This assignment pertains to chapter 15 and focuses on:
Rules of probability, conditional probability
Addition rule in Venn diagram, ‹or , ›and
Appearance of rules in tables
Appearance of rules in trees
Independence vs dependence of events
Appearance of independence in tables
Appearance of independence in trees
Trees as an aid to reliable calculation
Recitation
Assignment 11 - 17 - 09
Bayes' formula (formula aside, it is just the rules in action)
You will also see some classical models in play.
This assignment pertains to chapter 15 and focuses on:
Rules of probability, conditional probability
1. Addition
An aircraft
has
two‹orengines.
The left engine has
rule in Venn
diagram,
, ›and
Appearance
of rulesof
in tables
probability
0.0001
failing over the course of a flight. The
of rules
in trees 0.0002 of failing over the course of
right Appearance
engine has
probability
Independence vs dependence of events
a flight.
Failure
of
both engines
Appearance of independence
in tables will be catastrophic. What is
the probability
at least one
engine will not fail?
Appearance ofthat
independence
in trees
Trees as an aid to reliable calculation
Bayes' formula (formula aside, it is just the rules in action)
This
cannot be answered using the information given. The
You will also see some classical models in play.
reason is that there are four regions of the Venn Diagram and
we
three has
things:
1. know
An only
aircraft
two engines. The left engine has
0.0001
=
P(A)
P(left engine
fails)
probability 0.0001 of=failing
over the
course of a flight. The
0.0002has
= P(B)
= P(right
engine
right engine
probability
0.0002
offails)
failing over the course of
of probabilities
of the four
=1
a flight.sumFailure
of both engines
will regions
be catastrophic.
What is
# at least
we need
the engine
sum of these
the probability that
one
willthree
not fail?"
A›andB
A›B
A›B
A›B
fail beonly
A fails using
onlythe
B fails
neither fails
Thisboth
cannot
answered
information
given. The
(we knowisonly
that)there are four regions of the Venn Diagram and
reason
that
sum ofonly
thesethree
two is 0.0001
"
we#know
things:
à sum of
0.0002engine
â
0.0001
= these
P(A)two
= isP(left
fails)
# 0.0002 =sum
of all=four
probabilities
is 1fails)
"
P(B)
P(right
engine
sum of probabilities of the four regions = 1
a. Fill out a Venn
labeling
A, B,
the four
# Diagram,
we need the
sum of these
threeand identify
"
regions
with
A›and
B their labels
A›B as used above.
A›B
A›B
both fail
only A fails only B fails neither fails
(we know only that)
# sum of these two is 0.0001 "
à sum of these two is 0.0002 â
#
sum of all four probabilities is 1
"
a. Fill out a Venn Diagram, labeling A, B, and identify the four
regions with their labels as used above.
rec11-17-09.nb
3
b. What if A, B are independent events? We say that events
A, B are statistically independent if the conditional probability
P(B§if A) (that is, the conditional probability the right engine
fails if the left engine is known to have failed) remains at P(B).
If engine failures are independent the probability P(B) of failure
of the right engine is not changed if we know that the left
engine has failed. Suppose this is true in the present case.
We now have one more piece of information for a total of four.
b. What if A, B are independent events? We say that events
Use
rule P(A›B)
= P(A) P(B
§ A) to
A, B the
are multiplication
statistically independent
if the conditional
probability
determine
P(A›B),
probabilityprobability
that both engines
fail.
P(B§if A) (that
is, thetheconditional
the rightwill
engine
Place
in theisVenn
diagram
solveremains
for eachat of
the
fails ifthis
the value
left engine
known
to haveand
failed)
P(B).
three
remaining
probabilities
of
the
regions
of
the
diagram.
If engine failures are independent the probability P(B) of failure
P(B) - P(A›B)
=
of P(A›B)
the right=engine
is not changed
if we know that the left
P(A›B)
=
engine
has failed.
Suppose this is true in the present case.
