Matakuliah
Tahun
Versi
: A0064 / Statistik Ekonomi
: 2005
: 1/1
Pertemuan 6
Probabilitas-2
1
Learning Outcomes
Pada akhir pertemuan ini, diharapkan mahasiswa
akan mampu :
• Menghitung beberapa persoalan dalam
hitung peluang seperti peluang seragam,
bersyarat, total, dan Bayes
2
Outline Materi
• Irisan (Intersection) dan Gabungan
(Union)
• Peluang Seragam
• Peluang Bersyarat
• Hukum Probabilitas Total dan Teorama
Bayes
3
COMPLETE
2-4
BUSINESS STATISTICS
5th edi tion
2-4 Conditional Probability
Rules of conditional probability:
P( A B)  P( A  B) so P( A  B)  P( A B) P( B)
P( B)
 P( B A) P( A)
If events A and D are statistically independent:
P ( A D)  P ( A)
P ( D A)  P ( D)
McGraw-Hill/Irwin
so
P( A  D)  P( A) P( D)
Aczel/Sounderpandian
© The McGraw-Hill Companies, Inc., 2002
COMPLETE
2-5
BUSINESS STATISTICS
5th edi tion
Contingency Table - Example 2-2
Counts
AT& T
IBM
Total
Telecommunication
40
10
50
Computers
20
30
50
Total
60
40
100
Probabilities
AT& T
IBM
Telecommunication
.40
.10
.50
Computers
.20
.30
.50
Total
.60
.40
1.00
McGraw-Hill/Irwin
Total
Aczel/Sounderpandian
Probability that a project
is undertaken by IBM
given it is a
telecommunications
project:
P ( IBM  T )
P(T )
.10

.2
.50
P ( IBM T ) 
© The McGraw-Hill Companies, Inc., 2002
COMPLETE
2-6
BUSINESS STATISTICS
5th edi tion
2-5 Independence of Events
Conditions for the statistical independence of events A and B:
P ( A B )  P ( A)
P ( B A)  P ( B )
and
P ( A  B )  P ( A) P ( B )
P ( Ace  Heart )
P ( Heart )
1
1
 52 
 P ( Ace )
13 13
52
P ( Ace Heart ) 
P ( Ace  Heart ) 
McGraw-Hill/Irwin
P ( Heart  Ace )
P ( Ace )
1
1
 52   P ( Heart )
4
4
52
P ( Heart Ace ) 
4 13 1

 P ( Ace ) P ( Heart )
52 52 52
Aczel/Sounderpandian
© The McGraw-Hill Companies, Inc., 2002
COMPLETE
BUSINESS STATISTICS
2-7
5th edi tion
Independence of Events - Example 2-5
Events Television (T) and Billboard (B) are
assumed to be independent.
a) P ( T  B )  P ( T ) P ( B )
 0.04 * 0.06  0.0024
b) P ( T  B )  P ( T )  P ( B )  P ( T  B )
 0.04  0.06  0.0024  0.0976
McGraw-Hill/Irwin
Aczel/Sounderpandian
© The McGraw-Hill Companies, Inc., 2002
COMPLETE
BUSINESS STATISTICS
2-8
5th edi tion
Product Rules for Independent Events
The probability of the intersection of several independent events
is the product of their separate individual probabilities:
P( A  A  A  An )  P( A ) P( A ) P( A ) P( An )
1
2
3
1
2
3
The probability of the union of several independent events
is 1 minus the product of probabilities of their complements:
P( A  A  A  An )  1 P( A ) P( A ) P( A ) P( An )
1
2
3
1
2
3
Example 2-7:
(Q  Q  Q Q )  1  P(Q ) P(Q ) P(Q ) P(Q )
1
2
3
10
1
2
3
10
 1.9010  1.3487 .6513
McGraw-Hill/Irwin
Aczel/Sounderpandian
© The McGraw-Hill Companies, Inc., 2002
COMPLETE
BUSINESS STATISTICS
2-9
5th edi tion
2-6 Combinatorial Concepts
Consider a pair of six-sided dice. There are six possible outcomes
from throwing the first die {1,2,3,4,5,6} and six possible outcomes
from throwing the second die {1,2,3,4,5,6}. Altogether, there are
6*6=36 possible outcomes from throwing the two dice.
In general, if there are n events and the event i can happen in
Ni possible ways, then the number of ways in which the
sequence of n events may occur is N1N2...Nn.

