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Distribusi binomial berasal dari percobaan binomial yaitu suatu proses Bernoulli yang diulang sebanyak n kali dan saling bebas. Distribusi Binomial merupakan distribusi peubah acak diskrit. Secara langsung, percobaan binomial memiliki ciri-ciri sebagai berikut: – – – – percobaan tersebut dilakukan berulang-ulang sebanyak n kali setiap percobaan menghasilkan keluaran yang dapat dikategorikan sebagai gagal dan sukses probabilitas sukses p tetap konstan dari satu percobaan ke percobaan lain percobaan yang berulang adalah saling bebas

Ruang sampel A untuk percobaan E yang terdiri dari himpunan tak hingga tetapi masih terhitung dari titik – titik sampel: Jika S = Sukses dan G = Gagal E1 : S (sukses pada percobaan pertama) E2 : GS (gagal pada percobaan pertama dan sukses pada percobaan kedua) E3 : SG (sukses pada percobaan pertama, gagal pada percobaan kedua) E4 : GGS (gagal pada percobaan 1 dan 2, sukses pada percobaan ketiga) E5 : GSG (gagal pada percobaan 1 dan 3, sukses pada percobaan kedua) E6 : SGG (gagal pada percobaan 2 dan 3, sukses pada percobaan pertama)

En : SSS ...S GGG...G (sukses sebanyak x kali, gagal sebanyak n – x kali) x n x

Jika peluang sukses dinotasikan dengan p maka peluang gagal adalah q = 1 – p. Peubah acak X menyatakan banyaknya sukses dari n percobaan yang saling bebas. Maka peluang X pada masing – masing percobaan E adalah:

P(X) p P(X) qp pq P(X) pq

2 2

untuk E1 untuk E2 untuk E3

P(X) q p pq untuk E4 P(X) qpq pq 2 untuk E5 P(X) pqq pq 2 untuk E6 P(X) p x q n x untuk En

2 Dapat dilihat bahwa E2 dan E3 memberikan hasil yang sama. Jumlahnya , yaitu 1

jumlah semua titik sampel yang mungkin menghasilkan x = 1 yang sukses dan n – x = 2 – 1 = 1 yang gagal dari 2 percobaan. Begitupun untuk E4 , E5 , dan E6 juga

3 memberikan hasil yang sama. Jumlahnya , yaitu jumlah semua titik sampel yang 1

mungkin yang menghasilkan x = 1 yang sukses dan n – x = 3 – 1 = 2 yang gagal dari 3 percobaan. Secara umum, jumlah titik sampel yang mungkin untuk menghasilkan x sukses dan n – x gagal dalam n percobaan adalah banyaknya cara yang berbeda dalam

n mendistribusikan x sukses dalam barisan n percobaan, sehingga terdapat cara. x

Dan distribusi peluang atau Probability Mass Function (PMF) X dinyatakan pada definisi berikut:

n n P(X x) f ( x) p x q n x p x (1 p)n x x x

untuk x 1, 2,..., n dan 0 p 1 . Pembuktian distribusi Binomial merupakan suatu PMF. Bukti: Untuk membuktikan suatu peubah acak adalah PMF, maka harus ditunjukan: 1. 2. f ( x) 0

f ( x) 1 x Akan ditunjukkan distribusi binomial memenuhi kedua syarat di atas: 1. f ( x) 0

Karena 0 p 1 dan nilai kombinasi pasti positif maka f(x) pasti positif. 2.

f ( x) 1 x Menggunakan persamaan binomial Newton pada

f ( x) , akan diperoleh: x f ( x) x p (1 p) x x x 0

n

n

n x

( p (1 p)) n 1n 1

(1)

Dari 1 dan 2 dapat dikatakan bahwa distribusi binomial merupakan PMF.

