Transcript chapter 6 ppt

Chapter 6
Probability
Suppose two six-sided die is 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.
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
is
the
die
E’ and
E alsobe
denote
the
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.
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)
What is Probability?
Three different approaches to
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
how the
As the number ofNotice
repetitions
of relative
a chance
frequency of heads
experiment increase,
the
chance
that
the
approaches ½ the larger
relative frequencythe
of occurrence
for an event
number of trials!
will differ from the true probability by more
than any small number approaches 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 is that
measure
of the
strength
beliefdifferent
that a
different
people
couldofassign
particular outcome
will 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.
Probability Rules!
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.
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
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(A  B)
P(B|A) 
P(A)
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 . . .
Age 18-29
At Least
One Tattoo
18
Age 30-50
6
44
50
24
76
100
Totals
No Tattoo
Totals
32
50
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
18
No Tattoo
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
18
No Tattoo
Totals
How many
Age 18-29
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 jointprobability 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.
Independent Events
Two events
if knowing that one
If are
twoindependent
events are not
will occur
(or has occurred)
independent,
they aredoes
saidnot change the
probability
that
the otherevents.
occurs.
to be
dependent
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
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
we just multiply
the will have a
What isevents,
the probability
that a child
probabilities
hitch hiker’s
thumb? together.
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 = 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
In the
long run, with
89% of
will run
P(E1) = P(E2) = .98 P(E3) = such
.94 systems
P(E4) = .99
properly when tested
shortly after purchase.
What is the probability that none of these
components are defective?
P (E1  E 2  E 3  E 4 ) 
(.98)(.98)(.94)(.99) = .89
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
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:
the then
1)PickSampling
a card atwithout
random,replacement
replace the–card,
arecard
typically dependent events.
pickevents
a second
2) Pick a card at random, do NOT replace, then
replacement – the events
pickSampling
a secondwith
card.
are typically independent events.
If I pick two cards from a standard deck
without replacement,
is the
probability
Probability ofwhat
a spade
given
I drew a
that I select twospade
spades?
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
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,
P(L) = .00207
P(LC) = .99723
P(+|L) = .937denominator
P(-|L) =becomes
.063
C) = .97
C)P(LC).
P(+|LC) = .03
P(+|L)P(L)P(-|L
+ P(+|L
the
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 )
Estimating Probabilities
using Simulation
Simulation
1.
Design a method that uses a random mechanism
(such as a random number generator or table,
Simulation
provides
a means
of estimating
tossing
a coin
or die, etc.)
to represent
an
probabilities
when
wethe
are
unable to
observation.
Be sure
that
important
determine them
or it
ispreserved.
characteristics
of theanalytically
actual process
are
impractical to estimate them empirically by
2. Generate an observation using the method in step 1
observation.
and determine if 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