Pick 5 cards from a deck
of 52 - with replacement
 52*52*52*52*52=525
380,204,032 different possible
outcomes
McGraw-Hill/Irwin
Aczel/Sounderpandian

Pick 5 cards from a deck
of 52 - without
replacement
 52*51*50*49*48 =
311,875,200 different possible
outcomes
© The McGraw-Hill Companies, Inc., 2002
COMPLETE
2-10
BUSINESS STATISTICS
5th edi tion
More on Combinatorial Concepts
(Tree Diagram)
.
. ..
. . .
. .
.
Order the letters: A, B, and C
C
B
C
B
A
C
C
A
B
C
A
B
McGraw-Hill/Irwin
Aczel/Sounderpandian
A
B
A
.
..
..
.
ABC
ACB
BAC
BCA
CAB
CBA
© The McGraw-Hill Companies, Inc., 2002
COMPLETE
BUSINESS STATISTICS
2-11
5th edi tion
Factorial
How many ways can you order the 3 letters A, B, and C?
There are 3 choices for the first letter, 2 for the second, and 1 for
the last, so there are 3*2*1 = 6 possible ways to order the three
letters A, B, and C.
How many ways are there to order the 6 letters A, B, C, D, E,
and F? (6*5*4*3*2*1 = 720)
Factorial: For any positive integer n, we define n factorial as:
n(n-1)(n-2)...(1). We denote n factorial as n!.
The number n! is the number of ways in which n objects can
be ordered. By definition 1! = 1 and 0! = 1.
McGraw-Hill/Irwin
Aczel/Sounderpandian
© The McGraw-Hill Companies, Inc., 2002
COMPLETE
BUSINESS STATISTICS
2-12
5th edi tion
Permutations (Order is important)
What if we chose only 3 out of the 6 letters A, B, C, D, E, and F?
There are 6 ways to choose the first letter, 5 ways to choose the
second letter, and 4 ways to choose the third letter (leaving 3
letters unchosen). That makes 6*5*4=120 possible orderings or
permutations.
Permutations are the possible ordered selections of r objects out
of a total of n objects. The number of permutations of n objects
taken r at a time is denoted by nPr, where
n!
P

n r ( n  r )!
For example:
6
McGraw-Hill/Irwin
P
3
6!
6! 6 * 5 * 4 * 3 * 2 * 1
 
 6 * 5 * 4  120
(6  3)! 3!
3 * 2 *1
Aczel/Sounderpandian
© The McGraw-Hill Companies, Inc., 2002
COMPLETE
BUSINESS STATISTICS
2-13
5th edi tion
Combinations (Order is not Important)
Suppose that when we pick 3 letters out of the 6 letters A, B, C, D, E, and F
we chose BCD, or BDC, or CBD, or CDB, or DBC, or DCB. (These are the
6 (3!) permutations or orderings of the 3 letters B, C, and D.) But these are
orderings of the same combination of 3 letters. How many combinations of 6
different letters, taking 3 at a time, are there?
Combinations are the possible selections of r items from a group of n items  n
regardless of the order of selection. The number of combinations is denoted  r
and is read as n choose r. An alternative notation is nCr. We define the number
of combinations of r out of n elements as:
n!
 n

