T\u00ednh x\u00e1c su\u1ea5t l\u00f4 \u0111\u1ec1 xsmb ch\u00ednh x\u00e1c 100% t\u00ednh nhanh, th\u1eafng l\u1edbn<\/figcaption><\/figure>\nCh\u00fang ta ch\u1ec9 c\u00f3 duy nh\u1ea5t 1 c\u00e1ch t\u00ednh x\u00e1c su\u1ea5t l\u00f4 \u0111\u1ec1. Ch\u00ednh v\u00ec v\u1eady, anh em xem b\u00e0i vi\u1ebft n\u00e0y l\u00e0 \u0111\u1ea7y \u0111\u1ee7 cho v\u1ea5n \u0111\u1ec1 \u0111ang quan t\u00e2m.<\/p>\n
T\u1ea1i sao ph\u1ea3i t\u00ednh x\u00e1c su\u1ea5t l\u00f4 \u0111\u1ec1 ?<\/h3>\n
C\u0169ng nh\u01b0 thi \u0111\u1ea1i h\u1ecdc c\u00f3 t\u1ef7 l\u1ec7 ch\u1ecdi th\u00ec \u0111\u00e1nh l\u00f4 \u0111\u1ec1 c\u0169ng v\u1eady. V\u00ed d\u1ee5 anh em \u0111\u00e1nh 1 con \u0111\u1ec1 th\u00ec anh em \u0111ang c\u00f3 t\u1ec9 l\u1ec7 ch\u1ecdi l\u00e0 1\/10. Anh em \u0111\u00e1nh 1 con l\u00f4 mi\u1ec1n B\u1eafc s\u1ebd c\u00f3 t\u1ef7 l\u1ec7 ch\u1ecdi l\u00e0 1\/27.<\/p>\n
Nh\u01b0 v\u1eady, t\u1ef7 l\u1ec7 ch\u1ecdi n\u00e0y ch\u00ednh x\u00e1c x\u00e1c su\u1ea5t l\u00f4 \u0111\u1ec1 m\u00e0 anh em ph\u1ea3i \u0111\u1ed1i m\u1eb7t. T\u1eeb nh\u1eefng c\u00e1ch t\u00ednh \u0111\u00f3 ch\u00fang ta s\u1ebd t\u00ednh \u0111\u01b0\u1ee3c kh\u1ea3 n\u0103ng bao nhi\u00eau % s\u1ebd tr\u00fang v\u00e0 bao nhi\u00eau % s\u1ebd tr\u01b0\u1ee3t l\u00f4 \u0111\u1ec1.<\/p>\n
Sau khi \u0111\u00e3 t\u00ednh to\u00e1n c\u00e1c ki\u1ec3u. Anh em s\u1ebd c\u00f3 \u0111\u01b0\u1ee3c nh\u1eefng th\u00f4ng tin ch\u00ednh x\u00e1c. Nh\u1eb1m h\u1ed7 tr\u1ee3 t\u1ed1t nh\u1ea5t trong vi\u1ec7c \u0111\u00e1nh l\u00f4 \u0111\u1ec1 h\u00e0ng ng\u00e0y. Anh em s\u1ebd kh\u00f4ng c\u00f2n ph\u1ea3i suy ngh\u0129 v\u1ec1 v\u1ea5n \u0111\u1ec1 n\u00e0y n\u1eefa.<\/p>\n
1. C\u00e1ch t\u00ednh x\u00e1c su\u1ea5t tr\u00fang 1 con L\u00f4<\/h3>\n
Trong c\u00e1ch t\u00ednh x\u00e1c su\u1ea5t l\u00f4 \u0111\u1ec1 n\u00e0y ch\u00fang ta kh\u00f4ng t\u00ednh \u0111\u1ebfn vi\u1ec7c tr\u00fang m\u1ea5y nh\u00e1y. Ch\u00fang ta ch\u1ec9 t\u00ednh l\u00e0 Tr\u00fang L\u00f4.<\/p>\n
Nh\u01b0 v\u1eady, x\u00e1c su\u1ea5t \u0111\u1ec3 \u0111\u00e1nh tr\u00fang l\u00f4 = 1 \u2013 (x\u00e1c xu\u1ea5t t\u1ea1ch l\u00f4)<\/em> = 1 \u2013 0,99^27 (quay 27 s\u1ed1 \u0111\u1ec1u tr\u01b0\u1ee3t)<\/em> = 23,7657%<\/strong>.<\/p>\n2. C\u00e1ch t\u00ednh x\u00e1c su\u1ea5t tr\u00fang 1 con \u0110\u1ec1<\/h3>\n