WeP(A›B)
now have
more
of information
for a total of four.
= 1one
- sum
ofpiece
the other
three probabilities
Fill out the diagram, labeling all four events and indicating
Use probabilities
the multiplication
rule P(A›B) = P(A) P(B § A) to
their
in the diagram.
determine P(A›B), the probability that both engines will fail.
Place this value in the Venn diagram and solve for each of the
three remaining probabilities of the regions of the diagram.
P(A›B) = P(B) - P(A›B) =
P(A›B) =
P(A›B) = 1 - sum of the other three probabilities
Fill out the diagram, labeling all four events and indicating
their probabilities in the diagram.
c. Once you have the probabilities of each of the four regions
(from (b)) you can use the display above (a) to calculate P(at
least one engine does not fail). Shade the relevant region in the
diagram.
0.0001be
= P(A)
= P(leftusing
engine
This cannot
answered
thefails)
information given. The
P(B)are
= P(right
engine of
fails)
reason 0.0002
is that =there
four regions
the Venn Diagram and
sum
of probabilities
we know
only
three things: of the four regions = 1
need the
sum of these
"
0.0001 =#
P(A) =weP(left
engine
fails)three
A›0.0002
A›B
A›B fails) A›B
= P(B)
= P(right engine
andB
bothsum
fail of probabilities
only A failsof the
only
B fails
neither
four
regions
= 1 fails
(we know only that)#
we need the sum of these three
"
#
sum ofBthese two is
0.0001 "
A›
A›B
A›B
A›B
and
these two
is 0.0002 only
â B fails neither fails
bothàfailsum of only
A fails
"
(we#know only that)sum of all four probabilities is 1
# sum of these two is 0.0001 "
a. Fill out
a Venn
labeling
A, B, and identify the four
à sum
of theseDiagram,
two is 0.0002
â
regions
asprobabilities
used above.
# with their
sumlabels
of all four
is 1
"
2
rec11-17-09.nb
a. Fill out a Venn Diagram, labeling A, B, and identify the four
regions with their labels as used above.
b. What if A, B are independent events? We say that events
A, B are statistically independent if the conditional probability
P(B§if A) (that is, the conditional probability the right engine
fails
if theifleft
is known to have
failed)
P(B).
b. What
A, engine
B are independent
events?
Weremains
say thatatevents
If
are independent
probability
P(B)probability
of failure
A,engine
B are failures
statistically
independent the
if the
conditional
of
right engine is not changed if we know that the left
P(B§the
if A) (that is, the conditional probability the right engine
engine
has left
failed.
Suppose
this
true
in the
present
case.
fails if the
engine
is known
to is
have
failed)
remains
at P(B).
We
now have
one are
more
piece of information
for a P(B)
total of
If engine
failures
independent
the probability
of four.
failure
of the right engine is not changed if we know that the left
Use
thehasmultiplication
rulethis
P(A›B)
P(B § A)
to
engine
failed. Suppose
is true=inP(A)
the present
case.
determine
P(A›B),
thepiece
probability
that bothforengines
will
fail.
We now have
one more
of information
a total of
four.
Place this value in the Venn diagram and solve for each of the
three
remaining
probabilities
the regions= ofP(A)
the diagram.
Use the
multiplication
ruleof P(A›B)
P(B § A) to
P(A›B)P(A›B),
= P(B) - P(A›B)
=
determine
the probability
that both engines will fail.
Place
this value
P(A›B)
= in the Venn diagram and solve for each of the
three
remaining
of thethree
regions
of the diagram.
P(A›B)
= 1 -probabilities
sum of the other
probabilities
P(A›B)
= P(B)
- P(A›B)
= all four events and indicating
Fill
out the
diagram,
labeling
their
probabilities
in the diagram.
P(A›B)
=
P(A›B) = 1 - sum of the other three probabilities
Fill out the diagram, labeling all four events and indicating
their probabilities in the diagram.