C

  n r
 r
r!(n  r)!
For example:
6!
6!
6 * 5 * 4 * 3 * 2 * 1 6 * 5 * 4 120
 n




 20
   6 C3 
 r
3!(6  3)! 3!3! (3 * 2 * 1)(3 * 2 * 1) 3 * 2 * 1
6
McGraw-Hill/Irwin
Aczel/Sounderpandian
© The McGraw-Hill Companies, Inc., 2002
COMPLETE
BUSINESS STATISTICS
2-14
5th edi tion
Example: Template for Calculating
Permutations & Combinations
McGraw-Hill/Irwin
Aczel/Sounderpandian
© The McGraw-Hill Companies, Inc., 2002
COMPLETE
BUSINESS STATISTICS
2-15
5th edi tion
2-7 The Law of Total Probability and
Bayes’ Theorem
The law of total probability:
P( A)  P( A  B)  P( A  B )
In terms of conditional probabilities:
P( A)  P( A  B)  P( A  B )
 P( A B) P( B)  P( A B ) P( B )
More generally (where Bi make up a partition):
P( A)   P( A  B )
i
  P( AB ) P( B )
i
i
McGraw-Hill/Irwin
Aczel/Sounderpandian
© The McGraw-Hill Companies, Inc., 2002
COMPLETE
BUSINESS STATISTICS
2-16
5th edi tion
The Law of Total ProbabilityExample 2-9
Event U: Stock market will go up in the next year
Event W: Economy will do well in the next year
P(U W ) .75
P(U W )  30
P(W ) .80  P(W )  1.8 .2
P(U )  P(U W )  P(U W )
 P(U W ) P(W )  P(U W ) P(W )
 (.75)(.80)  (.30)(.20)
.60.06 .66
McGraw-Hill/Irwin
Aczel/Sounderpandian
© The McGraw-Hill Companies, Inc., 2002
COMPLETE
BUSINESS STATISTICS
2-17
5th edi tion
Bayes’ Theorem
•
•
Bayes’ theorem enables you, knowing just a little
more than the probability of A given B, to find the
probability of B given A.
Based on the definition of conditional probability
and the law of total probability.
P ( A  B)
P ( A)
P ( A  B)

P ( A  B)  P ( A  B )
P ( A B) P ( B)

P ( A B) P ( B)  P ( A B ) P ( B )
P ( B A) 
McGraw-Hill/Irwin
Aczel/Sounderpandian
Applying the law of total
probability to the denominator
Applying the definition of
conditional probability throughout
© The McGraw-Hill Companies, Inc., 2002
COMPLETE
BUSINESS STATISTICS
2-18
5th edi tion
Bayes’ Theorem - Example 2-10
•
A medical test for a rare disease (affecting 0.1% of the
population [ P( I )  0.001 ]) is imperfect:
When administered to an ill person, the test will indicate so
with probability 0.92 [ P(Z I )  .92  P(Z I )  .08 ]
• The event (Z I ) is a false negative
When administered to a person who is not ill, the test will
erroneously give a positive result (false positive) with
probability 0.04 [ P(Z I )  0.04  P(Z I )  0.96 ]
• The event (Z I ) is a false positive.
.
McGraw-Hill/Irwin
Aczel/Sounderpandian
© The McGraw-Hill Companies, Inc., 2002
COMPLETE
BUSINESS STATISTICS
2-19
5th edi tion
Example 2-10 (continued)
P ( I )  0.001
P( I  Z )
P( Z )
P( I  Z )

P( I  Z )  P( I  Z )
P( Z I ) P( I )

P( Z I ) P( I )  P( Z I ) P( I )
P( I Z ) 
P ( I )  0.999
P ( Z I )  0.92
P ( Z I )  0.04
McGraw-Hill/Irwin
(.92)( 0.001)
(.92)( 0.001)  ( 0.04)(.999)
0.00092
0.00092