Mean Jika X

B(n, p) ( X variabel random berdistribusi Binomial), maka nilai ekspektasi

dari X adalah

E ( X ) x.P( x) x x.P( X x) x 0 n

n

n x. p x (1 p) n x x 0 x n n x. p x (1 p) n x x 1 x n n! x. p x (1 p) n x x !(n x)! x 1 x. x 1 n

n.(n 1)! . p. p x 1 (1 p) n x x.( x 1)!(n x)!

np

(n 1)! p x 1 (1 p) n x x 1 ( x 1)!( n x )! n Misalkan m = n – 1 dan s = x – 1, maka persamaan di atas menjadi

E ( X ) np m! p s (1 p)m s s 0 s !(m s )! m Berdasarkan (1),

s !(m s)! p (1 p) s s 0

m

m!

m s

1 , maka

E ( X ) np s 0

m

m! p s (1 p) m s s !(m s )!

np.1 np

Sehingga didapat mean dari X, E ( X ) np

Variansi

Var ( X ) E ( X 2 ) ( E( X ))2

Dalam meencari Var(X), kita harus tahu nilai ekspektasi dari X2 :

E ( X 2 ) x 2 .P( X x) x 2 . x 0 x 0

n

n

n! p x (1 p) n x x !(n x)!

E ( X 2 ) x 2 .P( x) x x 2 .P ( X x ) x 0 n n x 2 . p x (1 p) n x x 0 x n n x 2 . p x (1 p) n x x 1 x n n! x2. p x (1 p) n x x !(n x)! x 1

n

x2. x 1 n

n

n.(n 1)! . p. p x 1 (1 p) n x x.( x 1)!(n x)! (n 1)! p x 1 (1 p) n x ( x 1)!(n x)!

np x. x 1

Misalkan m = n – 1 dan s = x – 1, maka persamaan di atas menjadi

E ( X 2 ) np ( s 1) s 0 m m m m! p s (1 p) m s np ( s 1) p s (1 p) m s s !(m s)! s 0 s

E ( X 2 ) np ( s 1) s 0

m

m! p s (1 p) m s s !(m s )!

m m np ( s 1) p s (1 p) m s s 0 s m m m s m m s np. s. p (1 p) 1. p s (1 p) m s s 0 s s 0 s np[ mp 1]

np[( n 1) p 1] np[ np p 1]

Var ( X ) E ( X 2 ) ( E ( X )) 2 np[np p 1] (np )2 (np )2 np 2 np (np )2 np np 2 np (1 p )

Sehingga didapat variansi dari X, Var ( X ) np(1 p)

Contoh: 1. Probabilitas bahwa sejenis komponen tertentu yang lolos uji kelayakan adalah ¾. Tentukan probabilitas dimana 2 dari 4 komponen yang selanjutnya diuji akan dinyatakan layak!

n n P(X x) f ( x) p x q n x p x (1 p)n x x x

p = ¾, q = 1 – ¾ = ¼ untuk x = 2

42

4 3 1 P( X 2) 2 4 4

2

6.

9 1 27 . 16 16 128

2. Berdasarkan data biro perjalanan PT Sentosa, yang khusus menangani perjalanan wisata turis mancanegara, 20% dari turis menyatakan sangat puas berkunjung ke Indonesia, 40% menyatakan puas, 25% menyatakan biasa saja, dan sisanya menyatakan kurang puas. Apabila kita bertemu dengan 5 orang dari peserta wisata turis mancanegara yang pernah menggunakan jasa biro perjalanan tersebut. Tentukan probabilitas: a. Tepat 2 diantaranya menyatakan biasa saja

b. Paling banyak 2 diantaranya menyatakan sangat puas

Jawab: n=5 a. p = 0.25, q = 1 – 0.25 = 0.75

5 2

5 1 3 P( X 2) 2 4 4

2

0.2637

b. p = 0.2, q = 1 – 0.2 = 0.8

P( X 2) P( X 0) P( X 1) P( X 2) 5 5 5 (0.2) 0 (0.8)5 (0.2)1 (0.8)51 (0.2) 2 (0.8)52 0 1 2 0.32768 0.40960 0.20480 0.94208…...

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