{"id":17977,"date":"2024-04-29T04:35:34","date_gmt":"2024-04-29T04:35:34","guid":{"rendered":"https:\/\/soicau3099.minhngocxoso.com\/ly-giai-tai-sao-can-tinh-xac-suat-khi-choi-lo-de\/"},"modified":"2024-04-29T04:35:34","modified_gmt":"2024-04-29T04:35:34","slug":"ly-giai-tai-sao-can-tinh-xac-suat-khi-choi-lo-de","status":"publish","type":"post","link":"https:\/\/sieubachthulo.com\/ly-giai-tai-sao-can-tinh-xac-suat-khi-choi-lo-de\/","title":{"rendered":"l\u00fd gi\u1ea3i t\u1ea1i sao c\u1ea7n t\u00ednh x\u00e1c su\u1ea5t khi ch\u01a1i l\u00f4 \u0111\u1ec1?"},"content":{"rendered":"
\u0110\u1ec3 gi\u1ea3i th\u00edch c\u00e2u h\u1ecfi t\u1ea1i sao c\u1ea7n t\u00ednh x\u00e1c su\u1ea5t khi ch\u01a1i l\u00f4 \u0111\u1ec1 v\u00e0 t\u1ea7m quan tr\u1ecdng c\u1ee7a x\u00e1c su\u1ea5t khi ch\u01a1i nh\u01b0 th\u1ebf n\u00e0o? H\u00e3y c\u00f9ng chuy\u00ean gia l\u00f4 \u0111\u1ec1 tr\u1ef1c tuy\u1ebfn ph\u00e2n t\u00edch ch\u00ednh x\u00e1c ngay sau \u0111\u00e2y!<\/p>\n
L\u00f4 \u0111\u1ec1 l\u00e0 t\u1eadp h\u1ee3p c\u1ee7a c\u00e1c l\u1ef1a ch\u1ecdn con s\u1ed1 sao cho tr\u00f9ng v\u1edbi 2 s\u1ed1 cu\u1ed1i c\u1ee7a gi\u1ea3i x\u1ed5 s\u1ed1. Tuy nhi\u00ean vi\u1ec7c th\u1eafng x\u1ed5 s\u1ed1 c\u0169ng t\u01b0\u01a1ng t\u1ef1 h\u00ecnh th\u1ee9c th\u1eafng l\u00f4 \u0111\u1ec3. V\u00e3n c\u00f3 ng\u01b0\u1eddi tr\u00fang l\u1edbn nh\u01b0ng kh\u1ea3 n\u0103ng \u0111\u1ebfn tay b\u1ea1n s\u1ebd kh\u00f4ng cao.<\/p>\n
Nh\u01b0 v\u1eady x\u00e1c su\u1ea5t trong l\u00f4 \u0111\u1ebb ch\u00ednh l\u00e0 t\u1ef7 l\u1ec7 % s\u1ebd ra c\u1ee7a con s\u1ed1 \u0111\u00f3 d\u1ef1a tr\u00ean s\u1ed1 l\u1ea7n trong m\u1ed7i k\u1ef3 quay.<\/p>\n
Ch\u00ednh nh\u1edd vi\u1ec7c t\u00ednh x\u00e1c su\u1ea5t ra l\u00f4 \u0111\u1ec3, ng\u01b0\u1eddi ch\u01a1i c\u00f3 th\u1ec3 n\u1eafm \u0111\u01b0\u1ee3c quy lu\u1eadt v\u1ec1 s\u1ed1. T\u1eeb \u0111\u00f3 ch\u1ecdn \u0111\u01b0\u1ee3c con s\u1ed1 n\u00e0o v\u1ec1 t\u1ef7 l\u1ec7 cao v\u00e0 s\u1ed1 n\u00e0o \u00edt v\u1ec1 nh\u1ea5t<\/p>\n
