ECON1202 Lecture 7
Probability II: Probability in action
UNSW Economics
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Agenda
1 Probability trees;
2 Rules of probability for independent events;
3 Conditional probability and Bayes’ Formula.
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Re-introducing Trees
• We looked at Probability Trees last time, as a diagram with levels.
A
B
C
A
1
2
3
1
2
A
A
B
C
B
C
3
B
C
• Consider drawing cards
from a pack, two
mutually exclusive
events associated with
drawing a card are,
Black
Pr(
Bla
ck)
Red
Pr(Red)
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By the cards...
• In fact, any number of mutually exclusive events can be
represented with a tree;
• Consider drawing cards, but this time, focussing on the suit,
♦
Pr
(♦
)
13
52
♥
P
r(
♥
)
13
52
♣
P
r(♣
)1352
♠
Pr(♠)
13
52
• With a 52 card pack, there are 13 cards of each suit, giving a
probability of each event occurring (drawing a card of that suit)
equal to 1352 .
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Definition: Multiplication Law
In general, if two events A and B can occur in sequence, then the
probability that both events have occurred is the same as asking what
is the probability that one event occurs, times the probability that the
other event occurs, given the first has occurred, or:
Pr(A∩ B) = Pr(A)Pr(B|A)
= Pr(B)Pr(A|B) .
A
Pr(A)
B
Pr(B|A)
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Example (Multiplication Law)
Suppose that the administration of the University of Coogee has been doing some
surveys of the timing of students dropping out of their five-year degree programs.
The results indicate that 25% drop out after first year, and then for the subsequent
three years, 10%, 8% and 4% of the students who stay each year will drop out. Given
these data, what is the probability that a first year student will drop out after 2, 3 and 4
years?
S10.75
S20.90
S30.92
S40.96
D40.04
D30.08
D20.10
D1
0.25
Applying the multiplication law:
Pr(S1 ∩D2) = Pr(S1)Pr(D2|S1) = (0.75)(0.10) = 0.075
Pr(S2 ∩D3) = Pr(S1 ∩ S2)Pr(D3|S2) = (0.75)(0.90)(0.08) = 0.054
Pr(S3 ∩D4) = Pr(S1 ∩ (S2 ∩ S3))Pr(D4|S3) = (0.75)[(0.90)(0.92)](0.04) = 0.025
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Conditional Probabilities
Often, we have a sequence of events, where the possible outcomes of
the second layer depend on the first layer.
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Example (Conditional Probability)
Given that a card picked from a full deck is red-suited, what is the probability
that it is diamond suited?
• We have the language of ‘given x,
what is the probability that y’, so
we can draw a tree:
• Once we pick a red suit, which
occurs with Pr(R) = 12 , we have
only two options: {♦,♥}, so we
can write the conditional
probability,
Pr(♦|R) =
1
2
.
R
Pr
(R
)
0.5
0
♦
Pr(♦
|R)
0.50
♥
Pr(♥|R)
0.50
B
Pr(B)
0.50 ♣
Pr(♣
|B)
0.50
♠
Pr(♠|B)
0.50
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Definition: Conditional
Probability
If two events, A1 and B1 can occur
in sequence, then the
CONDITIONAL PROBABILITY of
event B1 occurring, given that
event A1 has occurred, is expressed
as,
Pr(B1|A1) =
Pr(B1 ∩A1)
Pr(A1)
.
A1
Pr
(A
1
)
B1
Pr(B
1|A1
)
B2
Pr(B
2 |A
1 )
A2
Pr(A
2 )
B1
Pr(B
1|A
2)
B2
Pr(B
2 |A
2 )
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Multiple levels...
If there is more than one level, we
have some kind of order to the
events.
Caution! Probabilities given on
Probability Trees
When drawing Probability Trees, it
is conventional to give the
conditional probability of the
event occurring for all branches
deeper than the first level.
A
Pr
(A
)
C
Pr(C
|A)
D
Pr(D|A)
B
Pr(B)
E
Pr(E
|B)
F
Pr(F|B)
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Definition: Probability Tree
A probability tree shows one or
more events that can occur, by
using labels for each event (the
nodes) and branches that indicate
allowable paths through the event
tree. At each level, the associated
probabilities e.g. Pr(A1) . . . Pr(Ak)
must:
1 Be MUTUALLY EXCLUSIVE
events: no two events can
happen simultaneously;
2 Be exhaustive: the sum of the
probabilities given equals 1
(no possible events are
missing).
A1
Pr
(A
1
)
B1
Pr(B
1|A1
)
B2
Pr(B
2 |A
1 )
A2
Pr(A
2 )
B1
Pr(B
1|A
2)
B2
Pr(B
2 |A
2 )
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Suppose we were to draw a Probability Tree to represent the flipping
of a (fair coin) twice:
H1
Pr
(H
1
)
H2
Pr(H
2|H
1)
T2
Pr(T
2 |H
1 )
T1
Pr(T
1 )
H2
Pr(H
2|T1
)
T2
Pr(T
2 |T
1 )
In this case, we would actually have the scenario that,
Pr(H2|H1) = Pr(H2)
Pr(T2|H1) = Pr(T2)
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That is, the order doesn’t matter, or in other words, we are dealing
with independent events:
Definition: Independent Events
SupposeA and B are events with positive probabilities. If either
Pr(B|A) = Pr(B) or
Pr(A|B) = Pr(A) ,
then A and B are said to be INDEPENDENT EVENTS.
