T\u00ednh x\u00e1c su\u1ea5t \u0111\u1ec3 tr\u00fang 1 con l\u00f4 \u0111\u1ec1<\/b><\/span><\/h2>\n\u0110\u00e2y l\u00e0 <\/span>c\u00e1ch t\u00ednh x\u00e1c su\u1ea5t trong l\u00f4<\/b> \u0111\u1ec1 \u0111\u01a1n gi\u1ea3n nh\u1ea5t b\u1edfi b\u1ea1n ch\u1ec9 c\u1ea7n t\u00ecm cho 1 con l\u00f4 ho\u1eb7c 1 con \u0111\u1ec1. C\u1ee5 th\u1ec3:<\/span><\/p>\n\n- X\u00e1c su\u1ea5t tr\u00fang 1 con l\u00f4 = 1 \u2013 (x\u00e1c su\u1ea5t t\u1ea1ch l\u00f4) = 1 \u2013 0,99^27 (quay 27 s\u1ed1 \u0111\u1ec1u t\u1ea1ch) = 23,7657 %. Trong tr\u01b0\u1eddng h\u1ee3p n\u00e0y s\u1ebd kh\u00f4ng t\u00ednh \u0111\u1ebfn tr\u00fang 1 nh\u00e1y, 2 nh\u00e1y ho\u1eb7c 3 nh\u00e1y.<\/span><\/li>\n
- \u0110\u1ed1i v\u1edbi 1 con \u0111\u1ec1 th\u00ec kh\u00f4ng c\u1ea7n d\u00f9ng \u0111\u1ebfn <\/span>thu\u1eadt to\u00e1n t\u00ednh l\u00f4<\/i><\/b> \u0111\u1ec1 b\u1edfi x\u00e1c su\u1ea5t \u0111\u1ec3 tr\u00fang m\u1ed9t con lu\u00f4n l\u00e0 1\/100 = 1%.<\/span><\/li>\n<\/ul>\n
C\u00f3 th\u1ec3 th\u1ea5y, x\u00e1c su\u1ea5t \u0111\u1ec3 tr\u00fang 1 con l\u00f4 kh\u00e1 cao nh\u01b0ng \u0111i\u1ec1u \u0111\u00f3 c\u0169ng \u0111\u1ed3ng ngh\u0129a t\u1ef7 l\u01b0\u1ee3c c\u01b0\u1ee3c cho c\u00e1ch \u0111\u00e1nh n\u00e0y kh\u00e1 th\u1ea5p.<\/span><\/p>\nT\u00ednh x\u00e1c su\u1ea5t tr\u00fang l\u00f4 xi\u00ean<\/b><\/span><\/h2>\nTrong c\u00e1ch t\u00ednh <\/span>x\u00e1c su\u1ea5t tr\u00fang l\u00f4 xi\u00ean<\/i><\/b>, ch\u00fang ta s\u1ebd quan t\u00e2m t\u1edbi l\u00f4 xi\u00ean 2 v\u00e0 l\u00f4 xi\u00ean nh\u01b0 sau:<\/span><\/p>\nX\u00e1c su\u1ea5t tr\u00fang l\u00f4 xi\u00ean 2 <\/i><\/b><\/span><\/h3>\nCh\u00fang ta s\u1ebd chia ra l\u00e0m 2 tr\u01b0\u1eddng h\u1ee3p l\u00e0 kh\u00f4ng v\u1ec1 c\u1ea3 2 con v\u00e0 ch\u1ec9 v\u1ec1 1 con:<\/span><\/p>\n\n- X\u00e1c su\u1ea5t kh\u00f4ng v\u1ec1 2 con = 0,98^27 = 57,957 %.<\/span><\/li>\n
- X\u00e1c su\u1ea5t v\u1ec1 1 con duy nh\u1ea5t th\u00ec l\u1ea1i chia ra 2 tr\u01b0\u1eddng h\u1ee3p l\u00e0 con th\u1ee9 nh\u1ea5t v\u1ec1 th\u00ec con th\u1ee9 2 kh\u00f4ng v\u1ec1 v\u00e0 ng\u01b0\u1ee3c l\u1ea1i. Do \u0111\u00f3, x\u00e1c su\u1ea5t ch\u1ec9 v\u1ec1 1 con = 2 * x\u00e1c su\u1ea5t tr\u00fang 1 con l\u00f4 * x\u00e1c su\u1ea5t kh\u00f4ng tr\u00fang con n\u00e0o = 2 * 23,7657% * 0,99^27 = 36,238%.<\/span><\/li>\n