C\u1ee5 th\u1ec3 trong c\u00e1ch t\u00ednh x\u00e1c su\u1ea5t l\u00f4 \u0111\u1ec1 ch\u00ednh x\u00e1c ta c\u00f3 th\u1ec3 th\u1ea5y \u0111\u01b0\u1ee3c:<\/p>\n
Hi\u1ec7n nay vi\u1ec7c ch\u01a1i l\u00f4 \u0111\u1ec1 c\u1ea7n li\u00ean quan m\u1eadt thi\u1ebft \u0111\u1ebfn t\u00ednh x\u00e1c su\u1ea5t. N\u1ebfu b\u1ea1n kh\u00f4ng n\u1eafm \u0111\u01b0\u1ee3c ph\u01b0\u01a1ng ph\u00e1p v\u00e0 quy lu\u1eadt t\u00ednh n\u00e0y, kh\u1ea3 n\u0103ng t\u00ecm s\u1ed1 tr\u00fang g\u1ea7n nh\u01b0 l\u00e0 0%. C\u00f9ng t\u00ecm hi\u1ec3u ch\u00ednh x\u00e1c m\u1ed9t v\u00e0i c\u00e1ch t\u00ednh l\u00f4 \u0111\u1ec1 \u0111\u01b0\u1ee3c ti\u1ebft l\u1ed9 b\u1edfi chuy\u00ean gia l\u00f4 \u0111\u1ec1 nh\u01b0 sau:<\/p>\n
<\/p>\n
Ph\u01b0\u01a1ng ph\u00e1p n\u00e0y ngo\u00e0i vi\u1ec7c gi\u00fap b\u1ea1n t\u00ednh tr\u00fang 2 s\u1ed1 ho\u1eb7c 3 s\u1ed1. Ngo\u00e0i ra c\u00f2n nh\u1ea5n m\u1ea1nh v\u00e0o vi\u1ec7c tr\u00fang l\u00f4. B\u1edfi ph\u01b0\u01a1ng ph\u00e1p n\u00e0y ph\u1ea3i t\u00ednh 1 s\u1ed1 \u0111\u00fang th\u00ec m\u1edbi t\u00ednh ti\u1ebfp c\u00e1c s\u1ed1 sau.<\/p>\n
C\u00e1c ph\u00e9p t\u00ednh x\u00e1c su\u1ea5t \u0111\u1ec1u c\u00f3 s\u1ef1 li\u00ean k\u1ebft v\u1edbi nhau v\u00e0 l\u1ea5y s\u1ed1 \u0111\u1ea7u ti\u00ean l\u00e0m g\u1ed1c.<\/p>\n
V\u00ec trong l\u00f4 \u0111\u1ec3 ch\u1ec9 c\u00f3 hai k\u1ebft qu\u1ea3 duy nh\u1ea5t l\u00e0 tr\u00fang v\u00e0 tr\u01b0\u1ee3t. N\u00ean suy ra x\u00e1c su\u1ea5t tr\u00fang s\u1ebd l\u00e0 50%. N\u00ean suy ra t\u1ef7 l\u1ec7 50% \u0111\u01b0\u1ee3c \u00e1p d\u1ee5ng khi b\u1ea1n c\u00f3 hai l\u1ef1a ch\u1ecdn l\u00e0 ch\u1eb5n v\u00e0 l\u1ebb (t\u00e0i x\u1ec9u )<\/p>\n
\u0110\u1ea7u ti\u00ean n\u1ebfu mu\u1ed1n t\u00ednh x\u00e1c su\u1ea5t 1 l\u00f4 tr\u00fang, b\u1ea1n c\u1ea7n t\u00ednh l\u00f4 tr\u01b0\u1ee3t tr\u01b0\u1edbc ti\u00ean. Theo \u0111\u00f3, c\u00e1c s\u1ed1 c\u1ea7n s\u1ebd l\u00e0:<\/p>\n
