Matakuliah
Tahun
Versi
: I0134 – Metoda Statistika
: 2005
: Revisi
Pertemuan 08
Kejadian Bebas dan Bersyarat
1
Learning Outcomes
Pada akhir pertemuan ini, diharapkan mahasiswa
akan mampu :
• Mahasiswa dapat menggunakan kaidah
peluang aditif dan multiplikatif, kejadian
bebas, bersyarat dan jaidah Bayes.
2
Outline Materi
•
•
•
•
Kejadian bebas
Kejadian bersyarat
Peluang total
Kaidah Bayes
3
Basic Rules for Probability
•
Union - Probability of A or B or both
–
•
P( A  B)  n( A  B)  P( A)  P( B)  P( A  B)
n( S )
Mutually exclusive events:
P( A  CProbability
)  0 so P( A-Probability
C)  P( A)of
P
C) B
Conditional
A(given
–
P( A B)  P( A  B)
P( B)
Independent events:
P( A B)  P( A)
P( B A)  P( B)
4
Conditional Probability
Rules of conditional probability:
P( A B)  P( A  B)
P( B)
so P( A  B)  P( A 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)
so
P( A  D)  P( A) P( D)
5
Contingency Table
(Contoh Soal)
Counts
AT& T
IBM
Total
Telecommunication
40
10
50
Computers
20
30
50
Total
60
40
100
Probabilities
AT& T
IBM
Total
Telecommunication
.40
.10
.50
Computers
.20
.30
.50
Total
.60
.40
1.00
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 ) 
6
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 ) 
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
7
Independence of Events
(Contoh Soal)
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
8
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
9
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

Pick 5 cards from a
deck of 52 - without
deck of 52 - with
replacement
replacement
– 52*52*52*52*52=525
380,204,032 different
possible outcomes
– 52*51*50*49*48 =
311,875,200 different
possible outcomes
10
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
A
B
A
.
..
..
.
ABC
ACB
BAC
BCA
CAB
CBA
11
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.
12
Permutations
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 nPr.
n!
P

n r ( n  r )!
6
For example:
6!
6! 6 * 5 * 4 * 3 * 2 * 1
P
 
 6 * 5 * 4  120
(6  3)! 3!
3 * 2 *1
3
13
Combinations
Suppose that when we picked 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 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
14
The Law of Total Probability
and Bayes’ Theorem
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
15
The Law of Total Probability
Contoh Soal
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
16
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) 
Applying the law of total
probability to the denominator
Applying the definition of
conditional probability throughout
17
Bayes’ Theorem
(Contoh Soal)
•
A medical test for a rare disease (affecting 0.1%
P ( I )  0[.001
of the population
]) 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.
.
18
Contoh Soal (continued)
P ( I )  0.001
P ( I )  0.999
P ( Z I )  0.92
P ( Z I )  0.04
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 ) 
(.92)( 0.001)
(.92)( 0.001)  ( 0.04)(.999)
0.00092
0.00092


0.00092  0.03996
.04088
.0225

19
Contoh Soal (Tree Diagram)
Prior
Probabilities
Conditional
Probabilities
P( Z I )  0.92
P( I )  0001
.
P( I )  0.999
P( Z I )  008
.
P( Z I )  004
.
Joint
Probabilities
P( Z  I )  (0.001)(0.92) .00092
P( Z  I )  (0.001)(0.08) .00008
P( Z  I )  (0.999)(0.04) .03996
P( Z I )  096
.
P( Z  I )  (0.999)(0.96) .95904
20
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 )
1
1
Applying the law of total
probability to the denominator
1
i
P( A B ) P( B )

 P( A B ) P( B )
1
1
i
Applying the definition of
conditional probability throughout
i
21
Bayes’ Theorem Extended
(Contoh Soal)



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
Event A  Appreciation
P( A H )  0.70
P( A M )  0.40
P( A L)  0.20
22
Contoh Soal (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
23
Contoh Soal (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
P ( A  L)  ( 0.20)( 0.20)  0.04
P ( A  L)  ( 0.20)( 0.80)  0.16
24
• Selamat Belajar Semoga Sukses.
25