konsep-probabilitas-4
Transcript of konsep-probabilitas-4
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E DAA
BABA
1
:
• B , .
• , .
• D / (), .
2
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• E/ () .
• /. B . B ().
• / .
3
C :
D , 3 B .
= BBB, BB, BB, BB,B, B, B, .
() = 23 = 8. A
, A = BB, BB, BB. (A) = 3.
4
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D
• B A , A, (A), :
5
n
m
S n
An AP ==
)(
)()(
A :
• 0 ≤ (A) ≤ 1 , A 0 1
• (A) = 0, A
( ),
A 0. D
A .
• (A) = 1, A, A 1. D
A .
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C (1):
• . B ?
:
• = , B = B
• = , B,
B, BB
• A =
A = , B, B
,
•
74
3
)(
)()( ==
S n
An AP
C (2):
• 6 , 4 , 3
. B
,
: () , () .
:
• , = , C = , =
(). =
(). =
8
136
)()()( ==
S n M n M P
13
7
13
034
)(
)()()(
)(
)()( =
−+=
∩−+=
∪=∪
S n
T C nT nC n
S n
T C nT C P
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(1):
• B A B , A B A B .
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)()()()( B AP BP AP B AP ∩−+=∪
(2):
• B A, B, C , A, B, C :
10
)()()(
)()()()()(
C B APC BPC AP
B APC P BP APC B AP
∩∩+∩−∩−
∩−++=∪∪
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C :
A 2/3 4/9. B 1/4, A ?
:
B , B , :
A :
(∪ B) = () + (B) (∩ B)= 2/3 + 4/9 1/4
= 31/36
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D (
• B A B ,
:
)()()( BP AP B AP +=∪
)()()()( C P BP APC B AP ++=∪∪
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B A, B, C ,
:
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C :
• B 7 11 ?
:• B A 7, A =
(1,6), (6,1), (2,5), (5,2), (3,4), (4,3)• B B 11, B =
(5,6), (6,5)• 7 11
:(A∪ B) = (A) + (B) (A ∩ B)
= 6/36 + 2/36 0
= 8/36
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D :
• B A A
, :
)(1)'( AP AP −=
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C:
, A , .
:
A =
= (1,1), (2,2) , (3,3), (4,4), (5,5), (6,6)
(A) = 6/36
,
= (A):
(A) = 1 (A)
= 1 6/36
= 30/36
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D ():
• D
.
• D A B
, A
B
B
A.
• B A B , :
)(.)()( BP AP B AP =∩
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C:
• , ?
:
• = (,), (,), (,), (,)
• , A = 1 (A) = 2/4 =
= (,), (,)
B = 2 (B) = 2/4 =
= (,), (,)
A∩ B = 1 2
= (,) (A∩ B) =
• B A B : (A∩ B) = (A). (B)
= .
=
, A B .
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():
• A B A .
• D (BA) B A
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0)(,)(
)()( >
∩= AP jika
AP
B AP A BP
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C (1):
• 20 , 5 . B 2
. B
?
• :
A =
B =
= (A ∩ B)
(A ∩ B) = (A). (BA)
= 5/20 . 4/19
= 1/19
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C (2):
• B 100
()
(), : 20 , 30
, 40 , 10
.
• A ,
?
• A ,
?
• A ,
?
• A
, ?
20
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:
= , = , = , = .
• ,
• A ,
• A ,
• A ,
• A ,
21
20 40 60
30 10 40
50 50 100
67.060
40
10060
10040
)(
)()( ===
∩=
RP
RS P RS P
75.040
30
10040
10030
)(
)()( ===
∩=
W P
W J PW J P
40.050
20
100
50100
20
)(
)()( ===
∩=
J P
J RP J RP
20.050
10
10050
10010
)(
)()( ===
∩=
S P
S W PS W P
A B :
• A1, A2, A3
.
• B
.
22
S
A1 A2 A3
B
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B :
23
P(B) = P(B A 1). P(A1) + P(B A 2). P(A2) + P(B A 3). P(A3)
=
∑=
3
1
)().(i
ii AP A BP
disebut Hukum Probabilitas Total
• , A1, A2, A3, , A B
,
AB :
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∑=
=∩
=n
i ii
iii
i
AP A BP
AP A BP
BP
A BP B AP
1
)().(
)().(
)(
)()(
disebut Rumus Bayes (Aturan Bayes).
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C:
• 2 . 1
2 , 2 1 1
, 3 2 . D
A
1
..
• B ?
• B 2?
25
• ( ) =
• ( 2) =
26
)3().3()2().2()1().1()( M PP M PP M PP M P ++=
5.06
3
6
120.
3
1
2
1.
3
1
2
2.
3
1==
+=++=
33.03
1
636
1
63
21.
31
)(
)2().2()2( =====
M P
M PP M P