T\u1ef7 l\u1ec7 t\u1ea1ch con l\u00f4 trong c\u00e1ch t\u00ednh \u0111\u1ea7u ti\u00ean n\u00e0y l\u00e0 99%. \u00c1p d\u1ee5ng cho 100 s\u1ed1 v\u00e0 b\u1ea1n ch\u1ec9 \u0111\u00e1nh c\u00f3 1 con duy nh\u1ea5t n\u00ean ch\u1eafc ch\u1eafn l\u00e0 tr\u01b0\u1ee3t):<\/p>\n
V\u00ed d\u1ee5 t\u1ec9 l\u1ec7 tr\u00fang l\u00f4 l\u00e0 1 (m\u1ee9c ho\u00e0n h\u1ea3o). Khi t\u1ea1ch c\u00e0ng nhi\u1ec1u, t\u1ef7 l\u1ec7 c\u00e0ng g\u1ea7n v\u1ec1 1 nh\u1ea5t. V\u00ec v\u1eady, t\u1ef7 l\u1ec7 \u0111\u1ea1t b\u1eb1ng 1 s\u1ebd tr\u00fang l\u00f4. V\u1eady ta c\u00f3 c\u00f4ng th\u1ee9c chung nh\u01b0 sau:<\/p>\n
[1 \u2013 (99% ^ 27)] * 100 ~ 23,7%<\/p>\n
**Ch\u00fa th\u00edch:<\/p>\n
Nh\u01b0 v\u1eady b\u1ea1n s\u1ebd b\u1ea1n t\u00ednh \u0111\u01b0\u1ee3c s\u1ed1 l\u00f4 nh\u01b0 sau:<\/p>\n
C\u00f4ng th\u1ee9c t\u1ed5ng<\/p>\n
[1 \u2013 [{100-n}% ^ 27)] * 100 (v\u1edbi n l\u00e0 s\u1ed1 con l\u00f4 b\u1ea1n \u0111\u00e1nh)<\/p>\n
Ng\u01b0\u1eddi ch\u01a1i n\u00ean k\u1ebft h\u1ee3p m\u1ed9t c\u00e1ch th\u1ed1ng k\u00ea \u0111\u1ec3 n\u00e2ng t\u1ef7 l\u1ec7 tr\u00fang cao. C\u00f4ng th\u1ee9c tr\u00ean \u0111\u00e2y ch\u1ec9 mang t\u00ednh ch\u1ea5t tham kh\u1ea3o v\u1ec1 m\u1eb7t l\u00fd thuy\u1ebft. Tr\u00ean th\u1ef1c t\u1ebf, b\u1ea1n ch\u1ec9 c\u1ea7n bi\u1ebft th\u1ed1ng k\u00ea l\u00e0 \u1ed5n. Ch\u00fang t\u00f4i \u0111ang l\u00fd gi\u1ea3i cho b\u1ea1n th\u1ea5y Do \u0111\u00f3, b\u1ea1n c\u00f3 th\u1ec3 b\u1ecf qua c\u00f4ng th\u1ee9c tr\u00ean m\u00e0 ch\u1ec9 xem \u0111\u00e2y l\u00e0 th\u00f4ng tin n\u00ean tham kh\u1ea3o.<\/p>\n H\u00ecnh th\u1ee9c t\u00ednh x\u00e1c su\u1ea5t cho l\u00f4 \u0111\u01b0\u1ee3c \u01b0\u1edbc t\u00ednh nh\u01b0 sau :<\/p>\n Nh\u01b0 v\u1eady h\u1ea7u h\u1ebft c\u00e1c l\u00f4 tr\u01b0\u1ee3c s\u1ebd gi\u1ea3m t\u1eeb t\u1eeb ho\u1eb7c kh\u00f4ng c\u1ea7n t\u00ednh. B\u1ea1n ch\u1ec9 c\u1ea7n d\u1ef1a theo 100 s\u1ed1 l\u00e0 100% (hay 1) cho d\u1ec5. Nh\u01b0 v\u1eady, n\u1ebfu tr\u01b0\u1ee3t n con th\u00ec m\u1ee9c \u0111\u1ed9 gi\u1ea3m t\u01b0\u01a1ng \u1ee9ng n l\u1ea7n. C\u1ee9 nh\u01b0 v\u1eady, b\u1ea1n t\u00ednh gi\u1ea3m xu\u1ed1ng s\u1ebd cho ra x\u00e1c su\u1ea5t t\u01b0\u01a1ng \u1ee9ng.