Download Chapter 6 - Solanco School District Moodle

Chapter 6
Probability
How can the likelihood of an event be measured?
Chance Experiments and Events 6-1
• What is a chance experiment and how can those chances be
represented?
Suppose two six-sided die are rolled and
they both land on sixes.
Or a coin is flipped and it lands on heads.
Or record the color of the next 20 cars to
pass an intersection.
These would be examples of chance experiments.
Chance experiment – any activity or situation in
which there is uncertainty about which of two
or more plausible outcomes will result.
Suppose a six-sided die is rolled. The
possible outcomes are that the die could
land with 1 dot up or 2, 3, 4, 5, or 6 dots
up.
S = {1, 2, 3, 4, 5, 6}
The sum of the
“S”
stands
for
sample
space.
We
use
set notation
to list
This would be an example of probabilities
a sample
space.
of the
the outcomes of the sample space.
outcomes in the sample
space equals ONE.
Sample space - the collection of all possible
outcomes of a chance experiment
Suppose two coins are flipped. The sample
space would be:
S = {HH, HT, TH, TT}
Where H = heads and T = tails
We can also use a tree diagram to represent a sample
space.
H
H
T
T
H
T
We
HTfollow the
branches out to
show an
outcome.
Alternate Method—Charting:
What is the sample space for
flipping a coin four times?
Counting Principle
There are four choices to make
_____ _____ _____ _____ = ____ ways
SIMPLE WAY TO MAKE SURE GET ALL THE
OUTCOMES:
Since there are 16 possibilities
with two options on each
16÷2 = 8
heads then 8 tails
8÷2 = 4
heads then 4 tails repeat
4÷2 = 2
heads then 2 tails repeat
2÷2 = 1
head then 1 tail repeat
FLIP 1 FLIP 2 FLIP 3 FLIP 4
Rolling two distinct die (ie red and green)
—remember that a (1,2) is different than rolling (2,1)
1
2
3
4
5
6
1
(1, 1)
(1,2)
(1,3)
(1,4)
(1,5)
(1,6)
2
(2, 1)
(2,2)
(2,3)
(2,4)
(2,5)
(2,6)
3
(3, 1)
(3,2)
(3,3)
(3,4)
(3,5)
(3,6)
4
(4, 1)
(4,2)
(4,3)
(4,4)
(4,5)
(4,6)
5
(5, 1)
(5,2)
(5,3)
(5,4)
(5,5)
(5,6)
6
(6, 1)
(6,2)
(6,3)
(6,4)
(6,5)
(6,6)
--this can also be used for sums
+
1
2
3
4
5
6
1
2
3
4
5
6
7
2
3
4
5
6
7
8
3
4
5
6
7
8
9
4
5
6
7
8
9
10
5
6
7
8
9
10
11
6
7
8
9
10
11
12
Suppose a six-sided die is rolled. The
outcome that the die would land on an even
number would be
E = {2, 4, 6}
This would be an example of an event.
We typically use capital letters to denote an event.
Event - any collection of outcomes (subset)
from the sample space of a chance experiment
Suppose a six-sided die is rolled. The
event that the die would land on an even
number would be
E = {2, 4, 6}
The sum of the
probabilities of
The
superscript
“C”
What
would
the
event
that
the
die
E’ and
E alsobe
denote
theis
complementary
stands for complement
complement
of E
events equals
NOT landing on an
even number?
ONE.
EC = {1, 3, 5}
This is an example of complementary events.
Complement - Consists of all outcomes that are
not in the event
These complementary events can be shown
on a Venn Diagram.
E = {2, 4, 6} and EC = {1, 3, 5}
Let the circle represent event
E.
Let the rectangle represent
the sample space.
Let the shaded area represent
event not E.
Venn Diagrams
Shaded area represents the indicated statement
A or B
A and B
Disjoint Events
A’
Suppose a six-sided die is rolled. The event
that the die would land on an even number
would be E = {2, 4, 6}
The event that the die would land on a prime
number would be P = {2, 3, 5}
What would be the event E or P happening?
E or P = {2, 3, 4, 5, 6}
This is an example of the union of two
events.
The union of A or B - consists of all outcomes
that are in at least one of the two events, that
is, in A or in B or in both.
A or B  A  B
Consider a
The is
bride
marriage
or to the
This
similar
takes
allof
her
union
of
two
union
A and B.
stuff –& when
the
people
All of A and all of B
groom
takes
two
people
are put together!
all
his stuff
marry,
what&
they
putdo
it
do they
together!
with
their
possessions ?
This symbol means
And live happily
“union”
ever after!
Let’s revisit rolling a die and getting an even
or a prime number . . .
E or P = {2, 3, 4, 5, 6}
E or way
P would
be any this is with a Venn Diagram.
Another
to represent
number in either circle.
Even number
Why is the number 1
outside the circles?
6
1
Prime number
4
3
2
5
Suppose a six-sided die is rolled. The event that
the die would land on an even number would be
E = {2, 4, 6}
The event that the die would land on a prime
number would be P = {2, 3, 5}
What would be the event E and P happening?
E and P = {2}
This is an example of the intersection
of two events.
The intersection of A and B - consists of all
outcomes that are in both of the events
A and B  A  B
This symbol means
“intersection”
Let’s revisit rolling a die and getting an even
or a prime number . . .
E andbeP ONLY
= {2}
E and P would
the middle
To represent
thispart
withthat
a Venn Diagram:
the circles have in
common
4
6
1
3
2
5
Suppose a six-sided die is rolled.
Consider the following 2 events:
A = {2}
B = {6}
On a single die roll, is it possible for A
and B to happen at the same time?
These events are mutually exclusive.
Mutually exclusive (or disjoint) events -two
events have no outcomes in common; two
events that NEVER happen simultaneously
A Venn Diagram for the roll of a six-sided
die and the following two events:
A = {2} B = {6}
A and B are mutually
The intersection of A and
exclusive (disjoint) B is empty!
since they have no
outcomes in common
4
6
2
1
5
3
Practice with Venn Diagrams
On the following four slides you will find Venn
Diagrams representing the students at your
school.
Some students are enrolled in Statistics, some
in Calculus, and some in Computer Science.
For the next four slides, indicate what
relationships the shaded regions represent.
(use complement, intersection, and union)
Statistics
Calculus
Computer Science
Calculus or Computer Science
Statistics
Calculus
Computer Science
(Statistics or Computer Science) and not Calculus
Statistics
Calculus
Computer
Science
Com Sci
Statistics and Computer Science and not Calculus
Statistics
Calculus
Computer Science
Statistics and not (Computer Science or Calculus)
Summarization
• When you flip a coin twice, how
many ways can it land?
• a) 1