Caution! Special Multiplication Law
Notice, that if A and B are independent events then
Pr(A ∩ B) = Pr(A)Pr(B)
e.g., the probability of getting two heads in a row is just ( 12)(
1
2).
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The two-stage problem
Scenario: To study, or not to study...
Suppose that a large study is conducted for first-year students of
ECON1202. The purpose of the study is to investigate if the final
examination is a good test of whether students undertake ‘consistent
study’ in the course (that is, studying at least 4 hours outside
university hours, every week of session). Results from a post-course
survey indicate that 85% of students undertook consistent study, of
whom 81% pass; for the rest (those who don’t undertake consistent
study), only 54% pass. Suppose a student is randomly selected, and
turns out to have passed. What is the probability that this student
studied consistently?
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• This scenario is one where two-stages are at play;
• More than that, we have a contingent probability as before, but
this time, the unknown is in the first stage.
S
Pr
(S
)
0.8
5
P
Pr(P
|S)
0.81
F
Pr(F|S)
0.19
N
Pr(N)
0.15 P
Pr(P
|N)
0.54
F
Pr(F|N)
0.46
We want:
Pr(S|P) =
Prob.(Study AND Pass)
Prob.(Pass)
or, as a contingent probability
Pr(S|P) =
Pr(S ∩ P)
Pr(P)
with numbers,
Pr(S|P) =
(0.85)(0.81)
(0.85)(0.81) + (0.15)(0.54)
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• Notice, that this question asks things the other way around to our
normal thinking for conditional probability;
• Here, the sequence is ‘backwards’:
1 Rather than: ‘Prob. layer-2 event, given layer-1 event’;
2 We have: ‘Prob. layer-1 event, given layer-2 event’!
• This has a special name in probability theory ...
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Definition: Bayes’ Formula
SupposeA1 . . .An is an exhaustive list of n mutually exclusive events
that can occur for a given population S, and B is any event in S such
that Pr(B) > 0, then the conditional probability of some Ai, given that
event B has occurred is given by,
Pr(Ai|B) =
Pr(Ai)Pr(B|Ai)
Pr(A1)Pr(B|A1) + · · ·+ Pr(An)Pr(B|An)
, (1)
which is the general form of BAYES’ FORMULA.
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Or, in other words...
For Probability Trees, Bayes’ Formula is asking,
Pr(Ai|Bj) =
The Prob. of a path through Ai to Bj
The sum of all Prob.s for paths to Bj
A1
Pr
(A
1
)
B1
Pr(B
1|A1
)
B2
Pr(B
2 |A
1 )
A2
Pr(A
2 )
B1
Pr(B
1|A
2)
B2
Pr(B
2 |A
2 )
In two-layer problems, this reduces
to:
Pr(A1)Pr(B2|A1)
Pr(A1)Pr(B2|A1) + Pr(A2)Pr(B2|A2)
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Example (Applying Bayes’ Formula)
A car manufacturing plant has three car chassis production machines,M1,M2 and
M3. For historical reasons, a car rig on the production line has a 0.6, 0.3 and 0.1
probability of going to each machine respectively. Each of the three machines add a
chassis to a rig without fault with 0.55, 0.60 and 0.30 probabilities respectively,
what is the probability that a rig with a chassis fault came fromM2?
M1
0.6
OK
0.
55
F
0.45
M2
0.3
OK
0.
60
F
0.40
M3
0.1
OK
0.
30
F
0.70
Pr(M2|F) =
Pr(M2)Pr(F|M2)
Pr(M1)Pr(F|M1) + Pr(M2)Pr(F|M2) + Pr(M3)Pr(F|M3)
=
(0.3)(0.40)
(0.6)(0.45) + (0.3)(0.40) + (0.1)(0.70)
= 0.12/0.46 = 0.26
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Example (More Bayes’ Formula)
Suppose that the plant from the previous example restructure their car chassis
production line. Now they have just two machines doing 50% of the work each:
M1, which always adds the chassis to the rig without fault, and (the old)M2, which
has had a mid-operation check added to it which is rejecting 5% of the chassis due
to early faults, the other 95% go on to the final stage, where 90% of them are
faultless. Given that a complete chassis is faultless, what’s the probability it came
fromM2 now?
M1
0.50
OKf
1.00
M2
0.50
OK10.95
OKf0.90
F0.10
F0.05
Pr(M2|OKf ) =
Pr(M2)Pr(OK1|M2)Pr(OKf |OK1)
Pr(M1)Pr(OKf |M1) + Pr(M2)Pr(OK1|M2)Pr(OKf |OK1)
=
(0.50)(0.95)(0.90)
(0.50)(1.00) + (0.50)(0.95)(0.90)
=
0.4275
0.9275
= 0.46
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