Pertemuan 08 Kejadian Bebas dan Bersyarat

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Pertemuan 08Kejadian Bebas dan Bersyarat

• Matakuliah I0134 Metoda Statistika
• Tahun 2005
• Versi Revisi

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Learning Outcomes

• Pada akhir pertemuan ini, diharapkan mahasiswa
• akan mampu
• Mahasiswa dapat menggunakan kaidah peluang aditif
  dan multiplikatif, kejadian bebas, bersyarat dan
  jaidah Bayes.

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Outline Materi

• Kejadian bebas
• Kejadian bersyarat
• Peluang total
• Kaidah Bayes

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Basic Rules for Probability

• Union - Probability of A or B or both
• Mutually exclusive events
• Conditional Probability - Probability of A given
  B
• Independent events

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Conditional Probability
Rules of conditional probability
so
If events A and D are statistically independent
so
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Contingency Table (Contoh Soal)
Counts
AT T
IBM
Total
Probability that a project is undertaken by IBM
given it is a telecommunications project
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
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Independence of Events
Conditions for the statistical independence of
events A and B
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Independence of Events (Contoh Soal)
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Product Rules for Independent Events
The probability of the intersection of several
independent events is the product of their
separate individual probabilities
The probability of the union of several
independent events is 1 minus the product of
probabilities of their complements
Example 2-7
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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 6636 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 - without
  replacement
• 5251504948 311,875,200 different possible
  outcomes

• Pick 5 cards from a deck of 52 - with replacement
• 5252525252525 380,204,032 different possible
  outcomes

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More on Combinatorial Concepts(Tree Diagram)
.
.
Order the letters A, B, and C
C
.
.
.
ABC
B
.
.
.
.
C
B
ACB
.
.
A
C
A
.
.
BAC
.
B
C
A
C
BCA
.
.
A
B
CAB
B
A
CBA
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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 321 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? (654321 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.
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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
654120 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.
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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 regardless of the
order of selection. The number of combinations is
denoted and is read n choose r. An alternative
notation is nCr. We define the number of
combinations of r out of n elements as
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The Law of Total Probability and Bayes Theorem
In terms of conditional probabilities
More generally (where Bi make up a partition)
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The Law of Total ProbabilityContoh Soal
Event U Stock market will go up in the next
year Event W Economy will do well in the next
year
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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.

Applying the law of total probability to the
denominator
Applying the definition of conditional
probability throughout
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Bayes Theorem (Contoh Soal)

• A medical test for a rare disease (affecting 0.1
  of the population ) is
  imperfect
• When administered to an ill person, the test will
  indicate so with probability 0.92
  
• The event 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
  
• The event is a false positive.
  .

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Contoh Soal (continued)
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Contoh Soal (Tree Diagram)
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Bayes Theorem Extended

• Given a partition of events B1,B2 ,...,Bn

Applying the law of total probability to the
denominator
Applying the definition of conditional
probability throughout
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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
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Contoh Soal (continued)
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Contoh Soal (Tree Diagram)
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• Selamat Belajar Semoga Sukses.