
  C all the diamonds D all the hearts
• Events A, B, C and D are collectively exhaustive
• Events B, C and D are also collectively exhaustive

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Formal Probability

• Addition Rule
• For two disjoint events A and B, the probability
  that one or the other occurs is the sum of the
  probabilities of the two events.
• P(A or B) P(A) P(B), provided that A and B
  are disjoint.

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So, out of the six numbers that can show up on
top, we have four ways that we can roll either a
5 or an even number. The probability is given by
Probability of rolling a 5
Probability of rolling an even number
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Example Venn Diagram

• Venn diagram probabilities correspond to
  right-hand endpoints of probability tree

P(Washer and Dryer)
P(not Washer and Dryer)
P(Washer and not Dryer)
P(not Washer and not Dryer)
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Notice, if we want the probability of rolling a 5
or rolling a number greater than 3. There are
three numbers greater than 3 on a die and one of
them is the 5. We cannot count the 5 twice. The
probability is given by
Probability of rolling a 5
Probability of rolling the same 5
Probability of rolling a number greater than 3
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Stated mathematically the rule is given by
Thus, the probability of drawing a 3 or a club
from a standard deck of cards is
Cards with a 3
Card that is a 3 and a club
Cards with clubs
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Computing Compound Probability

• Probability of a compound event, A or B

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In order to calculate the probability that we
roll a Five given that we roll a pair of dice.
You must add up all of the possible combinations
of that occurring do not forget ORDER!!!!! Red
die versus green die
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• What Can Go Wrong?
• Beware of probabilities that dont add up to 1.
• To be a legitimate probability distribution, the
  sum of the probabilities for all possible
  outcomes must total 1.
• Dont add probabilities of events if theyre not
  disjoint.
• Events must be disjoint to use the Addition Rule.

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• Sometimes we are interested in a combination of
  outcomes (e.g., a die is rolled and comes up
  even).
• A combination of outcomes is called an event.
• When thinking about what happens with
  combinations of outcomes, things are simplified
  if the individual trials are independent.
• Roughly speaking, this means that the outcome of
  one trial doesnt influence or change the outcome
  of another.
• For example, coin flips are independent.

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General multiplication rule

• The probability that any two events, A and B,
  both occur is
• P(A and B) P(A)P(BA)
• This is the general multiplication rule.
• If A and B are independent then P(A and B)
  P(A)P(B)
• (A and B are independent when they have no
  influence on each others occurrence.)

• What is the probability of randomly drawing
  either an ace or heart from a pack of 52 playing
  cards? There are four aces in the pack and
  thirteen hearts.
• P(heartace) 1/4 P(ace) 4/52
• P(ace and heart) P(ace) P(heartace)
  (4/52)(1/4) 1/52
• Notice that heart and ace are independent events.

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• Computing Joint Probability
• The probability of a joint event, A and B

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If the set of crayons consists only of red,
yellow, and blue, the probability of picking red
is . The probability of tossing a die and
rolling a 5 is . Butthe probability of
picking red and rolling a 5 is given by
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This can be illustrated using a tree diagram.
Since there are three choices for the color and
six choices for the die, there are eighteen
different results. Out of these, only one gives
a combination of red and 5. Therefore, the
probability of picking a red crayon and rolling a
5 is given by
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The multiplication rule for independent events
can be stated asThis rule can be extended for
more than two independent events
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• The Law of Large Numbers
• The Law of Large Numbers (LLN) says that the
  long-run relative frequency of repeated
  independent events gets closer and closer to the
  true relative frequency as the number of trials
  increases.
• For example, consider flipping a fair coin many,
  many times. The overall percentage of heads
  should settle down to about 50 as the number of
  outcomes increases.

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You can flip a coin ten times and have heads come
up seven times, but this does not mean that the
probability is 0.7. The more times a coin is
flipped, the more certainty we have to determine
the probability of coming up heads.
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• Thanks to the LLN, we know that relative
  frequencies settle down in the long run, so we
  can officially give the name probability to that
  value.
• Probabilities must be between 0 and 1, inclusive.
• A probability of 0 indicates impossibility.
• A probability of 1 indicates certainty.

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• Formal Probability
• Multiplication Rule
• For two independent events A and B, the
  probability that both A and B occur is the
  product of the probabilities of the two events.
• P(A and B) P(A) x P(B), provided that A and B
  are independent.

