{"id":18407,"date":"2024-04-18T12:28:13","date_gmt":"2024-04-18T12:28:13","guid":{"rendered":"https:\/\/soicau3063.minhngocxoso.com\/?p=18407"},"modified":"2024-04-18T12:28:13","modified_gmt":"2024-04-18T12:28:13","slug":"nuoi-lo-de-theo-bai-toan-soi-cau-vip","status":"publish","type":"post","link":"https:\/\/soicaudexoso.com\/nuoi-lo-de-theo-bai-toan-soi-cau-vip\/","title":{"rendered":"nu\u00f4i l\u00f4 \u0111\u1ec1 theo b\u00e0i to\u00e1n soi c\u1ea7u vip"},"content":{"rendered":"

nu\u00f4i l\u00f4 \u0111\u1ec1 theo b\u00e0i to\u00e1n soi c\u1ea7u vip<\/h1>\n
\n

T\u00ednh x\u00e1c su\u1ea5t l\u00f4 \u0111\u1ec1 ch\u00ednh x\u00e1c v\u00e0 hi\u1ec7u qu\u1ea3 l\u00e0 k\u1ef9 n\u0103ng \u0111\u00f2i h\u1ecfi ng\u01b0\u1eddi ch\u01a1i ph\u1ea3i c\u00f3 nhi\u1ec1u kinh nghi\u1ec7m v\u00e0 luy\u1ec7n t\u1eadp th\u01b0\u1eddng xuy\u00ean. C\u00f3 nhi\u1ec1u ph\u01b0\u01a1ng ph\u00e1p t\u00ednh x\u00e1c su\u1ea5t l\u00f4 \u0111\u1ec1 c\u00e1c mi\u1ec1n nhanh ch\u00f3ng v\u00e0 c\u1ef1c k\u1ef3 hi\u1ec7u qu\u1ea3 b\u00ean d\u01b0\u1edbi. M\u1eddi c\u00e1c b\u1ea1n c\u00f9ng theo d\u00f5i.<\/p>\n

T\u00ednh x\u00e1c su\u1ea5t l\u00f4 \u0111\u1ec1 \u0111\u01b0\u1ee3c hi\u1ec3u nh\u01b0 th\u1ebf n\u00e0o \"\"<\/h2>\n

B\u00ecnh th\u01b0\u1eddng m\u1ecdi ng\u01b0\u1eddi \u0111\u00e1nh l\u00f4 \u0111\u1ec1 tr\u00ean m\u1ea1ng  v\u1edbi nhi\u1ec1u g\u00f3i tr\u00fang t\u1ef7 l\u1ec7 \u0103n r\u1ea5t cao nh\u01b0 1 \u0103n 99. H<\/strong>o\u1eb7c g\u00f3i ho\u00e0n % hoa h\u1ed3ng cao \u0103n 80 hoa h\u1ed3ng 19.5%<\/strong> , 70 hoa h\u1ed3ng 29.5%<\/strong>\u2026. Nh\u01b0ng h\u00f4m nay ch\u00fang t\u1ed1i c\u0169ng s\u1ebd b\u1ed5 sung nhi\u1ec1u ki\u1ebfn th\u1ee9c v\u1ec1 t\u00ednh x\u00e1c xu\u1ea5t l\u00f4 \u0111\u1ec1<\/strong> \u0111\u1ec3 cho c\u00e1c b\u1ea1n n\u1eafm \u0111\u01b0\u1ee3c ph\u1ea7n th\u1eafng nhi\u1ec1u h\u01a1n. V\u00ec v\u1eady c\u00e1c b\u1ea1n c\u1ea7n ph\u1ea3i xem c\u00e1ch t\u00ednh x\u00e1c xu\u1ea5t % t\u1ef7 l\u1ec7 tr\u00fang c\u1ee7a l\u00f4 c\u1ee7a \u0111\u1ec1 b\u00ean d\u01b0\u1edbi<\/p>\n

T\u00ednh x\u00e1c xu\u1ea5t tr\u00fang l\u00f4 \u0111\u1ec1 bao nhi\u00eau \u0111\u1ec3 soi c\u1ea7u l\u00f4 hi\u1ec7u qu\u1ea3.<\/h3>\n