Given set A and B below find P(A or B)

A ={ 1, 3, 5, 7, 9}
B ={ 1, 2, 3, 4, 6, 12}

• b) 2
• c) 4
• d) none of the above




c, d
a)
b)
c)
d)
{1, 3}
{1, 3, 5, 7, 9}
{1, 2, 3, 4, 6, 12}
{1, 2, 3, 4, 5, 6, 7, 9,12}
Homework Pg 289-291 3, 5, 8, 9,
10 chart for part a don’t do part e, 11, 12
6.2 Definition of Probability?
What is probability?
Activator:
take a coin from your pocket
What is the probability of getting a head on the flip of a fair coin?
How do you know?
This is called theoretical probability
Flip a coin twice. Record the number of heads and tails.
How many had exactly one head and one tail.
Flip a coin eight times. Record the number of heads and tails.
How many had exactly four heads and four tails.
This is called experimental probability
The Classical Approach
When the outcomes in a sample space are
equally likely, the probability of an event E,
denoted by P(E), is the ratio of the number of
outcomes favorable to E to the total number of
outcomes in the sample space.
favorable outcomes
P (E ) 
total outcomes
Examples: flipping a coin, rolling a die, etc.
On some football teams, the honor of calling the toss at the
beginning of the football game is determined by random selection.
Suppose this week a member of the 11-player offensive team will be
selected to call the toss. There are five interior linemen on the
offensive team.
If event L is defined as the event that an interior linemen is
selected to call the toss, what is probability of L?
P(L) = 5/11
Consider an archer shooting arrows at a target.
The probability of getting a bulls’ eye should be
the ratio of the area of the inner circle to the
area of the entire target.
What if a very experienced
archer were shooting the arrows?
Would the probability of a bull’s
eye still be the same?
The classical approach doesn’t work
for every situation.
The Relative Frequency Approach
The probability of event E, denoted by P(E), is
defined to be the value approached by the
relative frequency of occurrence of E in a very
long series of trials of a chance experiment.
Thus, if the number of trials is quite large,
number of times E occurs
P( E ) 
number of trials
Consider flipping a coin and recording the
relative frequency of heads.
When the number of
coin flips is small,
there is a lot of
variability in the
relative frequency of
“heads” (as shown in
this graph).
What do you notice in
the graph at the
right?
Consider flipping a coin and recording the
relative frequency of heads.
The graph at the
right shows the
relative frequency
when the coin is
flipped a large
number of times.
What do you notice in
this graph at the
right?
Law of Large Numbers (James Bernoulli)
Notice how the relative
As the number of repetitions offrequency
a chance experiment
increase,
the
of heads
chance that the relative frequency
of occurrence
for an event will differ
approaches
½ the larger
from the true probability by more
than
any
small
number
approaches
the number of trials!
0.
OR in other words, after a large
number of trials, the relative
frequency approaches the true
probability.
The Subjective Approach
Probability can be interpreted as a personal
The
problem
with
a subjective
approach
measure
of the
strength
of belief
that ais that
different
peoplewill
could
assign different
particular
outcome
occur.
probabilities to the same outcome based on their
subjective viewpoints.
Example: An airline passenger may report
that her probability of being placed on
standby (denied a seat) due to overbooking is
0.1. She arrived at this through personal
experience and observation of events.
6.3
Probability Rules!
What are the basic properties of probability?
Fundamental Properties of Probability
Property 1. Legitimate Values
For any event E,
0 < P(E) < 1
Property 2. Sample space
If S is the sample space,
P(S) = 1
Properties Continued . . .
Property 3. Addition
If two events E and F are disjoint,
P(E or F) = P(E) + P(F)
Property 4. Complement
For any event E,
P(E) + P(not E) = 1
Probabilities of Equally Likely Outcomes
Consider an experiment that can result in any one of N
possible outcomes.
Denote the simple events by O1, O2, …, ON. If these
simple events are equally likely to occur, then
1
1
1
, P (O2 )  , , P (ON ) 
1. P (O1 ) 
N
N
N
2. For any event E,
number of outcomes in E
P (E ) 
N
Suppose you roll a six-sided die once. Let E
be the event that you roll an even number.
P(E) = P(2 or 4 or 6) = 3/6
Number of outcomes in E
Over N
Addition Rule for Disjoint Events
If events E1, E2, . . ., Ek are disjoint (mutually exclusive)
events, then
P(E1 or E2 or . . . or Ek) =
P(E1) + P(E2) + . . . + P(Ek)
In words, the probability that any of these k disjoint
events occurs is the sum of the probabilities of the
individual events.
1) For any event E
0≤p(E)≤1
2) If S is the sample space for an experiment then
P(S) = 1
3) If two events are disjoint, P(E or F) = P(E) +P(F)
4) For any event E
P(E) + P(not E) = 1
P(E)
 =1 – P(not E) or P(not E) = 1- P(E)
A large auto center sells cars made by many different
manufacturers. Three of these are Honda, Nissan, and
Toyota. Consider a chance experiment that consist of
observing the make of the next car sold. Suppose that
P(H) = 0.25, P(N) = 0.18, P(T) = 0.14.
Are these disjoint events? yes
P(H or N or T) =
.25 + .18+ .14 = .57
P(not (H or N or T)) = 1 - .57 = .43

What is the probability of rolling a 6

At least a 6?

At most a 6?

A 3 or a 5?
+
1
2
3
4
5
6
1
2
3
4
5
6
7
2
3
4
5
6
7
8
3
4
5
6
7
8
9
4
5
6
7
8
9
10
5
6
7
8
9
10
11
6
7
8
9
10
11
12
Homework Pg 302-305 16, 21, 23, 24, 28, 29
Conditional probability
6-4
What is conditional probability and how can it be found?
Sometimes the knowledge that one event has occurred
changes our assessment of the likelihood that another event
occurs.
Consider the following example:
Suppose that 0.1% of all the individuals
in a population
have a certain disease. The presence of the disease is not
discernable from appearances, but there is a screening test
for the disease.
Let D = the event that a person has the disease
P(D) = .001
Disease example continued . . .
Suppose that 0.1% of all the individuals
in a population have a certain
Knowingdisease.
that event P, the
persontest
tested
positive,
has
80% of those with positive
results
actually
occurred, changes the
have the disease.
probability
of event
D, the
20% of those with positive
test results
actually
person
having
the disease,
do NOT have the disease
(false
positive)
from 0.001 to 0.80.
Read:
Probability that a person has the disease
Let P = “GIVEN”
the event
that a person tests
the person tests positive
positive for the disease
P(D|P) = 0.80
This is an example of conditional probability.
Conditional Probability
A probability that takes into account a given condition has occurred
P( E and F) P( E  F )
P( E | F ) 