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• Formal Probability
• Multiplication Rule
• Two independent events A and B are not disjoint,
  provided the two events have probabilities
  greater than zero

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• Formal Probability - Notation
• Notation alert
• We use the notation P(A or B) and P(A and B).
• Or you might see the following
• P(A ? B) instead of P(A or B)
• P(A ? B) instead of P(A and B)

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• What Can Go Wrong? (cont.)
• Dont multiply probabilities of events if theyre
  not independent.
• The multiplication of probabilities of events
  that are not independent is one of the most
  common errors people make in dealing with
  probabilities.
• Dont confuse disjoint and independentdisjoint
  events cant be independent.

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Multiplication Rule for Dependent Events
Dependent events are events that are not
independent. The occurrence of one event affects
the probability of the occurrence of other
events. An example of dependent events is
picking a card from a standard deck then picking
another card from the remaining cards in the deck.
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For instance, what is the probability of picking
two kings from a standard deck of cards? The
probability of the first card being a king is
. However, the probability of the second
card depends on whether or not the the first card
was a king.
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If the first card was a king then the probability
of the second card being a king is .
If the first card was not a king, the
probability of the second card being a king is
. Therefore, the selection of the first card
affects the probability of the second card.
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When we are looking at probability for two
dependent events we need to have notation to
express the probability for an event to occur
given that another event has already occurred.
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If A and B are the two events, we can express the
probability that B will occur if A has already
occurred by using the notation This notation
is generally read as the probability of B, given
A.
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The multiplication rule can now be expanded to
include dependent events. The rule now
readsOf course, if A and B are independent,
then
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As an example, in a group of 25 people 16 of them
are married and 9 are single. What is the
probability that if two people are randomly
selected from the group, they are both married?
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If A represents the first person chosen is
married and B represents the second person chosen
is married thenHere, is now the
event of picking another married person from the
remaining 15 married persons. The probability
for the selection made in B is affected by the
selection in A.
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Conditional probability

• Conditional probabilities reflect how the
  probability of an event can change if we know
  that some other event has occurred/is occurring.
• Example The probability that a cloudy day will
  result in rain is different if you live in Los
  Angeles than if you live in Seattle.
• Our brains effortlessly calculate conditional
  probabilities, updating our degree of belief
  with each new piece of evidence.
• The conditional probability of event B given
  event A is(provided that P(A) ? 0)

If A and B are independent, P(B A) P(B).
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Conditional Probability (continued)

• Example appliance store purchases
• Prob(Washer) 0.20
• Prob(Dryer) 0.25
• Prob(Washer and Dryer) 0.15
• Conditional probability of buying a Dryer given
  that they bought a Washer
• Prob(Dryer given Washer)
• Prob(Washer and Dryer)/Prob(Washer) 0.15/0.20
  0.75
• 75 of those buying a washer also bought a dryer
• Conditional probability of Washer given Dryer
• Prob(Washer and Dryer)/Prob(Dryer) 0.15/0.25
  0.60
• 60 of those buying a dryer also bought a washer

Watch the denominator!
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Independent Events

• Two events are Independent if information about
  one does not change the likelihood of the other
• Three equivalent ways to check independence
• Prob (A given B) Prob (A)
• Prob (B given A) Prob (B)
• Prob (A and B) Prob (A) ? Prob (B)
• Two events are Dependent if not independent
• e.g., Prob(Washer and Dryer) 0.15
• Prob (Washer) ? Prob (Dryer) 0.20 ? 0.25
  0.05
• Washer and Dryer are not independent
• They are dependent