4
rec11-17-09.nb
c. Once you have the probabilities of each of the four regions
(from (b)) you can use the display above (a) to calculate P(at
least one engine does not fail). Shade the relevant region in the
diagram.
c. Once you have the probabilities of each of the four regions
(from (b)) you can use the display above (a) to calculate P(at
least one engine does not fail). Shade the relevant region in the
diagram.
d. There is another way to determine the probability that at
least one engine does not fail. From the display above (a) see
that the probability at least one engine does not fail is just 1 P(both
fail)is= another
1 - P(A›B).
Obtain
by thisthe
means
the probability
d. There
way to
determine
probability
that at
that
least
one engine
doesfail.
not fail.
leastatone
engine
does not
From the display above (a) see
2.
has at
two
engines,
continued.
Asfail
in #1
thatAn
theaircraft
probability
least
one engine
does not
is we
just will
1assume
P(A)==1 0.0001
and P(B)
= 0.0002.
But this
we will
P(both fail)
- P(A›B).
Obtain
by this means
thetime
probability
not
engine
failures
events. Engine failthat assume
at least one
engine
doesare
notindependent
fail.
ures
areaircraft
often caused
by flocks
of birds
at airports.
anwe
engine
2. An
has two
engines,
continued.
As inIf#1
will
fails
it isP(A)
likely
due to this
means
ofwill
the
assume
= 0.0001
and cause
P(B) =which
0.0002.
Butthat
this failure
time we
left
engine should
raise theareprobability
forevents.
failure Engine
of the right
not assume
engine failures
independent
failengine.
Let uscaused
suppose
heofprobability
of right Ifengine
failures are often
by that
flocks
birds at airports.
an engine
ure
to P(B
§ A) means
= 0.07 that
when
A occurs.
failsrises
it is from
likelyP(B)
due =to0.0002
this cause
which
failure
of the
left engine should raise the probability for failure of the right
a.
UsingLet
the us
multiplication
= P(A)
P(Bengine
§ A) deterengine.
suppose thatrule
he P(A›B)
probability
of right
failmine
the from
probability
engines
fail§ A)
in this
dependent
ure rises
P(B) =both
0.0002
to P(B
= 0.07
when Asetup.
occurs.
d. There is another way to determine the probability that at
d. There is another way to determine the probability that at
least one engine does not fail. From the display above (a) see
least one engine does not fail. From the display above (a) see
that the probability at least one engine does not fail is just 1 that the probability at least one engine does not fail is just 1 P(both fail) = 1 - P(A›B). Obtain by this means the probability
P(both fail) = 1 - P(A›B). Obtain by this means the probability
that at least one engine does not fail.
that at least one engine does not fail.
2. An aircraft has two engines, continued. As in #1 we will
2. An aircraft has two engines, continued. As in #1 we will
assume P(A) = 0.0001 and P(B) = 0.0002. But this time we will
assume P(A) = 0.0001 and P(B) = 0.0002. But this time we will
not assume engine failures are independent events. Engine failnot assume engine failures are independent events. Engine failures are often caused by flocks of birds at airports. If an engine
ures are often caused by flocks of birds at airports. If an engine
fails it is likely due to this cause which means that failure of the
fails it is likely due to this cause which means that failure of the
left engine should raise the probability for failure of the right
left engine should raise the probability for failure of the right
engine. Let us suppose that he probability of right engine failengine. Let us suppose that he probability of right engine failure rises from P(B) = 0.0002 to P(B § A) = 0.07 when A occurs.
ure rises from P(B) = 0.0002 to P(B § A) = 0.07 when A occurs.
rec11-17-09.nb
5
a. Using the multiplication rule P(A›B) = P(A) P(B § A) detera. Using the multiplication rule P(A›B) = P(A) P(B § A) determine the probability both engines fail in this dependent setup.
mine the probability both engines fail in this dependent setup.
b. Re-do the entire Venn Diagram (the probabilities will differ
b. Re-do the entire Venn Diagram (the probabilities will differ
what they were in the independent case).
what they were in the independent case).
c. From (a) determine the probability that at least one engine
c. From (a) determine the probability that at least one engine
will not fail in this dependent setup. Is it larger or smaller than
will not fail in this dependent setup. Is it larger or smaller than
when the engine failures were independent?
when the engine failures were independent?
d. From the completed Venn Diagram determine
d. From the completed Venn Diagram determine
P(left engine fails § right engine fails)
P(left engine fails § right engine fails)
= P(A § B) = P(A›B) / P(B).
