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ECON1202 Lecture 7
Probability II: Probability in action
Agenda
1 Probability trees;
2 Rules of probability for independent events;
3 Conditional probability and Bayes’ Formula.
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)
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 .
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)
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
c©UNSW Economics
Conditional Probabilities
Often, we have a sequence of events, where the possible outcomes of
the second layer depend on the first layer.
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
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 )
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)
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).