Nhi\u1ec1u ng\u01b0\u1eddi ngh\u0129 x\u00e1c su\u1ea5t tr\u00fang l\u00f4 \u0111\u1ec1 l\u00e0 27% v\u00ec m\u1ed7i m\u1ed9t l\u1ea7n quay l\u00e0 c\u00f3 27 gi\u1ea3i trong t\u1ed5ng s\u1ed1 100 s\u1ed1. Nh\u01b0ng th\u1ef1c t\u1ebf \u1edf k\u1ebft qu\u1ea3 x\u1ed5 s\u1ed1 h\u00e0ng ng\u00e0y, c\u00f3 nhi\u1ec1u k\u1ebft qu\u1ea3 l\u1eb7p l\u1ea1i 2-3 l\u1ea7n. V\u00ed d\u1ee5: ng\u00e0y 28\/10\/2024, k\u1ebft qu\u1ea3 47 v\u1ec1 2 nh\u00e1y. \u0110\u1ec3 t\u00ednh \u0111\u01b0\u1ee3c x\u00e1c su\u1ea5t l\u00f4 \u0111\u1ec1, ta ph\u1ea3i t\u00ednh theo c\u00f4ng th\u1ee9c to\u00e1n h\u1ecdc x\u00e1c su\u1ea5t.<\/p>\n

T\u00ednh x\u00e1c xu\u1ea5t tr\u00fang 1 con l\u00f4<\/h3>\n

C\u00f4ng th\u1ee9c n\u00e0y ch\u1ec9 t\u00ednh x\u00e1c su\u1ea5t tr\u00fang l\u00f4, kh\u00f4ng t\u00ednh tr\u00fang 2 hay 3 nh\u00e1y.<\/p>\n

\u0110\u1ec3 t\u00ednh x\u00e1c su\u1ea5t tr\u00fang l\u00f4, ta t\u00ednh x\u00e1c su\u1ea5t t\u1ea1ch l\u00f4 tr\u01b0\u1edbc. Trong x\u1ed5 s\u1ed1 c\u00f3 27 k\u1ebft qu\u1ea3, 27 l\u1ea7n quay. X\u00e1c su\u1ea5t m\u1ed7i l\u1ea7n t\u1ea1ch l\u00f4 l\u00e0 99%. Nh\u01b0 v\u1eady t\u1ef7 l\u1ec7 t\u1ea1ch l\u00f4 khi quay \u0111\u1ebfn gi\u1ea3i cu\u1ed1i c\u00f9ng l\u00e0 0,99^27.<\/p>\n

T\u1eeb \u0111\u00f3, ta c\u00f3:<\/strong><\/p>\n

X\u00e1c su\u1ea5t tr\u00fang m\u1ed9t con l\u00f4 = 1 \u2013 x\u00e1c su\u1ea5t t\u1ea1ch l\u00f4 = 1 \u2013 0,99^27 = 1 \u2013 0,7623 = 23,7657%.<\/strong><\/p>\n

T\u00ednh \u0111\u01b0\u1ee3c x\u00e1c su\u1ea5t tr\u00fang m\u1ed9t con l\u00f4 k\u1ebft h\u1ee3p v\u1edbi kinh nghi\u1ec7m, ph\u01b0\u01a1ng ph\u00e1p soi l\u00f4 c\u1ea7u \u0111\u1ec3 t\u00ecm ra con s\u1ed1 ch\u00ednh x\u00e1c v\u00e0 hi\u1ec7u qu\u1ea3 nh\u1ea5t.<\/p>\n

T\u00ednh x\u00e1c su\u1ea5t tr\u00fang 1 con \u0111\u1ec1<\/h3>\n

\u0110\u1ec1 ch\u1ec9 v\u1ec1 \u0111\u00fang 1 con s\u1ed1 trong t\u1ed5ng 100 s\u1ed1, n\u00ean x\u00e1c su\u1ea5t tr\u00fang \u0111\u1ec1 l\u00e0 1%.<\/strong><\/p>\n

T\u00ednh x\u00e1c su\u1ea5t tr\u00fang l\u00f4 xi\u00ean 2<\/h3>\n

L\u00f4 xi\u00ean 2 t\u1ee9c l\u00e0 b\u1ea1n \u0111\u00e1nh 2 con l\u00f4 v\u00e0 b\u1eaft bu\u1ed9c ph\u1ea3i tr\u00fang c\u1ea3 2 con m\u1edbi \u0111\u01b0\u1ee3c t\u00ednh \u0103n. T\u1ef7 l\u1ec7 tr\u00fang l\u00f4 xi\u00ean 2 ch\u1eafc ch\u1eafn l\u00e0 th\u1ea5p, nh\u01b0ng t\u1ef7 l\u1ec7 c\u01b0\u1ee3c l\u1ea1i cao h\u01a1n nhi\u1ec1u so v\u1edbi l\u00f4 th\u00f4ng th\u01b0\u1eddng. \u0110\u1ec3 t\u00ednh x\u00e1c su\u1ea5t tr\u00fang l\u00f4 xi\u00ean 2, ta ph\u1ea3i t\u00ednh \u0111\u01b0\u1ee3c x\u00e1c su\u1ea5t t\u1ea1ch l\u00f4 xi\u00ean 2.<\/p>\n

