Dari setiap 100 kaleng susu hasil produksi yang dijual, 5 kaleng susu timbangannya tidak tepat. Jika kita beli 100 kaleng susu, berapa peluang: paling banyak 6 kaleng susu tidak tepat timbangannya,

Jawaban yang benar adalah 6.086.922.536 · (19^(94))/(20^(100)) Ingat konsep distribusi binomial P(X=x) = C(n,x) . p^x . q^(n-x) dengan p : peluang sukses q : peluang gagal q = 1- p Kombinasi adalah cara menyusun objek tanpa memperhatikan urutan: C(n,r) = (n!)/((n-r)!r!) Dari setiap 100 kaleng susu hasil produksi yang dijual, 5 kaleng susu timbangannya tidak tepat, peluang kaleng susu tidak tepat timbangannya adalah p = 5/100 p = 1/20 sehingga q = 1-p = 1-1/20 = 19/20 Jika kita beli 100 kaleng susu, berarti n=100, peluang paling banyak 6 kaleng susu tidak tepat timbangannya adalah P(X≤6) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6) Hitung satu per satu. P(X=0) = C(100,0) · (1/20)^0 · (19/20)^(100-0) = (100!)/((100-0)!·0!) · 1 · (19/20)^(100) = (100!)/(100!) · (19/20)^(100) = (19/20)^(100) = (19^(100))/(20^(100)) = 19^6 · (19^(94))/(20^(100)) = 47.045.881 · (19^(94))/(20^(100)) P(X=1) = C(100,1) · (1/20)^1 · (19/20)^(100-1) = (100!)/((100-1)!·1!) · 1/20 · (19/20)^(99) = (100!)/(99!·1!) · 1/20 · (19/20)^(99) = (100·99!)/(99!·1!) · 1/20 · (19/20)^(99) = 100 · 1/20 · (19^(99))/(20^(99)) = 100 · (19^(99))/(20^(100)) = 100 · 19^5 · (19^(94))/(20^(100)) = 247.609.900 · (19^(94))/(20^(100)) P(X=2) = C(100,2) · (1/20)^2 · (19/20)^(100-2) = (100!)/((100-2)!·2!) · 1/(20^2) · (19/20)^(98) = (100!)/(98!·2!) · 1/(20^2) · (19/20)^(98) = (100·99·98!)/(98!·2·1) · 1/(20^2) · (19/20)^(98) = 4.950 · 1/(20^2) · (19^(98))/(20^(98)) = 4.950 · (19^(98))/(20^(100)) = 4.950 · 19^4 · (19^(94))/(20^(100)) = 645.088.950 · (19^(94))/(20^(100)) P(X=3) = C(100,3) · (1/20)^3 · (19/20)^(100-3) = (100!)/((100-3)!·3!) · 1/(20^3) · (19/20)^(97) = (100!)/(97!·3!) · 1/(20^3) · (19/20)^(97) = (100·99·98·97!)/(97!·3·2·1) · 1/(20^3) · (19/20)^(97) = 161.700 · 1/(20^3) · (19^(97))/(20^(97)) = 161.700 · (19^(97))/(20^(100)) = 161.700 · 19^3 · (19^(94))/(25^(100)) = 1.109.100.300 · (19^(94))/(25^(100)) P(X=4) = C(100,4) · (1/20)^4 · (19/20)^(100-4) = (100!)/((100-4)!·4!) · 1/(20^4) · (19/20)^(96) = (100!)/(96!·4!) · 1/(20^4) · (19/20)^(96) = (100·99·98·97·96!)/(96!·4·3·2·1) · 1/(20^4) · (19/20)^(96) = 3.921.225 · 1/(20^4) · (19^(96))/(20^(96)) = 3.921.225 · (19^(96))/(20^(100)) = 3.921.225 · 19^2 · (19^(94))/(20^(100)) = 1.415.562.225 · (19^(94))/(20^(100)) P(X=5) = C(100,5) · (1/20)^5 · (19/20)^(100-5) = (100!)/((100-5)!·5!) · 1/(20^5) · (19/20)^(95) = (100!)/(95!·5!) · 1/(20^5) · (19/20)^(95) = (100·99·98·97·96·95!)/(95!·5·4·3·2·1) · 1/(20^5) · (19/20)^(95) = 75.287.520 · 1/(20^5) · (19^(95))/(20^(95)) = 75.287.520 · (19^(95))/(20^(100)) = 75.287.520 · 19^1 · (19^(94))/(20^(100)) = 1.430.462.880 · (19^(94))/(20^(100)) P(X=6) = C(100,6) · (1/20)^6 · (19/20)^(100-6) = (100!)/((100-6)!·6!) · 1/(20^6) · (19/20)^(94) = (100!)/(94!·6!) · 1/(20^6) · (19/20)^(94) = (100·99·98·97·96·95·94!)/(94!·6·5·4·3·2·1) · 1/(20^6) · (19/20)^(94) = 1.192.052.400 · 1/(20^6) · (19^(94))/(20^(94)) = 1.192.052.400 · (19^(94))/(20^(100)) = 1.192.052.400 · (19^(94))/(20^(100)) Sehingga P(X≤6) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6) = 47.045.881 · (19^(94))/(20^(100)) + 247.609.900 · (19^(94))/(20^(100)) + 645.088.950 · (19^(94))/(20^(100)) + 1.109.100.300 · (19^(94))/(25^(100)) + 1.415.562.225 · (19^(94))/(20^(100)) + 1.430.462.880 · (19^(94))/(20^(100)) + 1.192.052.400 · (19^(94))/(20^(100)) = (47.045.881 + 247.609.900 + 645.088.950 + 1.109.100.300 + 1.415.562.225 + 1.430.462.880 + 1.192.052.400) · (19^(94))/(20^(100)) = 6.086.922.536 · (19^(94))/(20^(100)) Jadi, peluang paling banyak 6 kaleng susu tidak tepat timbangannya adalah 6.086.922.536 · (19^(94))/(20^(100))