If independent, all three are true. Use any one
to check.
Not equal
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Permutations
A permutation is an arrangement of objects where
order is important. For instance the digits 1,2,
and 3 can be arranged in six different orders ---
123, 132, 213, 231, 312, and 321. Hence, there
are six permutations of the three digits. In
fact there are six permutations of any three
objects when all three objects are used.
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In general the number of permutations can be
derived from the Multiplication Principal. For
three objects, there are three choices for
selecting the first object. Then there are two
choices for selecting the second object, and
finally there is only one choice for the final
object. This gives the number of permutations for
three objects as 3 2 16.
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Now suppose that we have 10 objects and wish to
make arrangements by selecting only 3 of those
objects. For the first object we have 10
choices. For the second we have 9 choices, and
for the third we have 8 choices. So the number
of permutations when using 3 objects out of a
group of 10 objects is 10 9 8720.
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We can use this example to help derive the
formula for computing the number of permutations
of r objects chosen from n distinct objects r ?
n. The notation for these permutations is
and the formula is
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We often use factorial notation to rewrite this
formula. Recall that
And Using this notation we can rewrite the
Permutation Formula for as
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It is important to remember that in using this
formula to determine the number of
permutations 1. The n objects must be
distinct 2. That once an object is used it
cannot be repeated 3. That the order of
objects is important.
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Combinations
A combination is an arrangement of objects in
which order is not important. We arrange r
objects from among n distinct objects where r ?
n. We use the notation C(n, r) to represent this
combination. The formula for C(n, r) is given
by
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The Combination Formula is derived from the
Permutation Formula in that for a permutation
every different order of the objects is counted
even when the same objects are involved. This
means that for r objects, there will be r!
different order arrangements.
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So in order to get the number of different
combinations, we must divide the number of
permutations by r!. The result is the value we
get for C(n, r) in the previous formula.
Permutation
Combination
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Permutations of Repeated Objects
It is possible that in a group of objects some of
the objects may be the same. In taking the
permutation of this group of objects, different
orders of the objects that are the same will not
be different from one another.
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In other words if we look at the group of letters
in the word ADD and use D1 to represent the first
D, and D2 to represent the second, we can then
write the different permutations as AD1D2, AD2D1,
D1AD2, D2AD1, D1D2A, and D2D1A.
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But if we substitute the Ds back for the D1 and
D2, then AD1D2 and AD2D1 both appear as ADD, and
the six permutations become only three distinct
permutations. Therefore we will need to divide
the number of permutations by 2 to get the number
of distinct permutations.
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In permutations of larger groups of objects, the
division becomes a little more complicated. To
explain the process, let us look at the word
WALLAWALLA. This word has 4 As, 4 Ls, and 2
Ws.
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Consider that there are 10 locations for each of
these letters. These 10 locations will be filled
with 4 As, and since the As are all the same,
the order in which we place the As will not
matter. So if we are filling 10 locations with
4As the number of ways we can do this is C(10,
4).
Remember
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Once these 4 locations have been filled, there
remain 6 locations to fill with the 4 Ls. These
can be filled in C(6,4) ways, and the last 2
locations are filled with the Ws in C(2,2) ways.
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Finally, we multiply these together to get
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This leads to the general formula for
permutations involving n objects with n1 of one
kind, n2 of a second kind, and nk of a kth kind.
The number of permutations in this case is
where nn1n2nk.
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Counting other choices sometimes requires a bit
more reasoning to determine how many
possibilities there are.
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Suppose there are three cards that are each
marked with a different letter, A, B, or C. If
the cards are face down, and a person can pick
one, two or all three of the cards, what is the
possibility that the person will pick up the card
with the letter A on it?
?
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In this case there are three ways that one card
can be picked. Out of these there is only one
possibility of picking the A.
First way
Second way
Third way
A is picked!
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There are three ways of picking two cards. Out
of these three pairs, there are two that will
include the A.
First way
Second way
A is picked!
Third way
A is picked!
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There is only one way to pick all three cards,
and of course, if all three cards are picked, the
A will always be included.
A is picked!
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So there are a total of seven ways the cards can
be picked if the person can pick one, two, or all
three cards. Of these choices, four of them will
include the A, so the probability that the A will
be picked is
Possibilities of picking the A card
Total of ways to pick the three cards
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• When we want the probability of an event from a
  conditional distribution, we write P(BA) and
  pronounce it the probability of B given A has
  already occurred.
• A probability that takes into account a given
  condition is called a conditional probability.

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• It Depends To find the probability of the event
  B given the event A, we restrict our attention to
  the outcomes in A. We then find the fraction of
  those outcomes B that also occurred.
• Note P(A) cannot equal 0, since we know that A
  has occurred.

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• The General Multiplication Rule
• When two events A and B are independent, we can
  use the multiplication rule for independent
  events
• P(A and B) P(A) x P(B)
• However, when our events are not independent,
  this earlier multiplication rule does not work.
  Thus, we need the General Multiplication Rule.

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• The General Multiplication Rule (cont.)
• We encountered the general multiplication rule in
  the form of conditional probability.
• Rearranging the equation in the definition for
  conditional probability, we get the General
  Multiplication Rule
• For any two events A and B,
• P(A and B) P(A) x P(BA)
• or
• P(A and B) P(B) x P(AB)

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• Independence
• Independence of two events means that the outcome
  of one event does not influence the probability
  of the other.
• With our new notation for conditional
  probabilities, we can now formalize this
  definition
• Events A and B are independent whenever P(BA)
  P(B). (Equivalently, events A and B are
  independent whenever P(AB) P(A).)