V\u1ec1 x\u00e1c su\u1ea5t tr\u00fang \u0111\u1ec1 th\u00ec kh\u00f4ng c\u00f3 g\u00ec ph\u1ea3i t\u00ednh to\u00e1n nhi\u1ec1u c\u1ea3. B\u1edfi v\u00ec ai c\u0169ng bi\u1ebft x\u00e1c su\u1ea5t \u0111\u1ec3 \u0111\u00e1nh trung 1 con \u0111\u1ec1 l\u00e0 1\/100 = 1%<\/strong>.<\/p>\n3. C\u00e1ch t\u00ednh x\u00e1c su\u1ea5t tr\u00fang l\u00f4 xi\u00ean 2<\/h3>\n
L\u00f4 xi\u00ean 2 l\u00e0 ch\u00fang ta \u0111\u00e1nh 2 con. Nh\u01b0 v\u1eady ch\u00fang ta s\u1ebd c\u00f3 2 tr\u01b0\u1eddng h\u1ee3p trong c\u00e1ch t\u00ednh x\u00e1c xu\u1ea5t l\u00f4 \u0111\u1ec1 l\u00e0: Kh\u00f4ng v\u1ec1 c\u1ea3 2 con<\/strong> v\u00e0 ch\u1ec9 v\u1ec1 1 con<\/strong>.<\/p>\nX\u00e1c xu\u1ea5t \u0111\u1ec3 kh\u00f4ng v\u1ec1 c\u1ea3 2 con<\/strong> = 0,98^27 = 57,957 %<\/strong>.<\/p>\nX\u00e1c xu\u1ea5t ch\u1ec9 v\u1ec1 1 con:<\/strong> tr\u01b0\u1eddng h\u1ee3p n\u00e0y l\u1ea1i chia ra 2 tr\u01b0\u1eddng h\u1ee3p nh\u1ecf l\u00e0 con th\u1ee9 nh\u1ea5t v\u1ec1, con th\u1ee9 2 kh\u00f4ng v\u1ec1 v\u00e0 ng\u01b0\u1ee3c l\u1ea1i. N\u00ean x\u00e1c xu\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%.<\/p>\nX\u00e1c su\u1ea5t kh\u00f4ng tr\u00fang l\u00f4 xi\u00ean 2<\/strong> = 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%.<\/p>\n4. C\u00e1ch t\u00ednh x\u00e1c su\u1ea5t tr\u00fang l\u00f4 xi\u00ean 3<\/h3>\n
T\u01b0\u01a1ng t\u1ef1 v\u1edbi c\u00e1ch t\u00ednh x\u00e1c su\u1ea5t l\u00f4 \u0111\u1ec1 \u1edf l\u00f4 xi\u00ean 2. Ch\u00fang ta s\u1ebd c\u00f3 th\u00f4ng tin c\u1ee5 th\u1ec3 nh\u01b0 sau:<\/p>\n
X\u00e1c kh\u00f4ng tr\u00fang l\u00f4 xi\u00ean 3<\/b> = X\u00e1c su\u1ea5t kh\u00f4ng v\u1ec1 c\u1ea3 3 con + x\u00e1c su\u1ea5t ch\u1ec9 v\u1ec1 1 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%.<\/p>\n
=> X\u00e1c su\u1ea5t tr\u00fang l\u00f4 xi\u00ean 3<\/b> = 100% \u2013 98,527% = 1,473% ( d\u1ec5 h\u01a1n tr\u00fang \u0111\u1ec1 1 t\u00fd).<\/p>\n5. C\u00e1ch t\u00ednh x\u00e1c su\u1ea5t tr\u00fang \u0110\u1ec1 3 ch\u00e2n v\u00e0 L\u00f4 3 ch\u00e2n<\/h3>\n