- X\u00e1c su\u1ea5t kh\u00f4ng tr\u00fang l\u00f4 xi\u00ean 2 = X\u00e1c su\u1ea5t kh\u00f4ng v\u1ec1 2 con + x\u00e1c su\u1ea5t v\u1ec1 1 con duy nh\u1ea5t = 57,957% + 36,238% = 94,195%. T\u1eeb \u0111\u00f3 suy ra x\u00e1c su\u1ea5t tr\u00fang l\u00f4 xi\u00ean 2 = 100% \u2013 x\u00e1c su\u1ea5t kh\u00f4ng tr\u00fang l\u00f4 = 100 \u2013 94,195% = 5,8%.<\/span><\/li>\n<\/ul>\n
X\u00e1c su\u1ea5t tr\u00fang l\u00f4 xi\u00ean 3<\/i><\/b><\/span><\/h3>\n<\/p>\n
C\u00e1ch t\u00ednh n\u00e0y c\u0169ng t\u01b0\u01a1ng t\u1ef1 v\u1edbi l\u00f4 xi\u00ean 2 nh\u01b0ng tr\u01b0\u1eddng h\u1ee3p kh\u00f4ng tr\u00fang l\u00f4 xi\u00ean 3 s\u1ebd ph\u1ee9c t\u1ea1p h\u01a1n m\u1ed9t ch\u00fat, c\u00f2n l\u1ea1i kh\u00f4ng kh\u00e1c l\u00f4 xi\u00ean 2.<\/span><\/p>\nX\u00e1c su\u1ea5t kh\u00f4ng tr\u00fang l\u00f4 xi\u00ean 3 = X\u00e1c su\u1ea5t \u0111\u00e1nh tr\u01b0\u1ee3t c\u1ea3 3 con + x\u00e1c su\u1ea5t ch\u1ec9 v\u1ec1 1 trong 3 con + x\u00e1c su\u1ea5t ch\u1ec9 v\u1ec1 2 con = 0,97^27 + 3 * 23,7675% * 0,98^27 + 3 * 5,8% * 0,99^27 = 98,527%.<\/span><\/p>\nT\u1eeb \u0111\u00f3, x\u00e1c su\u1ea5t tr\u00fang l\u00f4 xi\u00ean 3 = 100% \u2013 98,527% = 1,473%. C\u00e1ch \u0111\u00e1nh n\u00e0y d\u1ec5 tr\u00fang \u0111\u1ec1 h\u01a1n m\u1ed9t ch\u00fat.<\/span><\/p>\nM\u1ed9t s\u1ed1 c\u00e1ch t\u00ednh x\u00e1c su\u1ea5t kh\u00e1c<\/b><\/span><\/h2>\nNgo\u00e0i c\u00e1c c\u00e1ch \u0111\u00e1nh l\u00f4 tr\u00ean th\u00ec ng\u01b0\u1eddi ch\u01a1i th\u01b0\u1eddng \u0111\u00e1nh l\u00f4 3 ch\u00e2n v\u00e0 \u0111\u00e1nh \u0111\u1ec1 3 ch\u00e2n. Hai c\u00f4ng th\u1ee9c t\u00ednh n\u00e0y c\u0169ng kh\u00e1 \u0111\u01a1n gi\u1ea3n, kh\u00f4ng ph\u1ee9c t\u1ea1p nh\u01b0 c\u00e1c c\u00f4ng th\u1ee9c tr\u00ean. C\u1ee5 th\u1ec3:<\/span><\/p>\n\n- X\u00e1c su\u1ea5t tr\u00fang \u0111\u1ec1 3 ch\u00e2n c\u0169ng kh\u00f4ng c\u1ea7n ph\u1ea3i t\u00ednh v\u00ec ai c\u0169ng bi\u1ebft x\u00e1c su\u1ea5t \u0111\u1ec3 tr\u00fang \u0111\u1ec1 3 ch\u00e2n l\u00e0 1\/1000 = 0,1 %.<\/span><\/li>\n
- X\u00e1c su\u1ea5t tr\u00fang l\u00f4 3 ch\u00e2n = 1 \u2013 x\u00e1c su\u1ea5t tr\u01b0\u1ee3t l\u00f4 3 ch\u00e2n = 1 \u2013 0,999^27 = 2,67%.<\/span><\/li>\n<\/ul>\n