118

• Independent ? Disjoint
• Disjoint events cannot be independent! Well, why
  not?
• Since we know that disjoint events have no
  outcomes in common, knowing that one occurred
  means the other didnt.
• Thus, the probability of the second occurring
  changed based on our knowledge that the first
  occurred.
• It follows, then, that the two events are not
  independent.
• A common error is to treat disjoint events as if
  they were independent, and apply the
  Multiplication Rule for independent eventsdont
  make that mistake.

119

• Depending on Independence
• Its much easier to think about independent
  events than to deal with conditional
  probabilities.
• It seems that most peoples natural intuition for
  probabilities breaks down when it comes to
  conditional probabilities.
• Dont fall into this trap whenever you see
  probabilities multiplied together, stop and ask
  whether you think they are really independent.

120

• Drawing Without Replacement
• Sampling without replacement means that once one
  individual is drawn it doesnt go back into the
  pool.
• We often sample without replacement, which
  doesnt matter too much when we are dealing with
  a large population.
• However, when drawing from a small population, we
  need to take note and adjust probabilities
  accordingly.
• Drawing without replacement is just another
  instance of working with conditional
  probabilities.

121

• Reversing the Conditioning
• Reversing the conditioning of two events is
  rarely intuitive.
• Suppose we want to know P(AB), but we know only
  P(A), P(B), and P(BA).
• We also know P(A and B), since
• P(A and B) P(A) x P(BA)
• From this information, we can find P(AB)

122

• Reversing the Conditioning
• Reversing the conditioning of two events is
  rarely intuitive.
• Suppose we want to know P(AB), but we know only
  P(A), P(B), and P(BA).
• We also know P(A and B), since
• P(A and B) P(A) x P(BA)
• From this information, we can find P(AB)

123

• Reversing the Conditioning (cont.)
• When we reverse the probability from the
  conditional probability that youre originally
  given, you are actually using Bayess Rule.

124
Bayess Theorem
Adding up the parts of A in all the Bs
Same Event
125
Bayess Theorem Using Contingency Table
Fifty percent of borrowers repaid their loans.
Out of those who repaid, 40 had a college
degree. Ten percent of those who defaulted had a
college degree. What is the probability that a
randomly selected borrower who has a college
degree will repay the loan?
126
Bayess Theorem Using Contingency Table
Repay
Repay
Total
College
.2
.05
.25
.3
.45
.75
College
1.0
.5
.5
Total
127
Bayes Theorem Example

• A drilling company has estimated a 40 chance of
  striking oil for their new well.
• A detailed test has been scheduled for more
  information. Historically, 60 of successful
  wells have had detailed tests, and 20 of
  unsuccessful wells have had detailed tests.
• Given that this well has been scheduled for a
  detailed test, what is the probability
• that the well will be successful?

128
Bayes Theorem Example
(continued)

• Let S successful well
• U unsuccessful well
• P(S) 0.4 , P(U) 0.6 (prior probabilities)
• Define the detailed test event as D
• Conditional probabilities
• P(DS) 0.6 P(DU) 0.2
• Goal is to find P(SD)

129
Bayes Theorem Example
(continued)
Apply Bayes Theorem

• So the revised probability of success, given that
  this well has been scheduled for a detailed test,
  is 0.667

130
Bayes Theorem Example
(continued)

• Given the detailed test, the revised probability
  of a successful well has risen to 0.667 from the
  original estimate of 0.4

Event Prior Prob. Conditional Prob. Joint Prob. Revised Prob.
S (successful) 0.4 0.6 (0.4)(0.6) 0.24 0.24/0.36 0.667
U (unsuccessful) 0.6 0.2 (0.6)(0.2) 0.12 0.12/0.36 0.333
Sum 0.36
131
Counting Principles
Sometimes determining probability depends on
being able to count the number of possible events
that can occur, for instance, suppose that a
person at a dinner can choose from two different
salads, five entrees, three drinks, and three
desserts. How many different choices does this
person have for choosing a complete dinner?
132
The Multiplication Principal for counting (which
is similar to the Multiplication Principle for
Probability) says that if an event consists of a
sequence of choices, then the total number of
choices is equal to the product of the numbers
for each individual choice.
133
If c1,c2, c3, ,cn, represent the number of
choices that can be made for each option then the
total number of choices is c1 c2 c3 cn For
our person at the dinner, the total number of
choices would then be 2 5 3 390 different
choices for combining salad, entrée, drink, and
dessert.
134
Counting Rules