= P(A § B) = P(A›B) / P(B).
e. If one engine were to fail, which would concern you most,
e. If one engine were to fail, which would concern you most,
failure of the left engine or failure of the right engine? Consult
failure of the left engine or failure of the right engine? Consult
P(B § A) versus P(A § B).
P(B § A) versus P(A § B).
rec11-17-09.nb
7
3. A box has colored balls [ 3R 4G 5Y ]. We will sample
3. A box has colored balls [ 3R 4G 5Y ]. We will sample
three balls without replacement and with equal probability on
three balls without replacement and with equal probability on
those then remaining in the box.
those then remaining in the box.
a. What is P(R1)?
a. What is P(R1)?
b. What are
b. What are
P(R2 § R1) =
P(R2 § R1) =
P(R2 § R1) =
P(R2 § R1) =
(from [2R
(from [2R
(from [3R
(from [3R
4G 5Y ])
4G 5Y ])
8other])
8other])
b. What is P(R2)? We may get it using the rules:
b. What is P(R2)? We may get it using the rules:
P(R2) = P(R1 R2) + P(R1 R2) - 0 (addition rule)
P(R2) = P(R1 R2) + P(R1 R2) - 0 (addition rule)
P(R1 R2) = P(R1) P(R2 § R1) (multiplication rule)
P(R1 R2) = P(R1) P(R2 § R1) (multiplication rule)
P(R1 R2) = P(R1) P(R2 § R1) (multiplication rule)
P(R1 R2) = P(R1) P(R2 § R1) (multiplication rule)
Make the calculations and see that you get P(R2) = P(R1)
Make the calculations and see that you get P(R2) = P(R1)
(establishing that "order of the deal does not matter) even
(establishing that "order of the deal does not matter) even
though draws are without replacement.
though draws are without replacement.
c. Use the definition of conditional probability to find P(R1 §
c. Use the definition of conditional probability to find P(R1 §
R2) = P(R1 R2) / P(R2) (its just the multiplication rule P(A›B) = P(A) P(B §
R2) = P(R1 R2) / P(R2) (its just the multiplication rule P(A›B) = P(A) P(B §
A) taking A = R2, B = R1).
A) taking A = R2, B = R1).
Calculation (c) above is an example of Bayes' Formula except
Calculation (c) above is an example of Bayes' Formula except
that we've not even had to see the formula. What has happened
that we've not even had to see the formula. What has happened
is that we've determined the probability of the "earlier" draw
is that we've determined the probability of the "earlier" draw
having been red based on our observation that the "later" draw
having been red based on our observation that the "later" draw
is red. We've reasoned from observation back to an underlying
is red. We've reasoned from observation back to an underlying
cause. Bayes (see chapter 15) wrote out the calculations in
cause. Bayes (see chapter 15) wrote out the calculations in
detail and sensed that he was onto something. It is greatly impordetail and sensed that he was onto something. It is greatly important that you too realize the power of his seemingly innocuous
tant that you too realize the power of his seemingly innocuous
calculation for (although you cannot be expected to see it) this
calculation for (although you cannot be expected to see it) this
idea establishes a type of "logical" engine that can place
idea establishes a type of "logical" engine that can place
"informed by experts" probabilities on things not known and
"informed by experts" probabilities on things not known and
update these probabilities in the light of new evidence. This is
update these probabilities in the light of new evidence. This is
called the "Bayesian Method." I have seen it used to restore
a. Using the multiplication rule P(A›B) = P(A) P(B § A) determine the probability both engines fail in this dependent setup.
b. Re-do the entire Venn Diagram (the probabilities will differ
what they were in the independent case).