0.00092  0.03996
.04088
.0225

Aczel/Sounderpandian
© The McGraw-Hill Companies, Inc., 2002
COMPLETE
2-20
BUSINESS STATISTICS
5th edi tion
Example 2-10 (Tree Diagram)
Prior
Probabilities
Conditional
Probabilities
P( Z I )  0.92
P( I )  0001
.
P( I )  0999
.
P( Z I )  008
.
Joint
Probabilities
P( Z  I )  (0.001)(0.92) .00092
P( Z  I )  (0.001)(0.08) .00008
P( Z I )  004
.
P( Z  I )  (0.999)(0.04) .03996
P( Z I )  096
.
P( Z  I )  (0.999)(0.96) .95904
McGraw-Hill/Irwin
Aczel/Sounderpandian
© The McGraw-Hill Companies, Inc., 2002
COMPLETE
2-21
BUSINESS STATISTICS
5th edi tion
Bayes’ Theorem Extended
•
Given a partition of events B1,B2 ,...,Bn:
P( A  B )
P( B A) 
P( A)
P( A  B )

 P( A  B )
P( A B ) P( B )

 P( A B ) P( B )
1
1
1
i
1
1
i
McGraw-Hill/Irwin
Applying the law of total
probability to the denominator
Applying the definition of
conditional probability throughout
i
Aczel/Sounderpandian
© The McGraw-Hill Companies, Inc., 2002
COMPLETE
BUSINESS STATISTICS
2-22
5th edi tion
Bayes’ Theorem Extended Example 2-11



An economist believes that during periods of high economic growth, the U.S.
dollar appreciates with probability 0.70; in periods of moderate economic
growth, the dollar appreciates with probability 0.40; and during periods of
low economic growth, the dollar appreciates with probability 0.20.
During any period of time, the probability of high economic growth is 0.30,
the probability of moderate economic growth is 0.50, and the probability of
low economic growth is 0.50.
Suppose the dollar has been appreciating during the present period. What is
the probability we are experiencing a period of high economic growth?
Partition:
H - High growth P(H) = 0.30
M - Moderate growth P(M) = 0.50
L - Low growth P(L) = 0.20
McGraw-Hill/Irwin
Aczel/Sounderpandian
Event A  Appreciation
P( A H )  0.70
P( A M )  0.40
P( A L)  0.20
© The McGraw-Hill Companies, Inc., 2002
COMPLETE
BUSINESS STATISTICS
2-23
5th edi tion
Example 2-11 (continued)
P( H  A)
P( H A) 
P( A)
P( H  A)

P( H  A)  P( M  A)  P( L  A)
P( A H ) P( H )

P ( A H ) P ( H )  P ( A M ) P ( M )  P ( A L) P ( L)
( 0.70)( 0.30)

( 0.70)( 0.30)  ( 0.40)( 0.50)  ( 0.20)( 0.20)
0.21
0.21


0.21 0.20  0.04 0.45
 0.467
McGraw-Hill/Irwin
Aczel/Sounderpandian
© The McGraw-Hill Companies, Inc., 2002
COMPLETE
2-24
BUSINESS STATISTICS
5th edi tion
Example 2-11 (Tree Diagram)
Prior
Probabilities
Conditional
Probabilities
P ( A H )  0.70
P ( H )  0.30
P ( A H )  0.30
P ( A M )  0.40
Joint
Probabilities
P ( A  H )  ( 0.30)( 0.70)  0.21
P ( A  H )  ( 0.30)( 0.30)  0.09
P ( A  M )  ( 0.50)( 0.40)  0.20
P ( M )  0.50
P ( A M )  0.60 P ( A  M )  ( 0.50)( 0.60)  0.30
P ( L)  0.20
P ( A L)  0.20
P ( A L)  0.80
McGraw-Hill/Irwin
P ( A  L )  ( 0.20)( 0.20)  0.04
P ( A  L)  ( 0.20)( 0.80)  0.16
Aczel/Sounderpandian
© The McGraw-Hill Companies, Inc., 2002
COMPLETE
BUSINESS STATISTICS
2-25
5th edi tion
2-8 Using Computer: Template for
Calculating the Probability
of at least one success
McGraw-Hill/Irwin
Aczel/Sounderpandian
© The McGraw-Hill Companies, Inc., 2002
Penutup
• Materi yang terdapat dalam Probabilitas
ini pada hakekatnya adalah pengetahuan
dasar bagi estimasi atau pengambilan
keputusan (statistik inferensial)
26