P(F)
P( F )
• Example
Consider the population to be all families with two children.
Assuming that there is an equal chance of having a boy or a girl
there are 4 possible outcomes. GG, BB, GB, BG (ordered by age)
1) What is the probability of choosing a family with two girls
given the family has at least one girl?
1
3
2) What is the probability that they have 2 girls given that the
1
oldest is a girl?
2
The article “Chances Are You Know Someone with a
Tattoo, and He’s Not a Sailor” (Associated Press, June
11, 2006) included results from a survey of adults aged
18 to 50. The accompanying data are consistent with
summary values given in the article.
At Least One
Tattoo
No Tattoo
Totals
Age 18-29
18
32
50
Age 30-50
6
44
50
24
76
100
Totals
Assuming these data are representative of
adult Americans and that an adult is
selected at random, use the given
information to estimate the following
probabilities.
Tattoo Example Continued . . .
At Least
One Tattoo
No Tattoo
Totals
Age 18-29
18
32
50
Age 30-50
6
44
50
24
76
100
Totals
What is the probability that a randomly
selected adult has a tattoo?
P(tattoo) = 24/100 = 0.24
Tattoo Example Continued . . .
Age 18-29
At Least
One Tattoo
No Tattoo
18
32
How many adults
in the
Age 30-50
6
44
are
18-76
Totals
24 ages
How sample
many adults
in the
Totals
This is50a
50
condition!
100
sample are 29?
ages 18-29
What isAND
the probability
that a randomly
have a tattoo?
selected adult has a tattoo if they are
between 18 and 29 years old?
P(tattoo|age 18-29) = 18/50 = 0.36
Tattoo Example Continued . . .
At Least
One Tattoo
No Tattoo
Totals
How many
Age 18-29
18
32
50 in
adults
Age 30-50
6
44
the50
sample
have
Totals
24 in the 76
100 a
How many adults
tattoo?
ages 18-29
Whatsample
is the are
probability
that a randomly
AND
haveisabetween
tattoo? 18 and 29 years
selected
adult
This is a
old if they have a tattoo?
condition!
P(age 18-29|tattoo) = 18/24 = 0.75
Sometimes the knowledge that one event has occurred does NOT
change our assessment of the likelihood that another event occurs.
Consider the genetic trait, hitch hiker’s
thumb, which is the ability to bend the
last joint of the thumb back at an angle
of 60° or more.
Whether or not an offspring has hitch hiker’s thumb is determined by
two random events: which gene is contributed by the father and which
gene is contributed by the mother. Which gene is contributed by the
father does NOT affect which gene is contributed by the mother
These are independent events.
Let’s consider a bank that offers different types of
loans:
The bank offers both adjustable-rate and fixed-rate loans on
single-family dwellings, condominiums and multifamily dwellings.
The following table, called a joint-probability table, displays
probabilities based upon the bank’s long-run loaning practices.
Single Family
Condo
Adjustable
.40
.21
.09
.70
Fixed
Total
.10
.50
.09
.30
.11
.20
.30
P(Adjustable loan) =
.70
Multifamily Total
Bank Loan’s Continued . . .
Single Family
Condo
Adjustable
.40
.21
.09
.70
Fixed
Total
.10
.50
.09
.30
.11
.20
.30
P(Adjustable loan) =
Multifamily Total
.70
P(Adjustable loan|Condo) =
.21/.30 = .70
Knowing that the loan is for a condominium does not
change the probability that it is an adjustable-rate
loan. Therefore, the event that a randomly selected
loan is adjustable and the event that a randomly
selected loan is for a condo are independent.
Example
In 1912, the titanic sank. Almost 1500 people died, most of them
men. Was that because a man was less likely to survive than a
woman or because men outnumbered women 3 to 1?
Female Survival Rate p(S|F)
Titanic Survival Data:
surviving female 344

 73%
female
470
Male
Female
Totals
Survived
367
344
711
Died
1364
126
1490
totals
1731
470
2201
Male Survival Rate
P(S|M)
surviving male 367