6
rec11-17-09.nb
c. From (a) determine the probability that at least one engine
will not fail in this dependent setup. Is it larger or smaller than
when the engine failures were independent?
d. From the completed Venn Diagram determine
P(left engine fails § right engine fails)
= P(A § B) = P(A›B) / P(B).
c. If
From
determine
probability
that at
least one
e.
one (a)
engine
were tothefail,
which would
concern
youengine
most,
will notoffail
thisengine
dependent
setup.of Is
largerengine?
or smaller
than
failure
theinleft
or failure
theit right
Consult
when§ A)
theversus
engineP(A
failures
P(B
§ B). were independent?
3. A box has colored balls [ 3R 4G 5Y ]. We will sample
d. From
thewithout
completed
Venn Diagram
determine
three
balls
replacement
and with
equal probability on
engine failsin§ the
right
engine fails)
thoseP(left
then remaining
box.
= P(A § B) = P(A›B) / P(B).
a. What is P(R1)?
e. If one engine were to fail, which would concern you most,
failure
of are
the left engine or failure of the right engine? Consult
b. What
P(B § A)P(R2
versus
P(A= § B).
§ R1)
(from [2R 4G 5Y ])
P(R2 § R1) =
(from [3R 8other])
3. A box has colored balls [ 3R 4G 5Y ]. We will sample
three
ballsis without
replacement
with
b. What
P(R2)? We
may get itand
using
theequal
rules: probability on
those
then= remaining
the box.
P(R2)
P(R1 R2) +inP(R1
R2) - 0 (addition rule)
P(R1 R2) = P(R1) P(R2 § R1) (multiplication rule)
a. What is P(R1)?
P(R1 R2) = P(R1) P(R2 § R1) (multiplication rule)
Make the calculations and see that you get P(R2) = P(R1)
b. What are
(establishing that "order of the deal does not matter) even
P(R2 § R1) =
(from [2R 4G 5Y ])
though draws are without replacement.
P(R2 § R1) =
(from [3R 8other])
Whatthe
is P(R2)?
Weofmay
get it using
the rules: to find P(R1 §
c.b. Use
definition
conditional
probability
P(R2)
= P(R1
+ P(R1
R2)
0 (addition
R2)
= P(R1
R2) /R2)
P(R2)
(its just
the-multiplication
rulerule)
P(A›B) = P(A) P(B §
P(R1AR2)
=BP(R1)
A) taking
= R2,
= R1). P(R2 § R1) (multiplication rule)
P(R1 R2) = P(R1) P(R2 § R1) (multiplication rule)
Make the calculations
see that ofyou
get Formula
P(R2) = except
P(R1)
Calculation
(c) above isand
an example
Bayes'
(establishing
deal doesWhat
not has
matter)
even
that
we've notthat
even"order
had to of
see the
the formula.
happened
though
without replacement.
is
that draws
we've are
determined
the probability of the "earlier" draw
having been red based on our observation that the "later" draw
is red. We've reasoned from observation back to an underlying
c. Use the
definition
of conditional
probability
to find P(R1in§
cause.
Bayes
(see chapter
15) wrote
out the calculations
R2) = and
P(R1
R2) / that
P(R2)
the multiplication
= P(A)
P(B §
detail
sensed
he (its
wasjustonto
something.ruleItP(A›B)
is greatly
imporA)
taking
= R2, too
B = R1)
.
tant
thatA you
realize
the power of his seemingly innocuous
calculation for (although you cannot be expected to see it) this
Calculation
(c) above
is anofexample
Bayes' Formula
idea
establishes
a type
"logical"of engine
that can except
place
that we've not
had probabilities
to see the formula.