<\/p>\n Trong c\u00e1ch t\u00ednh n\u00e0y, ng\u01b0\u1eddi ch\u01a1i c\u1ea7n chia ra 3 tr\u01b0\u1eddng h\u1ee3p ch\u00ednh. B\u1edfi l\u00fd do kho ch\u01a1i l\u00f4 xi\u00ean l\u00e0 ch\u01a1i c\u00f9ng l\u00fac 2 con s\u1ed1. N\u00ean c\u00f9ng l\u00fac c\u1ea7n t\u00ednh x\u00e1c su\u1ea5t cho hai s\u1ed1 \u0111\u00f3 trong tr\u01b0\u1eddng h\u1ee3p:<\/p>\n T\u1eeb tr\u01b0\u1eddng h\u1ee3p chia ra tr\u00ean \u0111\u00e2y, ta c\u00f3 c\u00f4ng th\u1ee9c t\u00ednh l\u00f4 xi\u00ean tr\u00fang = 100 \u2013 94 (%). T\u01b0\u01a1ng \u1ee9ng v\u1edbi m\u1ee9c t\u1ef7 l\u1ec7 x\u1ea5p x\u1ec9 b\u1eb1ng 6<\/p>\n V\u1eady 6 % v\u1ed1n kh\u00f4ng ph\u1ea3i l\u00e0 t\u1ef7 l\u1ec7 t\u01b0\u01a1ng \u0111\u1ed1i nh\u1ecf. N\u00ean suy ra gi\u00e1 tr\u1ecb tr\u00fang l\u00f4 xi\u00ean 2 th\u01b0\u1eddng s\u1ebd cao. T\u1eeb \u0111\u00e2y b\u1ea1n c\u00f3 th\u1ec3 ch\u1ecdn 2 s\u1ed1 c\u00f3 x\u00e1c su\u1ea5t hay xu\u1ea5t hi\u1ec7n nh\u1ea5t v\u00e0 gh\u00e9p th\u00e0nh 1 xi\u00ean v\u00e0 \u0111\u00e1nh k\u1ebft h\u1ee3p<\/p>\n T\u01b0\u01a1ng t\u01b0 c\u00e1ch t\u00ednh x\u00e1c su\u1ea5t l\u00f4 xi\u00ean 2. Nh\u01b0ng v\u1edbi h\u00ecnh th\u1ee9c t\u00ednh x\u00e1c su\u1ea5t tr\u00fang l\u00f4 xi\u00ean 3 th\u01b0\u1eddng kh\u00f4ng tr\u00fang nhi\u1ec1u. N\u00ean c\u1ea7n \u00e1p d\u1ee5ng c\u00e1ch t\u00ednh ph\u1ee9c t\u1ea1p h\u01a1n. T\u1ea1i \u0111\u00e2y c\u0169ng chia th\u00e0nh 3 tr\u01b0\u1eddng h\u1ee3p t\u01b0\u01a1ng t\u1ef1:<\/p>\n V\u1eady x\u00e1c su\u1ea5t kh\u00f4ng tr\u00fang l\u00f4 xi\u00ean 3 l\u00e0:<\/p>\n A + B + C = (0,97^27) + (3 * 0,2376 * 0,98^27) + (3 * 0,058 * 0,99^27) = 0,9851 g\u1ea7n b\u1eb1ng 98,51%<\/p>\n Ta c\u00f3 c\u00f4ng th\u1ee9c t\u00ednh l\u00f4 xi\u00ean 3 tr\u00fang = 100 \u2013 98,51 = 1,49 (%).