 21%
male
1731
Overall Survival Rate P(S|on the ship)
survived
711

 32%
on the ship 2201
Lore is likely to be correct. Women and children first!
Homework Pg 311-315 30, 32, 35, 37, 38, 40, 42
Independence
6-5
What are independent events and how are their probabilities found?
Independent Events
Two events are independent if knowing that one will occur (or
If two
not
has occurred) does
notevents
changeare
the
probability that the other
independent, they are said
occurs.
to be dependent events.
Two events, E and F, are said to be independent if P(E|F) = P(E).
If P(E|F) = P(E), it is also true that P(F|E) = P(F).
Multiplication Rule for Two Independent Events
Since:
P( E  F )
P( E | F ) 
P( F )
P( E  F )  P( E | F ) P( F )
Two events E and F are independent, if and only if,
P(E  F)  P(E)  P(F)
Hitch Hiker’s Thumb
Revisited
Suppose that there is a 0.10 probability that a
parent will pass along the hitch hiker’s thumb
gene toSince
theirthese
offspring.
are independent
What isevents,
the probability
that a child
we just multiply
the will have a
hitch hiker’s
thumb? together.
probabilities
P(H+ fromThis
momwould
AND happen
H+ fromifdad)
=
the mother
contributes a hitch hiker’s gene (H+)
0.1
× 0.1if=the
0.01father contributes a
AND
hitch hiker’s gene (H+).
Multiplication Rule for k Independent
Events
Events E1, E2, . . ., Ek are independent if
knowledge that some number of the events have
occurred does not change the probabilities that
any particular one or more of the other events
occurred.
P (E1  E2  ...  Ek )  P (E1 )  P (E2 )  ...  P (Ek )
This relationship remains valid if one or more of
the events are replaced by their complement
(not E).
Suppose that a desktop computer system consist of a monitor, a mouse, a
keyboard, the computer processor itself, and storage devices such as a disk
drive. Most computer system problems due to manufacturer defects occur
soon in the system’s lifetime. Purchasers of new computer systems are
advised to turn their computers on as soon as they are purchased and then to
let the computer run for a few hours to see if any problems occur.
Let
E1 = event that a newly purchased monitor is not defective
E2 = event that a newly purchased mouse is not defective
E3 = event that a newly purchased disk drive is not defective
E4 = event that a newly purchased processor is not defective
Suppose the four events are independent with
P(E1) = P(E2) = .98
P(E3) = .94
P(E4) = .99
Let
E1 =
E2 =
E3 =
E4 =
event
event
event
event
that
that
that
that
a
a
a
a
newly
newly
newly
newly
purchased
purchased
purchased
purchased
monitor is not defective
mouse is not defective
disk drive is not defective
processor is not defective
Suppose the four events are independent
with
In the long
run, 89% of
such
systems
will run
P(E1) = P(E2) = .98 P(E3) = .94 P(E
)
=
.99
4
properly
when tested
shortly after purchase.
What is the probability that none of these components are
defective?
P (E1  E2  E 3  E 4 ) 
(.98)(.98)(.94)(.99) = .89
Let
E1 =
E2 =
E3 =
E4 =
event
event
event
event
that
that
that
that
a
a
a
a
newly
newly
newly
newly
purchased
purchased
purchased
purchased
monitor is not defective
mouse is not defective
disk drive is not defective
processor is not defective
Suppose the four events are independent with
P(E1) = P(E2) = .98 P(E3) = .94 P(E4) = .99
What is the probability that all these components will run properly
except the monitor?
C
P (E1
 E2  E3  E 4 ) 