What
happened
"informed
by even
experts"
on things
nothas
known
and
is that these
we've probabilities
determined in
thethe
probability
of the
"earlier"This
draw
update
light of new
evidence.
is
having the
been"Bayesian
red basedMethod."
on our observation
thatit the
"later"
draw
called
I have seen
used
to restore
is red. We've
reasoned
observation
back(UNC
to an 1990)
underlying
images
from the
faultyfrom
Hubble
Telescope
(the
cause. not
Bayes
(seebeing
chapter
out the
calculations
in
things
known
the 15)
true wrote
star fields
as they
should be
detail andand
sensed
that he was
onto
something.
It is
greatly imporimaged)
to determine
how
many
extremely
expensive
older
tant thattoyou
the power
of his
seemingly
innocuous
rockets
testtoo
(to realize
destruction)
in order
to establish
reliability
of
calculation
(although
you cannot
expected
to in
seethe
it) fleet
this
the
fleet, theforthings
not known
being be
which
rockets
ideafaulty).*
establishes a type of "logical" engine that can place
are
*Experts
are consulted
"informing us"
as to the probabilities
might place
different forms
"informed
by experts"
probabilities
on they
things
not onknown
and
of true star fields being the actual ones, or different patterns of fault that might be present among
update
inand
thecalculations
light ofarenew
evidence.
This
is
the
fleet of these
rockets. probabilities
Massive math models
then brought
to bear which
calculate
the "updated"
probabilities of
different images
correctitones
conditional
on what
called
the "Bayesian
Method."
I being
havetheseen
used
to restore
Hubble reports. We then pick the conditionally most probable true star field and print it out. It
images
from
thehavefaulty
Hubble
Telescope
is involved but
hey, we
computers.
Guess what?
They fixed (UNC
the Hubble1990)
and those(the
early
restorations
(fromknown
when it was
faulty) were
things not
being
the really
truegood.
star fields as they should be
4.
Weather.
forecast how
says many
there is
80% chance
of rainolder
Sat
imaged)
and to Adetermine
extremely
expensive
and
60%tochance
rain Sun. in order to establish reliability of
rockets
test (toofdestruction)
the fleet, the things not known being which rockets in the fleet
a.
you believe these events are independent (assume East
areDo
faulty).*
Lansing
the locale)?
*Experts are is
consulted
"informing us" as to the probabilities they might place on different forms
8
rec11-17-09.nb
of true star fields being the actual ones, or different patterns of fault that might be present among
the fleet of rockets. Massive math models and calculations are then brought to bear which calculate the "updated" probabilities of different images being the correct ones conditional on what
Hubble
reports.were
We then
pick the conditionally
probable
b.
If they
independent
you most
could
findtrue star field and print it out. It
is involved but hey, we have computers. Guess what? They fixed the Hubble and those early
P(rain
(decimal)
restorations (from
whenboth
it was days)
faulty) were
really good.=
4. Weather.
A forecast
says there
P(no rain
on the weekend)
= is 80% chance of rain Sat
and 60%
rainwhole
Sun. Venn:
What
the chance
heck, doofthe
the fleet, the things not known being which rockets in the fleet
are faulty).*
a.
Do you believe these events are independent (assume East
*Experts are consulted "informing us" as to the probabilities they might place on different forms
of
true star fields
being
the actual ones, or different patterns of fault that might be present among
Lansing
is the
locale)?
the fleet of rockets. Massive math models and calculations are then brought to bear which calculate the "updated" probabilities of different images being the correct ones conditional on what
rec11-17-09.nb
Hubble reports. We then pick the conditionally most probable true star field and print
it out. It9
is involved but hey, we have computers. Guess what? They fixed the Hubble and those early
b.
If
they
were
independent
you
could
find
restorations (from when it was faulty) were really good.