L\u01b0u \u00fd, \u0111\u1ec3 gia t\u0103ng t\u1ef7 l\u1ec7 chi\u1ebfn th\u1eafng th\u00ec ng\u01b0\u1eddi ch\u01a1i n\u00ean k\u1ebft h\u1ee3p \u00e1p d\u1ee5ng c\u00e1ch t\u00ednh x\u00e1c su\u1ea5t c\u00f9ng v\u1edbi c\u00e1c b\u00ed k\u00edp \u0111\u00e1nh l\u00f4 hi\u1ec7u qu\u1ea3. <\/span><\/p>\nHy v\u1ecdng th\u00f4ng qua c\u00e1c <\/span>c\u00e1ch t\u00ednh x\u00e1c su\u1ea5t trong l\u00f4 \u0111\u1ec1<\/b> tr\u00ean, ng\u01b0\u1eddi ch\u01a1i s\u1ebd \u0111\u01b0a ra \u0111\u01b0\u1ee3c ph\u00e1n \u0111o\u00e1n ch\u00ednh x\u00e1c nh\u1ea5t \u0111\u1ec3 gi\u00e0nh chi\u1ebfn th\u1eafng. Ch\u00fac b\u1ea1n th\u00e0nh c\u00f4ng trong s\u1ef1 nghi\u1ec7p l\u00f4 \u0111\u1ec1 c\u1ee7a m\u00ecnh nh\u00e9!<\/span><\/p>\n<\/div>\n<\/article>\n<\/section>\n<\/p>\n
C\u1ea6U \u0110\u1eb8P CAO C\u1ea4P<\/span><\/strong><\/h4>\n
- \n
- X\u00e1c su\u1ea5t tr\u00fang 1 con l\u00f4 = 1 \u2013 (x\u00e1c su\u1ea5t t\u1ea1ch l\u00f4) = 1 \u2013 0,99^27 (quay 27 s\u1ed1 \u0111\u1ec1u t\u1ea1ch) = 23,7657 %. Trong tr\u01b0\u1eddng h\u1ee3p n\u00e0y s\u1ebd kh\u00f4ng t\u00ednh \u0111\u1ebfn tr\u00fang 1 nh\u00e1y, 2 nh\u00e1y ho\u1eb7c 3 nh\u00e1y.<\/span><\/li>\n
- \u0110\u1ed1i v\u1edbi 1 con \u0111\u1ec1 th\u00ec kh\u00f4ng c\u1ea7n d\u00f9ng \u0111\u1ebfn <\/span>thu\u1eadt to\u00e1n t\u00ednh l\u00f4<\/i><\/b> \u0111\u1ec1 b\u1edfi x\u00e1c su\u1ea5t \u0111\u1ec3 tr\u00fang m\u1ed9t con lu\u00f4n l\u00e0 1\/100 = 1%.<\/span><\/li>\n<\/ul>\n
C\u00f3 th\u1ec3 th\u1ea5y, x\u00e1c su\u1ea5t \u0111\u1ec3 tr\u00fang 1 con l\u00f4 kh\u00e1 cao nh\u01b0ng \u0111i\u1ec1u \u0111\u00f3 c\u0169ng \u0111\u1ed3ng ngh\u0129a t\u1ef7 l\u01b0\u1ee3c c\u01b0\u1ee3c cho c\u00e1ch \u0111\u00e1nh n\u00e0y kh\u00e1 th\u1ea5p.<\/span><\/p>\n
T\u00ednh x\u00e1c su\u1ea5t tr\u00fang l\u00f4 xi\u00ean<\/b><\/span><\/h2>\n
Trong c\u00e1ch t\u00ednh <\/span>x\u00e1c su\u1ea5t tr\u00fang l\u00f4 xi\u00ean<\/i><\/b>, ch\u00fang ta s\u1ebd quan t\u00e2m t\u1edbi l\u00f4 xi\u00ean 2 v\u00e0 l\u00f4 xi\u00ean nh\u01b0 sau:<\/span><\/p>\n
X\u00e1c su\u1ea5t tr\u00fang l\u00f4 xi\u00ean 2 <\/i><\/b><\/span><\/h3>\n
Ch\u00fang ta s\u1ebd chia ra l\u00e0m 2 tr\u01b0\u1eddng h\u1ee3p l\u00e0 kh\u00f4ng v\u1ec1 c\u1ea3 2 con v\u00e0 ch\u1ec9 v\u1ec1 1 con:<\/span><\/p>\n
