{"id":17952,"date":"2024-03-28T11:41:39","date_gmt":"2024-03-28T11:41:39","guid":{"rendered":"https:\/\/soicauso3cang.com\/?p=17952"},"modified":"2024-03-28T11:41:39","modified_gmt":"2024-03-28T11:41:39","slug":"chia-se-bai-toan-danh-dan-lo-nam-chac-phan-thang","status":"publish","type":"post","link":"https:\/\/soicauso3cang.com\/chia-se-bai-toan-danh-dan-lo-nam-chac-phan-thang\/","title":{"rendered":"chia s\u1ebb b\u00e0i to\u00e1n \u0111\u00e1nh d\u00e0n l\u00f4 n\u1eafm ch\u1eafc ph\u1ea7n th\u1eafng"},"content":{"rendered":"
<\/p>\n\n\n
Xin ch\u00e0o c\u00e1c b\u1ea1n, h\u00f4m nay s\u1ebd chia s\u1ebb cho c\u00e1c b\u1ea1n v\u1ec1 c\u00e1c ph\u00e9p to\u00e1n trong l\u00f4 \u0111\u1ec1 c\u1ef1c k\u00ec hay v\u00e0 ch\u00ednh x\u00e1c nh\u00e9. D\u01b0\u1edbi \u0111\u00e2y l\u00e0 2 b\u00e0i to\u00e1n \u0111\u00e1nh l\u00f4 v\u00e0 \u0111\u00e1nh d\u00e0n l\u00f4, c\u00e1c b\u1ea1n c\u00f3 th\u1ec3 v\u1eadn d\u1ee5ng ch\u01a1i d\u00e0n l\u00f4 4 con, d\u00e0n l\u00f4 8 con \u0111\u1ec1u \u0111\u01b0\u1ee3c nh\u00e9, c\u00f9ng theo d\u00f5i nh\u00e9:<\/p>\n\n\n\n
Lu\u1eadt ch\u01a1i l\u00f4 \u0111\u1ea1i lo\u1ea1i nh\u01b0 sau: S\u00e1ng, b\u1ea1n \u0111\u1eb7t c\u01b0\u1ee3c 1 con s\u1ed1 trong ph\u1ea1m vi t\u1eeb 00 \u0111\u1ebfn 99 v\u00e0 m\u1ed9t s\u1ed1 \u0111i\u1ec3m l\u00f4, a \u0111i\u1ec3m ch\u1eb3ng h\u1ea1n. M\u1ed7i \u0111i\u1ec3m l\u00f4 ph\u1ea3i chi ph\u00ed 23000 (\u0111\u1ed3ng). D\u00e0n l\u00f4 g\u1ed3m 27 con l\u00f4 (l\u00e0 2 ch\u1eef s\u1ed1 cu\u1ed1i c\u1ee7a c\u00e1c s\u1ed1 t\u1ea1i c\u00e1c gi\u1ea3i X\u1ed5 s\u1ed1 c\u1ee7a Nh\u00e0 n\u01b0\u1edbc ph\u00e1t h\u00e0nh).<\/p>\n\n\n\n
N\u1ebfu c\u00f3 k con l\u00f4 tr\u00f9ng v\u1edbi s\u1ed1 b\u1ea1n \u0111\u00e3 \u0111\u1eb7t c\u01b0\u1ee3c th\u00ec s\u1ed1 ng\u01b0\u1eddi ta g\u1ecdi l\u00e0 b\u1ea1n \u0111\u00e3 tr\u00fang \u201ck nh\u00e1y\u201d. M\u1ed7i \u201cnh\u00e1y\u201d, ch\u1ee7 l\u00f4 tr\u1ea3 cho b\u1ea1n 80000.a (\u0111\u1ed3ng).<\/p>\n\n\n\n
Gi\u1ea3 s\u1eed anh b\u1ea1n c\u1ee7a ch\u00fang ta \u0111\u00e1nh 1 \u0111i\u1ec3m l\u00f4. Anh ta chi h\u1ebft 23000. D\u1ec5 th\u1ea5y \u0111\u00e1nh l\u00f4 l\u00e0 m\u1ed9t ph\u00e9p th\u1eed Bernoulli.<\/p>\n\n\n\n
X\u00e1c su\u1ea5t \u0111\u1ec3 anh ta tr\u00fang \u0111\u00fang k nh\u00e1y (k=0,1,\u2026,27) l\u00e0 Ck27(0,01)k(0,99)27\u2212k<\/p>\n\n\n\n
N\u1ebfu tr\u00fang, anh ta \u0111\u01b0\u1ee3c 80000.k (\u0111\u1ed3ng). Nh\u01b0 v\u1eady l\u00e3i :<\/p>\n\n\n\n
80000.k\u201323000 (\u0111\u1ed3ng)<\/p>\n\n\n\n
V\u1eady trung b\u00ecnh anh ta l\u00e3i:<\/p>\n\n\n\n
E=\u2211k=027[Ck27.(0,01)k(0,99)27\u2212k.(80k\u221223)]<\/p>\n\n\n\n
=80\u2211k=127[kCk27.(0,01)k(0,99)27\u2212k.]\u221223<\/p>\n\n\n\n
=80.27100\u2211k=026[Ck26.(0,01)k(0,99)26\u2212k.]\u221223<\/p>\n\n\n\n
=8.2,7\u201323=\u22121,4 (ngh\u00ecn \u0111\u1ed3ng)<\/p>\n\n\n\n