<\/p>\n V\u1eady c\u00f3 k\u1ebft lu\u1eadn g\u00ec? N\u1ebfu ch\u01a1i l\u00f4 xi\u00ean 3 kh\u1ea3 n\u0103ng tr\u00fang c\u1ee7a b\u1ea1n r\u1ea5t th\u1ea5p. N\u1ebfu ch\u01a1i 100 l\u1ea7n s\u1ebd ch\u1ec9 tr\u00fang 1 l\u1ea7n. Do \u0111\u00f3, s\u1ef1 l\u1ef1a ch\u1ecdn l\u00f4 xi\u00ean 3 lu\u00f4n c\u1ea7n c\u00e2n nh\u1eafc. Ch\u1ec9 n\u00ean ch\u1ecdn khi b\u1ea1n l\u00e0 ng\u01b0\u1eddi c\u00f3 kh\u1ea3 n\u0103ng t\u00ednh to\u00e1n t\u1ed1t v\u00e0 s\u1ef1 may m\u1eafn th\u00ec kh\u1ea3 n\u0103ng th\u1eafng m\u1edbi cao<\/p>\n T\u1ea5t nhi\u00ean c\u00e0ng ch\u01a1i v\u1edbi l\u00f4 xi\u00ean 4 s\u1ebd c\u00e0ng kh\u00f4ng th\u1ec3 mang v\u1ec1 chi\u1ebfn th\u1eafng cho b\u1ea1n. V\u00ec v\u1eady, n\u1ebfu tr\u00fang \u0111\u01b0\u1ee3c 1 l\u00f4 xi\u00ean 4 t\u1ee9c l\u00e0 b\u1ea1n r\u1ea5t may m\u1eafn, th\u1eadm ch\u00ed may h\u01a1n c\u1ea3 ng\u01b0\u1eddi tr\u00fang Vietlott<\/p>\n <\/p>\n N\u1ebfu b\u1ea1n \u0111ang c\u1ea7n \u0111\u00e1nh l\u00f4 trong nhi\u1ec1u ng\u00e0y li\u00ean ti\u1ebfp. C\u1ea7n \u01b0\u1edbc l\u01b0\u1ee3ng ch\u00ednh x\u00e1c x\u00e1c su\u1ea5t tr\u00fang 1 l\u1ea7n \u0111\u1ea1t g\u1ea7n 100%.<\/p>\n V\u1edbi d\u00e2n ch\u01a1i l\u00f4 \u0111\u1ec1 \u0111\u1ec1 th\u00ec c\u00f4ng th\u1ee9c t\u00ednh x\u00e1c su\u1ea5t l\u00f4 \u0111\u1ec1 n\u00e0y thu\u1ed9c v\u1ec1 ph\u1ea1m tr\u00f9 t\u00ednh to\u00e1n theo to\u00e1n h\u1ecdc. C\u1ee5 th\u1ec3:<\/p>\n G\u1ecdi A ch\u00ednh l\u00e0 bi\u1ebfn c\u1ed1 trong n ng\u00e0y v\u00e0 \u00e1m ch\u1ec9 k\u1ebft qu\u1ea3 ta kh\u00f4ng tr\u00fang l\u1ea7n n\u00e0o<\/p>\n G\u1ecdi B l\u00e0 bi\u1ebfn c\u1ed1 mang v\u1ec1 kh\u1ea3 n\u0103ng tr\u00fang cho ng\u01b0\u1eddi ch\u01a1i \u00edt nh\u1ea5t 1 l\u1ea7n<\/p>\n X\u00e1c su\u1ea5t \u0111\u1ec3 tr\u01b0\u1ee3t trong 27 l\u00f4 trong d\u00e0n l\u00f4 m\u1ed9t ng\u00e0y l\u00e0 0,99^27<\/p>\n P(A) t\u01b0\u01a1ng \u1ee9ng 0,99^27n<\/p>\n