(.02)(.98)(.94)(.99) = .018
Suppose I will pick two cards from a standard deck. This can be
done two ways:
Sampling without replacement – the
1)Pick a card at random, replace the card, then pick a second
events are typically dependent events.
card
2) Pick a card at random, do NOT replace, then pick a second
Sampling with replacement – the events
card.
are typically independent events.
If I pick two cards from a standard deck without replacement,
Probability of a spade given I drew a
what is the probability that I select two spades?
spade on the first card.
Are the events E1 = first card is a spade and E2 =
second card is a spade independent? NO
P(E1 and E2) =
P(E1) × P(E2|E1) =
1 12 1


4 51 17
Suppose the manufacturer of a certain
brand of light bulbs made 10,000 of
these bulbs and 500 are defective.
You randomly pick a package of two
such bulbs off the shelf of a store. What is the
probability that both bulbs are defective?
Are the events E1 = the first bulb is defective
and E2 = the second bulb is defective
independent?
To answer this question, let’s explore
probabilities
of these of
twoselecting
events? a
Whatthe
would
be the probability
defective light bulb? 500/10,000 = .05
Light Bulbs Continued . . .
If a would
random
of size n of
is taken
from
What
besample
the probability
selecting
aa
population
ofbulb?
size N, then the outcomes of
defective
light
selecting successive items from the population
500/10,000
= .05
without replacement
can
be treated as
independent when the sample size n is at most
Having selected one defective bulb, what is the
5% of the population size N.
probability of selecting another without
replacement?
499/9999 = .0499
These values are so close to each other that
when rounded to three decimal places they are
both .050. For all practical purposes, we can
treat them as being independent.
Light Bulbs Continued . . .
What is the probability that both
bulbs are defective?
Are the selections independent?
We can assume independence.
P(defective  defective) 
(0.05)(0.05) = .0025
Homework Pg 321-324 43, 46, 47, 50, 51, 54, 55
Some General probability rules
What are some basic or general probability rules?
6-6
General Rule for Addition
Since the intersection is added in twice, we
For any twosubtract
events Eout
andthe
F, intersection.
P (E  F )  P (E )  P (F )  P (E  F )
E
F
Musical styles other than rock and
pop are becoming more popular. A
survey of college students finds that
the probability they like country
music is .40. The probability that they
liked jazz is .30 and that they liked
both is .10. What is the probability
that they like country or jazz?
P (Country  Jazz ) 
.4 + .3 -.1 = .6
Here is a process to use when calculating the
union of two or more events.
P (E  F )
Ask yourself, “Are the
events mutually exclusive?”
In some problems, the
Yes intersection of the two events is
No
given (see previous example).
P (E )  P (F )
In some problems,
P (E )  P (F )  P (E  F )
the intersection of
the two events is not
given, but we know
If independent
that the events are
independent.
P (E )  P (F )
Suppose two six-sided dice are rolled (one
white and one red). What is the probability
that the white die lands on 6 or the red
Howdie
can
lands on 1?
you find
Let
A = white die landing on 6
the
probability
of A and B?
B = red die landing on 1
Are A and B disjoint?NO, independent events
cannot be disjoint
P (A  B )  P (A)  P (B )  P (A  B )
1 1  1  1  11
     