N\u1ebfu \u0111\u00e1nh l\u00f4 xi\u00ean 2 m\u00e0 ch\u1ec9 v\u1ec1 m\u1ed9t con ho\u1eb7c kh\u00f4ng v\u1ec1 con n\u00e0o th\u00ec \u0111\u1ec1u t\u00ednh l\u00e0 t\u1ea1ch.<\/p>\n

X\u00e1c su\u1ea5t kh\u00f4ng v\u1ec1 con n\u00e0o = 0,98^27 = 57,957%.<\/strong><\/p>\n

X\u00e1c su\u1ea5t ch\u1ec9 v\u1ec1 1 con = 2 * x\u00e1c su\u1ea5t tr\u00fang 1 l\u00f4 * x\u00e1c su\u1ea5t kh\u00f4ng tr\u00fang l\u00f4 = 2 * 23,7657% * 0,99^27 = 36,238%.<\/strong><\/p>\n

X\u00e1c su\u1ea5t kh\u00f4ng tr\u00fang l\u00f4 xi\u00ean 2 = X\u00e1c su\u1ea5t kh\u00f4ng v\u1ec1 c\u1ea3 2 con + x\u00e1c su\u1ea5t ch\u1ec9 v\u1ec1 1 con = 57,957% + 36,238% = 94,195%.<\/strong><\/p>\n

X\u00e1c su\u1ea5t tr\u00fang l\u00f4 xi\u00ean 2 = 100% \u2013 X\u00e1c su\u1ea5t kh\u00f4ng tr\u00fang = 100 \u2013 94,195% = 5,8%.<\/strong><\/p>\n

\"\"<\/p>\n

T\u00ednh x\u00e1c su\u1ea5t tr\u00fang l\u00f4 xi\u00ean 3<\/h3>\n

T\u01b0\u01a1ng t\u1ef1 nh\u01b0 c\u00e1ch \u0111\u00e1nh l\u00f4 xi\u00ean 2, nh\u01b0ng t\u1ef7 l\u1ec7 tr\u00fang l\u1ea1i th\u1ea5p v\u00e0 t\u1ef7 l\u1ec7 c\u01b0\u1ee3c l\u1ea1i cao h\u01a1n r\u1ea5t nhi\u1ec1u. V\u1ec1 c\u00e1ch t\u00ednh x\u00e1c su\u1ea5t tr\u00fang l\u00f4 xi\u00ean 3 c\u0169ng t\u01b0\u01a1ng t\u1ef1 nh\u01b0 l\u00f4 xi\u00ean 2.<\/p>\n

X\u00e1c su\u1ea5t kh\u00f4ng tr\u00fang l\u00f4 xi\u00ean 3 = X\u00e1c su\u1ea5t kh\u00f4ng v\u1ec1 c\u1ea3 3 con + x\u00e1c su\u1ea5t ch\u1ec9 v\u1ec1 2 con + x\u00e1c su\u1ea5t ch\u1ec9 v\u1ec1 1 con = 0,97^27 + 3 * 5,8% * 0,99^27 + 3 * 23,7675% * 0,98^27 = 98,527%<\/p>\n

X\u00e1c su\u1ea5t tr\u00fang l\u00f4 xi\u00ean 3 = 100% \u2013 98,527% = 1,473% ( cao h\u01a1n tr\u00fang \u0111\u1ec1 1 t\u00fd).<\/p>\n

B\u00e0i to\u00e1n nu\u00f4i l\u00f4 \u0111\u1ec1<\/h3>\n

Gi\u1ea3 s\u1eed trong n ng\u00e0y li\u00ean ti\u1ebfp ta nu\u00f4i l\u00f4. \u01af\u1edbc l\u01b0\u1ee3ng s\u1ed1 ng\u00e0y n \u0111\u1ec3 x\u00e1c su\u1ea5t ta tr\u00fang 1 l\u1ea7n \u0111\u1ea1t g\u1ea7n 100%.<\/p>\n

\u2013 G\u1ecdi A<\/strong> l\u00e0 bi\u1ebfn c\u1ed1 trong n ng\u00e0y, ta kh\u00f4ng tr\u00fang l\u1ea7n n\u00e0o<\/p>\n

\u2013 G\u1ecdi B<\/strong> l\u00e0 bi\u1ebfn c\u1ed1 ta tr\u00fang \u00edt nh\u1ea5t 1 l\u1ea7n<\/p>\n

X\u00e1c su\u1ea5t \u0111\u1ec3 ta tr\u01b0\u1ee3t c\u1ea3 27 con l\u00f4 trong d\u00e0n l\u00f4 c\u1ee7a m\u1ed9t ng\u00e0y l\u00e0: 0,99^27<\/p>\n