- \n
- X\u00e1c su\u1ea5t kh\u00f4ng v\u1ec1 2 con = 0,98^27 = 57,957 %.<\/span><\/li>\n
- X\u00e1c su\u1ea5t v\u1ec1 1 con duy nh\u1ea5t th\u00ec l\u1ea1i chia ra 2 tr\u01b0\u1eddng h\u1ee3p l\u00e0 con th\u1ee9 nh\u1ea5t v\u1ec1 th\u00ec con th\u1ee9 2 kh\u00f4ng v\u1ec1 v\u00e0 ng\u01b0\u1ee3c l\u1ea1i. Do \u0111\u00f3, x\u00e1c su\u1ea5t ch\u1ec9 v\u1ec1 1 con = 2 * x\u00e1c su\u1ea5t tr\u00fang 1 con l\u00f4 * x\u00e1c su\u1ea5t kh\u00f4ng tr\u00fang con n\u00e0o = 2 * 23,7657% * 0,99^27 = 36,238%.<\/span><\/li>\n
- X\u00e1c su\u1ea5t kh\u00f4ng tr\u00fang l\u00f4 xi\u00ean 2 = X\u00e1c su\u1ea5t kh\u00f4ng v\u1ec1 2 con + x\u00e1c su\u1ea5t v\u1ec1 1 con duy nh\u1ea5t = 57,957% + 36,238% = 94,195%. T\u1eeb \u0111\u00f3 suy ra x\u00e1c su\u1ea5t tr\u00fang l\u00f4 xi\u00ean 2 = 100% \u2013 x\u00e1c su\u1ea5t kh\u00f4ng tr\u00fang l\u00f4 = 100 \u2013 94,195% = 5,8%.<\/span><\/li>\n<\/ul>\n
X\u00e1c su\u1ea5t tr\u00fang l\u00f4 xi\u00ean 3<\/i><\/b><\/span><\/h3>\n
<\/p>\n
C\u00e1ch t\u00ednh n\u00e0y c\u0169ng t\u01b0\u01a1ng t\u1ef1 v\u1edbi l\u00f4 xi\u00ean 2 nh\u01b0ng tr\u01b0\u1eddng h\u1ee3p kh\u00f4ng tr\u00fang l\u00f4 xi\u00ean 3 s\u1ebd ph\u1ee9c t\u1ea1p h\u01a1n m\u1ed9t ch\u00fat, c\u00f2n l\u1ea1i kh\u00f4ng kh\u00e1c l\u00f4 xi\u00ean 2.<\/span><\/p>\n
X\u00e1c su\u1ea5t kh\u00f4ng tr\u00fang l\u00f4 xi\u00ean 3 = X\u00e1c su\u1ea5t \u0111\u00e1nh tr\u01b0\u1ee3t c\u1ea3 3 con + x\u00e1c su\u1ea5t ch\u1ec9 v\u1ec1 1 trong 3 con + x\u00e1c su\u1ea5t ch\u1ec9 v\u1ec1 2 con = 0,97^27 + 3 * 23,7675% * 0,98^27 + 3 * 5,8% * 0,99^27 = 98,527%.<\/span><\/p>\n
T\u1eeb \u0111\u00f3, x\u00e1c su\u1ea5t tr\u00fang l\u00f4 xi\u00ean 3 = 100% \u2013 98,527% = 1,473%. C\u00e1ch \u0111\u00e1nh n\u00e0y d\u1ec5 tr\u00fang \u0111\u1ec1 h\u01a1n m\u1ed9t ch\u00fat.<\/span><\/p>\n
M\u1ed9t s\u1ed1 c\u00e1ch t\u00ednh x\u00e1c su\u1ea5t kh\u00e1c<\/b><\/span><\/h2>\n
Ngo\u00e0i c\u00e1c c\u00e1ch \u0111\u00e1nh l\u00f4 tr\u00ean th\u00ec ng\u01b0\u1eddi ch\u01a1i th\u01b0\u1eddng \u0111\u00e1nh l\u00f4 3 ch\u00e2n v\u00e0 \u0111\u00e1nh \u0111\u1ec1 3 ch\u00e2n. Hai c\u00f4ng th\u1ee9c t\u00ednh n\u00e0y c\u0169ng kh\u00e1 \u0111\u01a1n gi\u1ea3n, kh\u00f4ng ph\u1ee9c t\u1ea1p nh\u01b0 c\u00e1c c\u00f4ng th\u1ee9c tr\u00ean. C\u1ee5 th\u1ec3:<\/span><\/p>\n