6 6  6  6  36
General Rule for Multiplication
For any two events E and F,
P (E  F )  P (E | F )  P (F )
Here is a process to use when calculating the
intersection of two or more events.
P (A  B )
Yes
P (A)  P (B )
Ask yourself, “ Are these
events independent?”
No
P (A)  P (B | A)
There are seven girls and eight boys in a math
class. The teacher selects two students at
random to answer questions on the board. What is
the probability that both students are girls?
Are these events independent?
NO
7 6
P(G1  G2 ) 

 .2
15 14
Light Bulbs Revisited . . .
A certain brand of light bulbs are defective five
percent of the time. You randomly pick a package
of two such bulbs off the shelf of a store. What
is the probability that exactly one bulb is
defective?
Let D1 = first light bulb is defective
D2 = second light bulb is defective

P (exactly one defective)  P D1  D
C
2
  D
C
1
 D2
= (.05)(.95) + (.95)(.05) = .095

An electronics store sells DVD players made by
one of two brands. Customers can also purchase
Thiswarranties
can happenfor
in one
two
ways: The
extended
the of
DVD
player.
following
areextended
given:
1) Theyprobabilities
purchased the
warranty
player
Let B1 = eventand
thatBrand
brand 11 DVD
is purchased
OR2 is purchased
B2 = event that brand
2) E
They
purchased the extended warranty
= event that extended warranty is purchased
and Brand 2 DVD player
P(B1) = .7
P(B2) = .3
P(E|B1) = .2
P(E|B2) = .4
If a DVD customer is selected at random, what
is the probability that they purchased the
extended warranty?
E  E  B1   E  B2 
DVD Player Continued . . .
Let
B1 = event that brand 1 is purchased
B2 = event that brand 2 is purchased
E = event that extended warranty is purchased
P(B1) = .7
P(B2) = .3
P(E|B1) = .2
P(E|B2) = .4
If a DVD customer
selected
at events
random, what
Theseisare
disjoint
is the probability that they purchased the
extendedUse
warranty?
the General Multiplication Rule:
E  E  B1   E  B2 
P E   P E  B1   P E  B2 
P E   P E | B1   P (B1 )  P E | B2   P (B2 )
DVD Player Continued . . .
Let
B1 = event that brand 1 is purchased
B2 = event that brand 2 is purchased
E = event that extended warranty is purchased
P(B1) = .7
P(B2) = .3
P(E|B1) = .2
P(E|B2) = .4
If a DVD customer is selected at random, what
is the probability that they purchased the
extended warranty?
P E   P E | B1   P (B1 )  P E | B2   P (B2 )
P(E) = (.2)(.7) + (.4)(.3) = .26
This is an example of the Law of Total Probabilities.
Law of Total Probabilities
If B1 and B2 are disjoint events with
probabilities P(B1) + P(B2) = 1, for any event E
P (E )  P (E  B1 )  P (E  B2 )
 P (E | B1 )  P (B1 )  P (E | B2 )  P (B2 )
More generally B1, B2, …, Bk are disjoint events with
probabilities P(B1) + P(B2) + … + P(Bk) = 1, for any event E
P (E )  P (E  B1 )  P (E  B2 )  ...  P (E  Bk )
 P (E | B1 )  P (B1 )  P (E | B2 )  P (B2 )
 ...  P (E | Bk )  P (Bk )
Bayes Rule (Theorem)
• A formula discovered by the Reverend Thomas
Bayes, an English Presbyterian minister, to
solve what he called “converse” problems.
Let’s examine the following problem before
looking at the formula . . .
Lyme’s disease is the leading tick-borne disease in the
United States and England. Diagnosis of the disease is
difficult and is aided by a test that detects particular
antibodies in the blood. The article, “Laboratory
Consideration in the Diagnosis and Management of Lyme
Borreliosis”, American Journal of Clinical Pathology, 1993,
used the following notations:
+ represents a positive result on a blood test
- represents a negative result on a blood test
L represents the patient actually has Lymes
LC represents the patient doesn’t have Lymes
The article gave the following probabilities:
P(L) = .00207
P(LC) = .99723
P(+|L) = .937
P(-|L) = .063
P(+|LC) = .03
P(-|LC) = .97
Lyme’s Disease Continued . . .
The article gave the following probabilities:
P(L) = .00207
P(LC) = .99723
P(+|L) = .937
P(-|L) = .063
P(+|LC) = .03
P(-|LC) = .97
Bayes’s converse problem poses this question:
“Given that a patient test positive, what is the
probability that he or she really has the
disease?”
written: P(L|+)
This question is of primary concern in
medical diagnosis problems!
Lyme’s Disease Continued . . .
The article gave the following probabilities:
Using
the Law of Total
Probabilities, the
P(L) = .00207
P(LC) = .99723
P(+|L) = .937denominator
P(-|L) =becomes
.063
C)P(LC).
C) = .97
P(+|L)P(L)P(-|L
+ P(+|L
P(+|LC) = .03
Bayes reasoned as follows:
P (L   )
P (L |  ) 
P ( )
Substitute values:
PSince
(  | L )  P (L )