P(rain both
days) (decimal)
4. Weather.
A forecast
says there=is 80% chance of rain Sat
rainofon
theSun.
weekend) =
and 60%P(no
chance
rain
What the heck, do the whole Venn:
a. Do you believe these events are independent (assume East
Lansing is the locale)?
b. If they were independent you could find
P(rain both days) (decimal) =
P(no rain on the weekend) =
What the heck, do the whole Venn:
P(rain both days) (decimal) =
P(no rain on the weekend) =
What the heck, re-do the whole Venn:
10
rec11-17-09.nb
c. If the events are not independent but instead
P(rain Sun § rain Sat) = 70%
find
P(rain both days) (decimal) =
P(no rain on the weekend) =
What the heck, re-do the whole Venn:
d. I arrive from out of town to find rain Sunday. What is the
conditional probability P(rain Sat § rain Sun)?
5. The Tree and Bayes' Method. A potential oilfield is
thought to have probability .3 of having oil in commercially
viable amount. Call this event OIL. A test can be performed
(involves seismic readings etc.) and the following claims are
made for the response of this test to OIL, or "no OIL" (i.e. OIL,
d. I arrive denoted
from outOIL
ofc):
town to find rain Sunday. What is the
sometimes
conditionalP(probability
P(rain
Sat
§ rain
Sun)?
§ OIL) = .17
(17%
false
negatives)
P(+ § no OIL) = .08 (8% false positives)
c. If the events are not independent but instead
P(rain Sun § rain Sat) = 70%
find
P(rain both days) (decimal) =
P(no rain on the weekend) =
What the heck, re-do the whole Venn:
c. If the events are not independent but instead
P(rain Sun § rain Sat) = 70%
find
P(rain both days) (decimal) =
P(no rain on the weekend) =
What the heck, re-do the whole Venn:
5.
Theout
Tree
and
Bayes'forMethod.
A potential
oilfield15)
is
a. Lay
a tree
diagram
this information
(see chapter
thought
to have probability .3 of having oil in commercially
with
branching
viable amount. Call this event OIL. A test can be performed
(involves seismic
etc.) andP(OIL+)
the following
P(+readings
§ OIL) = .83
= .3 .83claims are
made
for=the
P(OIL)
.3 response of this test to OIL, or "no OIL" (i.e. OIL,
c
sometimes denoted
): = .17
P(- §OIL
OIL)
P(OIL-) =
P(- § OIL) = .17 (17% false negatives)
P(+ § no OIL) = .08 (8% false positives)
P(+ § no OIL) = .08 P(noOIL+) =
a.P(no
Lay
out =a .7
tree diagram for this information (see chapter 15)
OIL)
with branching P(- § no OIL) = .92 P(noOIL-) =
d. I arrive from out of town to find rain Sunday. What is the
conditional probability P(rain Sat § rain Sun)?
rec11-17-09.nb
11
5. The Tree and Bayes' Method. A potential oilfield is
thought to have probability .3 of having oil in commercially
viable amount. Call this event OIL. A test can be performed
(involves seismic readings etc.) and the following claims are
made
for thefrom
response
this test
to OIL,
"no OIL"
(i.e.isOIL,
d. I arrive
out ofoftown
to find
rain or
Sunday.
What
the
c
sometimes
OILP(rain
):
conditionaldenoted
probability
Sat § rain Sun)?
P(- § OIL) = .17 (17% false negatives)
P(+ § no OIL) = .08 (8% false positives)
5. The Tree and Bayes' Method. A potential oilfield is
a.
Lay out
a treeprobability
diagram for.3this
(see
chapter 15)
thought
to have
of information
having oil in
commercially
with
viablebranching
amount. Call this event OIL. A test can be performed
(involves seismic readings etc.) and the following claims are
P(+ § OIL)
= .83
P(OIL+)
= .3
.83 (i.e. OIL,
made for the response
of this
test to OIL,
or "no
OIL"
P(OIL) = .3denoted OILc):
sometimes
P(- § OIL)
.17 false negatives)
P(OIL-) =
P(- § OIL)
= .17 =(17%
P(+ § no OIL) = .08 (8% false positives)
P(+diagram
§ no OIL)
.08
a. Lay out a tree
for= this
P(no
OIL)
=
.7
with branching
P(- § no OIL) = .92
P(+ § OIL) = .83
P(OIL) = .3
P(- § OIL) = .17
P(noOIL+)(see
= chapter 15)
information
P(noOIL-) =
P(OIL+) = .3 .83
P(OIL-) =
b. Fill out the complete Venn Diagram using the endpoint probabilities typed inP(+
bold§ no
above.