- \n
- X\u00e1c su\u1ea5t tr\u00fang \u0111\u1ec1 3 ch\u00e2n c\u0169ng kh\u00f4ng c\u1ea7n ph\u1ea3i t\u00ednh v\u00ec ai c\u0169ng bi\u1ebft x\u00e1c su\u1ea5t \u0111\u1ec3 tr\u00fang \u0111\u1ec1 3 ch\u00e2n l\u00e0 1\/1000 = 0,1 %.<\/span><\/li>\n
- X\u00e1c su\u1ea5t tr\u00fang l\u00f4 3 ch\u00e2n = 1 \u2013 x\u00e1c su\u1ea5t tr\u01b0\u1ee3t l\u00f4 3 ch\u00e2n = 1 \u2013 0,999^27 = 2,67%.<\/span><\/li>\n<\/ul>\n
L\u01b0u \u00fd, \u0111\u1ec3 gia t\u0103ng t\u1ef7 l\u1ec7 chi\u1ebfn th\u1eafng th\u00ec ng\u01b0\u1eddi ch\u01a1i n\u00ean k\u1ebft h\u1ee3p \u00e1p d\u1ee5ng c\u00e1ch t\u00ednh x\u00e1c su\u1ea5t c\u00f9ng v\u1edbi c\u00e1c b\u00ed k\u00edp \u0111\u00e1nh l\u00f4 hi\u1ec7u qu\u1ea3. <\/span><\/p>\n
Hy v\u1ecdng th\u00f4ng qua c\u00e1c <\/span>c\u00e1ch t\u00ednh x\u00e1c su\u1ea5t trong l\u00f4 \u0111\u1ec1<\/b> tr\u00ean, ng\u01b0\u1eddi ch\u01a1i s\u1ebd \u0111\u01b0a ra \u0111\u01b0\u1ee3c ph\u00e1n \u0111o\u00e1n ch\u00ednh x\u00e1c nh\u1ea5t \u0111\u1ec3 gi\u00e0nh chi\u1ebfn th\u1eafng. Ch\u00fac b\u1ea1n th\u00e0nh c\u00f4ng trong s\u1ef1 nghi\u1ec7p l\u00f4 \u0111\u1ec1 c\u1ee7a m\u00ecnh nh\u00e9!<\/span><\/p>\n<\/div>\n<\/article>\n<\/section>\n
<\/p>\n
C\u1ea6U \u0110\u1eb8P CAO C\u1ea4P<\/span><\/strong><\/h4>\n
- X\u00e1c su\u1ea5t tr\u00fang l\u00f4 3 ch\u00e2n = 1 \u2013 x\u00e1c su\u1ea5t tr\u01b0\u1ee3t l\u00f4 3 ch\u00e2n = 1 \u2013 0,999^27 = 2,67%.<\/span><\/li>\n<\/ul>\n
- X\u00e1c su\u1ea5t v\u1ec1 1 con duy nh\u1ea5t th\u00ec l\u1ea1i chia ra 2 tr\u01b0\u1eddng h\u1ee3p l\u00e0 con th\u1ee9 nh\u1ea5t v\u1ec1 th\u00ec con th\u1ee9 2 kh\u00f4ng v\u1ec1 v\u00e0 ng\u01b0\u1ee3c l\u1ea1i. Do \u0111\u00f3, x\u00e1c su\u1ea5t ch\u1ec9 v\u1ec1 1 con = 2 * x\u00e1c su\u1ea5t tr\u00fang 1 con l\u00f4 * x\u00e1c su\u1ea5t kh\u00f4ng tr\u00fang con n\u00e0o = 2 * 23,7657% * 0,99^27 = 36,238%.<\/span><\/li>\n
- \u0110\u1ed1i v\u1edbi 1 con \u0111\u1ec1 th\u00ec kh\u00f4ng c\u1ea7n d\u00f9ng \u0111\u1ebfn <\/span>thu\u1eadt to\u00e1n t\u00ednh l\u00f4<\/i><\/b> \u0111\u1ec1 b\u1edfi x\u00e1c su\u1ea5t \u0111\u1ec3 tr\u00fang m\u1ed9t con lu\u00f4n l\u00e0 1\/100 = 1%.<\/span><\/li>\n<\/ul>\n