C
C
P (  |PL(L) P()L) P (P(
 |LL
) )  P (L )
we can use P(+|L)
P(L) for
.937×(.00207
) the numerator.

.937(.00207)  .03(.99793)
 .0596
Bayes Rule (Theorem)
If B1 and B2 are disjoint events with
probabilities P(B1) + P(B2) = 1, for any event E
P (E | B1 )P (B1 )
P (B1 | E ) 
P (E | B1 )P (B1 )  P (E | B2 )P (B2 )
More generally B1, B2, …, Bk are disjoint events with
probabilities P(B1) + P(B2) + … + P(Bk) = 1, for any event E
P (E | Bi )P (Bi )
P (Bi | E ) 
P (E | B1 )P (B1 )  P (E | B2 )P (B2 )  ...  P (E | Bk )P (Bk )
defective
Smoke Detectors
A large real estate manager purchased 50,000 smoke
detectors to comply with new city ordinances.
She purchased 25,000 from company A, 15,000 from
company B, and 10,000 from company C.
.04
not defective
A
.96
defective
.5
.05
.3
It is known that some of the smoke detectors
malfunction and go off spontaneously. It is also
known that 4% of the detectors produced by
company A are defective, 5% of the detectors
produced by company B are defective, and 6% of the
detectors produced by company C are defective.
A call is received by the management office that one
of the detectors is malfunctioning. Find the
probability that it was produced by company A.
B
.95
.06
.2
not defective
defective
C
.94
not defective
Let A={produced by company A}
D={smoke detector is defective}
P(produced by co. A given that it is defect.)
P( A  D)
P( A | D) 
P( D)
P ( A | D) 
.5(.04)
.5(.04)  .3(.05)  .2(.06)
Example pg 279 #6.27
Two shipping services offer overnight delivery of parcels, and both promise delivery before 10AM. A mail-order
catalog company ships 30% of its overnight packages using shipping service 1 and 70% using service 2. Service 1
fails to meet the 10AM delivery promise 10% of the time, whereas service 2 fails to deliver by 10 AM 8% of the
time. Suppose you made a purchase from this company and were expecting your package by 10 AM but it is late.
Which service is more likely to have been used—ie who should you call to see where your important package is?
S1 = service 1 S2 = service 2 L = late
We know P(S1) = .3 P(S2) = .7
P(L|S1) = .1
P(L|S2) = .08
Since you know the package is late you need to find P(S1|L) and P(S2|L)
P( S1 | L) 
P( L | S1 ) P( S1 )
(.1)(. 3)
.03


 .3488
P( L | S1 ) P( S1 )  P( L | S 2 ) P( S 2 ) (.1)(. 3)  (.08)(. 7) .086
P ( S 2 | L) 
P( L | S 2 ) P( S 2 )
(.08)(.7)
.056


