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PROBABILITY

PROBABILITY. Probability Concept. There are 3 (three) ways of interpreting probabilities : Axiomatic Objective Subjective. Classical (Equally Likely). Empirical (Relative Frequency). Axiomatic. Defini si :.

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PROBABILITY

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  1. PROBABILITY

  2. Probability Concept • There are 3 (three) ways of interpreting probabilities : • Axiomatic • Objective • Subjective Classical (Equally Likely) Empirical (Relative Frequency)

  3. Axiomatic Definisi: Misal S adalah sample space dan Peluang terjadi peristiwa A adalah P(A). Maka berlaku : • P(A) > 0 • P(S) = 1 • If A1, A2, A3, … are mutually exclusive events then

  4. OBJECTIVEClassical (Equally Likely) Jika suatu eksperimen dengan sampel space S berisi N titik sample yang mempunyai kesempatan yang sama (equally likely) dan MEE, kemudian jika n diantaranya adalah peristiwa A maka peluang peluang terjadinya peristiwa A dinotasikanP(A), didefinisikan sebagai :

  5. OBJECTIVE Empirical Perhitunganpeluangberdasarkanpendekatanempirisadalahatasdasarpengertianfrekuensirelatifnya. Jika sebuah eksperimen dimana terjadi peristiwa A terjadi n kali, dari serangkaian kejadian N. Jika kejadian N makin makin besar maka peluang terjadi peristiwa A adalah frekuensi relatif dari peristiwa A yaitu :

  6. Contoh : Undian dengan sebuah mata uang yang homogen 1.000 kali, misalnya didapat muka G sebanyak 519 kali. Maka frekuensi relatif muka G = 0,519. Bila dilakukan 2.000 kali maka didapat muka G sebanyak 1.020 kali. Frekuensi relatifnya = 0,510. Jika dilakukan 5.000 kali didapat muka G = 2.530, maka frekuensi relatifnya = 0,506. Jika proses demikian diteruskan, nilai frekuensi relatifnya lambat laun makin dekat kepada sebuah bilangan yang merupakan peluang untuk muka G. Dalam hal ini bilangan tersebut adalah 0,5.

  7. Subjective Peluang P(A) adalah sebuah ukuran dari derajat kepercayaan seseorang mengenai suatu persitiwa A

  8. Probability Theorems Theorem 1 : P(Φ) = 0 Theorem 2 : If A ϵS P(A) < 1 then implies : 0 < P(A) < 1

  9. Probability Theorems Theorem 3 : If A, B ϵSthen Proof : B A

  10. If B A, then P(A – B) = P(A) – P(B) If B A, then A ∩ B = B • Corollary :  Probability Theorems • Corollary, theorem 3 : B A Proof :  P(A – B) = P(A) – P(A ∩ B) = P(A) – P(B) S=A B Proof : let A = S Theorem 3  P(S– B) = P(S) – P(S∩ B)

  11. Theorem 4 : if A, B S, then Probability Theorems P(AυB) = P(A) + P(B) – P(A ∩ B) Proof : A υB = (A – B) υ(B – A) υ(A ∩ B) P(A υB) = P(A – B)+P(B – A)+P(A ∩ B) B A = (A – B) υ(A ∩ B) A B = (B – A) υ(A ∩ B) P(A υB) = P(A – B)+P(B – A)+P(A ∩ B)  P(A υB) = P(A) - P(A ∩ B) + P(B) - P(A ∩ B) +P(A ∩ B) P(AυB) = P(A) + P(B) – P(A ∩ B)

  12. Corollary 2 : if B = and A and B are disjoint, then Probability Theorems • Corollary 1 : P(A υ B) < P(A) + P(B) Corollary 1 can be extended to an arbitrary of events υA j :

  13. Example Suppose a students is taking two mathematics courses ( I, II ). Let A be the event that he passes course I and B be the event that he passes course II. He feels that P(A)= 0,8 ; P(B)= 0,9 ; and P(A⋂B)= 0,75 • Describe an appropriate sample space for the experiment • Using Venn diagrams, pictorially represent S • Describe in words the events : • Find the probabilities of the events in part (c) i. ii. iii. iv.

  14. Example Solution : We use the ordered pair ( x1, x2 ) to represent passing or failing courses I and II respectively. Let xi = 1 designate passing and xi = 0 failing a. Then S= { (1,1),(1,0),(0,1),(0,0)} Example S b.

  15. Example c. (i) : passing at least one course ; alternatively, passing course I or course II or both (includes regions 1,2 and 3) (ii) : failing at least one course ; alternatively, failing course I or course II or both (includes regions 1,2 and 4) (iii) : passing course I and failing course II ( region 1) (iv) : failing both courses; alternatively, passing neither course (region 4)

  16. d. (i) P(A⋃B) = P(A)+ P(B) – P(A⋂B) = 0,8 + 0,9 – 0,75 P(A⋃B) = 0,95 (ii) P(Ā⋃ ) = P(Ā) + P( ) – P(Ā⋂ ) = 0,2 + 0,1 - P(Ā⋂ ) = Ā⋂ P( ) = P (Ā⋂ ) P(Ā⋂ ) = 1 – P(A⋃B) = 1 – 0,95 = 0,05 P(Ā⋃ ) = 0,2 + 0,1 – 0,05 = 0,25 Example

  17. Example (iii) P(A⋂ ) = ? P(A) = P(A⋂B) + P( A ⋂) P(A⋂ ) = 0,8 – 0,75 = 0,05 (iv) P( ) = 1 - P(A⋃B) = 1 – 0,95 = 0,05

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