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DISTRIBUSI BINOMIAL

DISTRIBUSI BINOMIAL. BERNOULLI PROCESS. The experiment consists of n repeated trials. Each trial results in an outcome that may be classified as a success or a failure. The probability of success, denoted by p , remains constant from trial to trial. The repeated trials are independent.

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DISTRIBUSI BINOMIAL

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  1. DISTRIBUSI BINOMIAL

  2. BERNOULLI PROCESS • The experiment consists of n repeated trials. • Each trial results in an outcome that may be classified as a success or a failure. • The probability of success, denoted by p, remains constant from trial to trial. • The repeated trials are independent.

  3. BINOMIAL DISTRIBUTION A Bernoulli trial can result in a success with probability p and a failure with probability q = 1-p. Then the probability distribution of the binomial random variable X, the number of successes in n independent trials, is

  4. CONTOH SOALDISTRIBUSI BINOMIAL (1) • The probability that a certain kind of component will survive a given shock test is ¾. Find the probability that exactly 2 of the next 4 components tested survive!

  5. CONTOH SOALDISTRIBUSI BINOMIAL (2) • The probability that a patient recovers from a rare blood disease is 0.4. If 15 people are known to have contracted this disease, what is the probability that exactly 5 survive?

  6. LATIHAN (1) • Sebuahprodusenobatbatukmemberikanpernyataanbahwaobatbatuknya 90% efektifdalammenyembuhkanpenyakitbatuk. Apabila 7 orangdenganbatukserupadiberikanobatdariprodusenitu, tentukanpeluangnya (a) tepat 3 orangdiantaranyasembuh (b) tepat 5 orangdiantaranyasembuh (c) semuanyasembuh

  7. LATIHAN (2) • Seorangpemain basket memilikipeluang 0,7 untukmemasukkan bola kedalamkeranjang. Apabilaiamelakukan 10 lemparanberturutan, tentukanpeluang (a) tepat 6 bola berhasilmasukkeranjang (b) tepat 3 bola berhasilmasukkeranjang (c) takkurangdari 8 bola masukkeranjang (d) takada bola yang masuk

  8. LATIHAN (3) • In a certain city district the need for money to buy drugs is given as the reason for 75% of all thefts. Find the probability that among the next 5 theft cases reported in this district, (a) exactly 2 resulted from the need for money to buy drugs; (b) at most 3 resulted from the need for money to buy drugs.

  9. LATIHAN (4) • A traffic control engineer reports that 75% of the vehicles passing through a checkpoint are from within the state. What is the probability that morethan 2of the next 9 vehicles are from out of the state?

  10. DISTRIBUSI POISSON

  11. POISSON PROBABILITY EXPERIMENT • The random variable is the number of times some event occurs during a defined interval. • The probability of the event is proportional to the size of the interval. • The intervals do not overlap and are independent.

  12. POISSON DISTRIBUTION • The probability distribution of the Poisson random variable X, representing the number of outcomes occuring in a given time interval or specified region denoted by t, is given by: • e  0,718281828

  13. CONTOH SOALDISTRIBUSI POISSON (1) • Rata-rata banyaknya nasabah yang masuk ke dalam antrian bagian teller suatu bank setiap menitnya adalah 2. Tentukan peluang dalam 1 menit datang 3 nasabah ke dalam antrian bagian teller tersebut!

  14. CONTOH SOALDISTRIBUSI POISSON (2) Seorang sekretaris rata-rata menerima panggilan telepon sebanyak 3 buah dalam setiap 20 menit. Tentukan peluang dalam 1 jam berikutnya ia menerima 7 buah panggilan telepon. Jawab: λ = 0,15/menit, t = 60 menit λt = 0,15 . 60 = 9

  15. CONTOH SOALDISTRIBUSI POISSON (3) • Banyaknya kata yang salah ejaan dalam suatu surat kabar adalah 3 dalam tiap 4 halaman. Tentukan peluang dalam 10 halaman surat kabar tersebut terdapat kurang dari 4 kata salah ejaan. • λ = 0,75/halaman t = 10 halaman λt = 0,75 . 10 = 7,5 • P[X<4] = p(0;7,5) + p(1;7,5) + p(2;7,5) + p(3;7,5) = 0,0006 + 0,0041 + 0,0156 + 0,0389 = 0,0592.

  16. CONTOH SOALDISTRIBUSI POISSON (4) • Pada contoh soal sebelumnya, tentukan peluang dalam 10 halaman surat kabat tersebut terdapat lebih dari 3 kata salah ejaan. • P[X>3] = p(4;7,5) + p(5;7,5) + p(6;7,5) + p(7;7,5) + ... = 1 – [p(0;7,5) + p(1;7,5) + p(2;7,5) + p(3;7,5)] = 1 – 0,0592 = 0,9408.

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