 .6512
P( L | S1 ) P( S1 )  P( L | S 2 ) P( S 2 ) (.1)(.3)  (.08)(.7) .086
So you should call service two to see where the package is because even though they have a lower overall percentage rate (.08), they
have 70% of the company’s business so 65% of the late packages from this company are from service 2.
(it’s a lot of math this time, but if you are constantly having to call it would be beneficial to know who to call first and save time
overall by only making 1 call 65% of the time)
Homework Pg 332-336 59, 61, 63, 66, 67, 73, 74 (same
as 73 then the comparison)
6-7 Estimating Probabilities using Simulation
How do you accurately estimate probabilities?
Simulation
1.
Design a method that uses a random mechanism
(such as a random number generator or table,
tossing
a coin
or die, etc.)
to represent
an
Simulation
provides
a means
of estimating
observation.
Be sure
that
important
probabilities
when
wethe
are
unable to
characteristics
of theanalytically
actual process
areispreserved.
determine them
or it
estimate using
themthe
empirically
2. impractical
Generate anto
observation
method inby
step 1
and determine ifobservation.
the outcome of interest has
occurred.
3. Repeat step 2 a large number of times
4. Calculate the estimated probability by dividing the
number of observations of the outcome of interest
by the total number of observations generated.
Suppose that couples who wanted children were to
continue having children until a boy was born.
Would this change the proportion of boys in the
population?
We will use simulation to estimate the proportion
of boys in the population if couples were to
continue having children until a boy was born.
1) We will use a single-random digit to represent a
child, where odd digits represent a male birth
and even digits represent a female birth.
2) Select random digits from a random digit table
until a male is selected and record the number
of boys and girls.
3) Repeat step 2 a large number of times.
Boy Simulation Continued . . .
Continue
this
process
large
number
of
Below
are four
rows
from athe
random
digit
times
100 trials).
table at the
back(at
of least
our textbook.
Calculate the proportion of boys out of the
Row
number of children born.
6 0 9 3 8 7 6 7 9 9 5 6 2 5 6 5 8 4 2 6 4
7 4 1 0 1 0 2 2 0 4 7 5 1 1 9 4 7 9 7 5 1
Notice that with only 10 trials, the
8 6 4 7proportion
3 6 3 4 5of1boys
2 3 is
1 10/22,
1 8 0 which
0 4 8 2 0
9 8 0 2 8 7 9 3 is8 close
4 0 4to20.5!
0 8 9 1 2 3 3 2
Trial 1: girl, boy
Trial 5: boy
Trial 9: girl, boy
Trial 2: boy
Trial 6: boy
Trial 3: girl, boy
Trial 7: boy
Trial 4: girl, boy
Trial 8: girl, girl, boy
Trial 10: girl, girl,
girl, girl, girl,
girl, boy
Example:
A professor wants to simulate the possible scores on a 20 question true/false quiz when
the students are guessing. Rather than finding 500 students to write down 20 answers
and then scoring each paper.
You could place 20 balls 10 red/10 blue in a box, allow blue to represent true, make a selection, replace the
ball and repeat 20 times “per quiz”, record the # of corrects. Repeat 1000 times and construct a table of
probability.
Why 20 balls with 10 of each?
to make it random but equally likely to pick either color—in case one had a slightly different feel
OR
You could use a random number table or generator allowing evens to be correct and odds incorrect. If using
a random number table use one digit at a time (pg. 760-761) on the calculator select MATH PRB5
RNDINT(starting point, endingpoint, # of items)
Example:
Sam makes field goals 82% of the time. How could you simulate 20 trials
of making at least two of the the next three field goals.
Describe your method:
Homework Pg 344-346 78, 81, 82, 84
REVIEW:
What Can Go Wrong?
• Beware of probabilities that don’t add to 1.
• Don’t add probabilities of events if they are not disjoint.
• Don’t multiply probabilities of events if they’re not
independent.
• Don’t confuse disjoint and independent.
• Reversing the conditioning: If you have P(A|B) but want
the reverse probability P(B|A), you need to find P(A and
B) and P(A) and use the formula (Baye’s Rule).
• Don’t use a simple probability rule when a general one is
appropriate.
• Don’t assume independence without reason to believe it.
• Samples are drawn without replacement unless you are
told otherwise.
• Know how to formally prove that two events A and B are
independent.
definition of independent events
P(E|F) = P(E)
To prove independence show by cross
multiplying P(E|F): P(E∩F)=P(E|F)P(F)
Or the def of independent
P(E∩F) =P(E)P(F)
Review
Homework Pg 349-351 87, 88, 92, 93, 95, 98, 101, 102