P(B) t\u01b0\u01a1ng \u1ee9ng 1\u2212P(A) = 1\u22120,99^27n<\/p>\n P(B) > m \u21d4 0,99^27n < 1\u2212m \u21d4 n. V\u1eady log0,99(1\u2212m)\/27<\/p>\n Trong \u0111\u00f3, m = 0,999 (t\u1ef7 l\u1ec7 99,99%), ta c\u00f3: n > 33,9. T\u1ee9c l\u00e0 trong th\u1eddi gian 34 ng\u00e0y, x\u00e1c su\u1ea5t ta \u0111\u00e1nh tr\u00fang cao nh\u1ea5t l\u00e0 99,99%. Ngh\u0129a l\u00e0 ng\u01b0\u1eddi ch\u01a1i c\u1ea7n c\u1ed1 g\u1eafng nu\u00f4i \u00edt nh\u1ea5t 1 con l\u00f4 n\u00e0o \u0111\u00f3. Sau th\u1eddi gian 34 ng\u00e0y th\u00ec t\u1ef7 l\u1ec7 th\u1eafng l\u00f4 n\u00e0y ch\u1eafc ch\u1eafn s\u1ebd c\u1ef1c kh\u1ee7ng. \u0110i\u1ec1u quan tr\u1ecdng l\u00e0 b\u1ea1n c\u1ea7n \u0111\u1ea3m b\u1ea3o \u0111\u1ee7 s\u1ed1 v\u1ed1n cho \u0111\u1ebfn kh con l\u00f4 n\u00e0y v\u1ec1 hay kh\u00f4ng.<\/p>\n Tr\u00ean \u0111\u00e2y l\u00e0 chia s\u1ebb v\u1ec1 c\u00e2u h\u1ecfi t\u1ea1i sao c\u1ea7n t\u00ednh x\u00e1c su\u1ea5t khi ch\u01a1i l\u00f4 \u0111\u1ec1. \u0110\u01b0\u1ee3c bi\u1ebft, khi b\u1ea1n n\u1eafm \u0111\u01b0\u1ee3c c\u00e1ch t\u00ednh n\u00e0y, b\u1ea1n s\u1ebd bi\u1ebft \u0111\u01b0\u1ee3c kh\u1ea3 n\u0103ng v\u1ec1 s\u1ed1 c\u00f3 t\u1ef7 l\u1ec7 bao nhi\u00eau. T\u1eeb \u0111\u00f3 k\u1ebft h\u1ee3p v\u1edbi nhi\u1ec1u ph\u01b0\u01a1ng ph\u00e1p kh\u00e1c nhau nh\u01b0 th\u1ed1ng k\u00ea l\u00f4 t\u00f4 \u0111\u1ec3 t\u00ecm ra con s\u1ed1 may m\u1eafn<\/p>\n <\/p>\n<\/div>\nC\u00e1ch t\u00ednh x\u00e1c su\u1ea5t theo h\u00ecnh th\u1ee9c l\u00f4<\/h3>\n
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C\u00e1ch t\u00ednh x\u00e1c su\u1ea5t tr\u00fang l\u00f4 xi\u00ean 2<\/h3>\n
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C\u00e1ch t\u00ednh x\u00e1c su\u1ea5t tr\u00fang l\u00f4 xi\u00ean 3<\/h3>\n
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M\u1ed9t b\u00e0i to\u00e1n nu\u00f4i l\u00f4 \u0111i\u1ec3n h\u00ecnh<\/h3>\n
\u0111\u00e1nh l\u00e0 tr\u00fang c\u1ea7u mi\u1ec1n b\u1eafc<\/h3>\n