OIL) = .08 P(noOIL+) =
P(no OIL) = .7
P(- § no OIL) = .92 P(noOIL-) =
b. Fill out the complete Venn Diagram using the endpoint probabilities typed in bold above.
P(+ § OIL) = .83
P(OIL+) = .3 .83
P(OIL) = .3
12
rec11-17-09.nb
P(- § OIL) = .17
P(OIL-) =
b. Fill out the complete Venn Diagram using the endpoint probabilities typed in bold above.
P(+ § no OIL) = .08 P(noOIL+) =
P(no OIL) = .7
P(- § no OIL) = .92 P(noOIL-) =
b. Fill out the complete Venn Diagram using the endpoint probabilities typed in bold above.
c. Read off the value P(+) = P(OIL+) + P((noOIL) +).
d. Use the definition P(OIL § +) = P(OIL+) / P(+) to find the
conditional probability that OIL is present if the test yields a
positive result. This is Bayes' idea.*
*You could use the formula in chapter 15 but can just as easily get it from the tree as we just did.
Bayesians (people who specialize in the method) are much assisted by the formula in the many
c. Read off the value P(+) = P(OIL+) + P((noOIL) +).
derivations and manipulations they have to make, but we don't do that here.
e. Find P(OIL § -). Does it make sense that it should be less
than P(OIL § +)?
d. Use the definition P(OIL § +) = P(OIL+) / P(+) to find the
conditional
probability
present
theconsult
test yields
a
f. Contemplate
the factthat
thatOIL
we is
might
haveif to
expert
positive
This
Bayes'
opinion result.
in order
to iscome
upidea.*
with numbers for P(OIL), P(- §
OIL),
P(+ § no OIL). That does not make them correct but it can
*You could use the formula in chapter 15 but can just as easily get it from the tree as we just did.
be a very
making
asmany
to
Bayesians
(peoplepromising
who specialize approach
in the method) to
are much
assistedthe
by thedecision
formula in the
derivations
they have
make, butItweseparates
don't do that here
.
whetherand
ormanipulations
not we drill
thetofield.
assumptions
from
e. Finddeductions
P(OIL § -).
Does from
it make
sense
that itattention
should on
be getless
logical
flowing
them,
focusing
than the
P(OIL
ting
best§ +)?
judgment possible on each input to the calculation.
Repeated in many drilling operations this can be a more perfectible process than "seat of the pants."
d. Use the definition P(OIL § +) = P(OIL+) / P(+) to find the
conditional probability that OIL is present if the test yields a
positive result. This is Bayes' idea.*
*You could use the formula in chapter 15 but can just as easily get it from the tree rec11-17-09.nb
as we just did.
13
Bayesians (people who specialize in the method) are much assisted by the formula in the many
derivations and manipulations they have to make, but we don't do that here.
e. Find P(OIL § -). Does it make sense that it should be less
than P(OIL § +)?
f. Contemplate the fact that we might have to consult expert
opinion in order to come up with numbers for P(OIL), P(- §
OIL), P(+ § no OIL). That does not make them correct but it can
be a very promising approach to making the decision as to
whether or not we drill the field. It separates assumptions from
logical deductions flowing from them, focusing attention on getting the best judgment possible on each input to the calculation.
Repeated in many drilling operations this can be a more perfectible process than "seat of the pants."
In[1]:=
Out[1]=
.3 µ .17 ê H.3 µ .17 + .7 µ .92L
0.07338129496402877`