• Rules for counting the number of possible
  outcomes
• Counting Rule 1
• If any one of k different mutually exclusive
  and collectively exhaustive events can occur on
  each of n trials, the number of possible
  outcomes is equal to

kn
135
Counting Rules
(continued)

• Counting Rule 2
• If there are k1 events on the first trial, k2
  events on the second trial, and kn events on
  the nth trial, the number of possible outcomes is
• Example
• You want to go to a park, eat at a restaurant,
  and see a movie. There are 3 parks, 4
  restaurants, and 6 movie choices. How many
  different possible combinations are there?
• Answer (3)(4)(6) 72 different possibilities

(k1)(k2)(kn)
136
Counting Rules
(continued)

• Counting Rule 3
• The number of ways that n items can be arranged
  in order is
• Example
• Your restaurant has five menu choices for lunch.
  How many ways can you order them on your menu?
• Answer 5! (5)(4)(3)(2)(1) 120 different
  possibilities

n! (n)(n 1)(1)
137
Permutations
A permutation is an arrangement of objects where
order is important. For instance the digits 1,2,
and 3 can be arranged in six different orders ---
123, 132, 213, 231, 312, and 321. Hence, there
are six permutations of the three digits. In
fact there are six permutations of any three
objects when all three objects are used.
138
In general the number of permutations can be
derived from the Multiplication Principal. For
three objects, there are three choices for
selecting the first object. Then there are two
choices for selecting the second object, and
finally there is only one choice for the final
object. This gives the number of permutations for
three objects as 3 2 16.
139
Now suppose that we have 10 objects and wish to
make arrangements by selecting only 3 of those
objects. For the first object we have 10
choices. For the second we have 9 choices, and
for the third we have 8 choices. So the number
of permutations when using 3 objects out of a
group of 10 objects is 10 9 8720.
140
We can use this example to help derive the
formula for computing the number of permutations
of r objects chosen from n distinct objects r ?
n. The notation for these permutations is
and the formula is
141
We often use factorial notation to rewrite this
formula. Recall that
And Using this notation we can rewrite the
Permutation Formula for as
142
Counting Rules

• Counting Rule 4
• Permutations The number of ways of arranging X
  objects selected from n objects in order is
• Example
• Your restaurant has five menu choices, and three
  are selected for daily specials. How many
  different ways can the specials menu be ordered?
• Answer
• different possibilities

143
It is important to remember that in using this
formula to determine the number of
permutations 1. The n objects must be
distinct 2. That once an object is used it
cannot be repeated 3. That the order of
objects is important.
144
Combinations
A combination is an arrangement of objects in
which order is not important. We arrange r
objects from among n distinct objects where r ?
n. We use the notation C(n, r) to represent this
combination. The formula for C(n, r) is given
by
145
The Combination Formula is derived from the
Permutation Formula in that for a permutation
every different order of the objects is counted
even when the same objects are involved. This
means that for r objects, there will be r!
different order arrangements.
146
So in order to get the number of different
combinations, we must divide the number of
permutations by r!. The result is the value we
get for C(n, r) in the previous formula.
Permutation
Combination
147
Permutations of Repeated Objects
It is possible that in a group of objects some of
the objects may be the same. In taking the
permutation of this group of objects, different
orders of the objects that are the same will not
be different from one another.
148
In other words if we look at the group of letters
in the word ADD and use D1 to represent the first
D, and D2 to represent the second, we can then
write the different permutations as AD1D2, AD2D1,
D1AD2, D2AD1, D1D2A, and D2D1A.
149
But if we substitute the Ds back for the D1 and
D2, then AD1D2 and AD2D1 both appear as ADD, and
the six permutations become only three distinct
permutations. Therefore we will need to divide
the number of permutations by 2 to get the number
of distinct permutations.
150
In permutations of larger groups of objects, the
division becomes a little more complicated. To
explain the process, let us look at the word
WALLAWALLA. This word has 4 As, 4 Ls, and 2
Ws.
151
Consider that there are 10 locations for each of
these letters. These 10 locations will be filled
with 4 As, and since the As are all the same,
the order in which we place the As will not
matter. So if we are filling 10 locations with
4As the number of ways we can do this is C(10,
4).
Remember
152
Once these 4 locations have been filled, there
remain 6 locations to fill with the 4 Ls. These
can be filled in C(6,4) ways, and the last 2
locations are filled with the Ws in C(2,2) ways.
153
Finally, we multiply these together to get