V\u1eady trung b\u00ecnh anh ta l\u1ed7 1400 (\u0111\u1ed3ng).<\/p>\n\n\n\n
R\u00f5 r\u00e0ng l\u00e0 \u0111\u00e1nh l\u00f4 l\u1ed7 \u00edt h\u01a1n \u0111\u00e1nh \u0111\u1ec1<\/p>\n\n\n\n Anh ta \u0111\u00e1nh n con \u0111\u1ec1 kh\u00e1c nhau. M\u1ed7i con \u0111\u00e1nh 1 (ngh\u00ecn \u0111\u1ed3ng). V\u1ecb chi l\u00e0 chi ph\u00ed h\u1ebft n (ngh\u00ecn \u0111\u1ed3ng)<\/p>\n\n\n\n X\u00e1c su\u1ea5t anh ta tr\u00fang l\u00e0 0,01n. N\u1ebfu tr\u00fang, anh ta \u0111\u01b0\u1ee3c 70 (ngh\u00ecn \u0111\u1ed3ng). T\u1ee9c l\u00e0 l\u00e3i: 70\u2013n (ngh\u00ecn \u0111\u1ed3ng)<\/p>\n\n\n\n X\u00e1c su\u1ea5t anh ta tr\u01b0\u1ee3t l\u00e0 (1\u22120,01n). N\u1ebfu tr\u00fang, anh ta \u0111\u01b0\u1ee3c l\u00e3i: \u2212n (ngh\u00ecn \u0111\u1ed3ng).<\/p>\n\n\n\n V\u1eady trung b\u00ecnh anh ta l\u00e3i:<\/p>\n\n\n\n Ed(n)=0,01n.(70\u2212n)\u2013n(1\u20130,01n)=\u22120,3n<\/p>\n\n\n\n maxn\u2208[0;100]Ed(n)=Ed(0)=0<\/p>\n\n\n\n V\u1eady l\u00e3i nh\u1ea5t khi \u0111\u00e1nh \u0111\u1ec1 l\u00e0 \u0111\u00e1nh 0 con.<\/p>\n\n\n\n H\u1eebm. Th\u1ebf th\u00ec n\u00f3i l\u00e0m g\u00ec. Anh ta l\u1ea7m b\u1ea7m v\u00e0 \u0111i \u0111\u00e1nh l\u00f4.<\/p>\n\n\n\n Anh ta \u0111\u00e1nh n con l\u00f4, m\u1ed7i con 1 \u0111i\u1ec3m. Th\u1ebf l\u00e0 chi h\u1ebft 23n (ngh\u00ecn \u0111\u1ed3ng).<\/p>\n\n\n\n X\u00e1c su\u1ea5t anh ta tr\u00fang \u0111\u00fang k nh\u00e1y (k=0,1,2,\u2026,27) l\u00e0<\/p>\n\n\n\n Ck27(n100)k(100\u2212n100)27\u2212k<\/p>\n\n\n\n N\u1ebfu tr\u00fang, anh ta \u0111\u01b0\u1ee3c 80k (ngh\u00ecn \u0111\u1ed3ng). T\u1ee9c l\u00e0 l\u00e3i: 80k\u201323n (ngh\u00ecn \u0111\u1ed3ng).<\/p>\n\n\n\n T\u1eeb \u0111\u00f3, trung b\u00ecnh anh ta l\u00e3i:<\/p>\n\n\n\n El(n)=\u2211k=027[Ck27(n100)k(100\u2212n100)27\u2212k.(80k\u221223n)]=\u22127n5<\/p>\n\n\n\n maxn\u2208[0;100]El(n)=El(0)=0<\/p>\n\n\n\n Do \u0111\u00f3 l\u00e3i nh\u1ea5t khi \u0111\u00e1nh l\u00f4 l\u00e0 \u0111\u00e1nh 0 con.<\/p>\n\n\n\n N\u00f3i cho d\u1ec5 hi\u1ec3u th\u00ec vi\u1ec7c \u0111\u00e1nh l\u00f4 \u0111\u1ec1 theo ph\u01b0\u01a1ng ph\u00e1p x\u00e1c su\u1ea5t th\u1ed1ng k\u00ea ch\u1ec9 c\u00f3 l\u1ee3i \u1edf ph\u00eda nh\u00e0 c\u00e1i, tuy nhi\u00ean \u0111\u00f3 ch\u1ec9 l\u00e0 nh\u1eadn \u0111\u1ecbnh phi\u1ebfn di\u1ec7n c\u1ee7a m\u1ed9t ng\u01b0\u1eddi m\u00e0 th\u00f4i. Trong th\u1ef1c t\u1ebf \u0111\u00e1nh l\u00f4 \u0111\u1ec1 y\u1ebfu t\u1ed1 may m\u1eafn v\u00e0 chi\u1ebfn thu\u1eadt r\u1ea5t quan tr\u1ecdng, v\u00ec n\u1ebfu kh\u00f4ng th\u00ec l\u0129nh v\u1ef1c n\u00e0y \u0111\u00e3 \u201cd\u1eb9p ti\u1ec7m\u201d l\u00e2u r\u1ed3i ch\u1ee9 kh\u00f4ng tr\u1edf n\u00ean b\u00f9ng n\u1ed5 v\u1ec1 s\u1ed1 l\u01b0\u1ee3ng ng\u01b0\u1eddi ch\u01a1i nh\u01b0 ng\u00e0y nay.<\/p>\n\n\n <\/p>\r\n\u2013 B\u00e0i to\u00e1n: \u0110\u00e1nh d\u00e0n l\u00f4<\/h2>\n\n\n\n
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