X\u00e1c xu\u1ea5t tr\u00fang \u0111\u1ec1 3 ch\u00e2n<\/strong> th\u00ec kh\u00f4ng ph\u1ea3i t\u00ednh v\u00ec ai c\u0169ng bi\u1ebft x\u00e1c xu\u1ea5t \u0111\u1ec3 tr\u00fang \u0111\u1ec1 3 ch\u00e2n l\u00e0 1\/1000 = 0,1 %.<\/p>\nX\u00e1c xu\u1ea5t tr\u00fang l\u00f4 3 ch\u00e2n<\/strong> = 1 \u2013 x\u00e1c xu\u1ea5t t\u1ea1ch l\u00f4 3 ch\u00e2n = 1 \u2013 0,999^27 = 2,67 %.<\/p>\nNu\u00f4i L\u00f4 nh\u01b0 th\u1ebf n\u00e0o t\u1ed1t nh\u1ea5t ?<\/h3>\n
Trong hai b\u00e0i vi\u1ebft tr\u01b0\u1edbc ch\u00fang ta \u0111\u00e3 n\u00f3i v\u1ec1 kinh nghi\u1ec7m nu\u00f4i l\u00f4 v\u00e0 c\u00e1ch nu\u00f4i l\u00f4 khung 7 ng\u00e0y. N\u1ebfu anh em n\u00e0o ch\u01b0a xem, h\u00e3y xem ngay \u0111\u1ec3 bi\u1ebft c\u00e1ch v\u00e0o ti\u1ec1n lu\u00f4n c\u00f3 l\u00e3i.<\/p>\n
Ngo\u00e0i ra, \u0111\u1ec3 t\u00ednh x\u00e1c su\u1ea5t l\u00f4 \u0111\u1ec1 lu\u00f4n c\u00f3 l\u00e3i th\u00ec anh em h\u00e3y \u01b0\u1edbc l\u01b0\u1ee3ng c\u1ed1 ng\u00e0y \u0111\u00e1nh \u0111\u1ec3 tr\u00fang \u0111\u1ea1t \u0111\u01b0\u1ee3c t\u1ef7 l\u1ec7 100%.<\/p>\n
Nh\u01b0 v\u1eady, ch\u00fang ta c\u00f3 th\u1ec3 d\u1ef1a v\u00e0o c\u00e1ch t\u00ednh x\u00e1c su\u1ea5t b\u00ean tr\u00ean \u0111\u1ec3:<\/p>\n
G\u1ecdi A l\u00e0 bi\u1ebfn c\u1ed1 trong n ng\u00e0y, ta kh\u00f4ng tr\u00fang l\u1ea7n n\u00e0o. G\u1ecdi B l\u00e0 bi\u1ebfn c\u1ed1 ta ch\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. Do \u0111\u00f3: P(A) = 0,99^27n.<\/p>\n
P(B) = 1\u2212P(A) = 1\u22120,99^27n<\/p>\n
P(B) > m \u21d4 0,99^27n < 1\u2212m \u21d4 n > log0,99(1\u2212m)\/27 V\u1edbi m = 0,9999, ta c\u00f3: n > 33,9, t\u1ee9c l\u00e0 trong v\u00f2ng 34 ng\u00e0y th\u00ec x\u00e1c su\u1ea5t ta \u0111\u00e1nh tr\u00fang \u00edt nh\u1ea5t 1 l\u1ea7n l\u00e0 99,99%.<\/p>\n
K\u1ebft Th\u00fac<\/h3>\n
M\u1ed7i ph\u01b0\u01a1ng ph\u00e1p soi c\u1ea7u \u0111\u1ec1u mang l\u1ea1i kh\u1ea3 n\u0103ng chi\u1ebfn th\u1eafng r\u1ea5t cao. Nh\u01b0ng \u0111\u1ec3 anh em v\u00e0o ti\u1ec1n h\u1ee3p l\u00fd c\u0169ng nh\u01b0 l\u1ef1a ch\u1ecdn s\u1ed1 ng\u00e0y (N) \u0111\u00e1nh h\u1ee3p l\u00fd. B\u00e0i vi\u1ebft n\u00e0y admin \u0111\u00e3 mang \u0111\u1ebfn cho anh em c\u00e1i nh\u00ecn t\u1ed5ng quan nh\u1ea5t v\u1ec1 c\u00e1ch t\u00ednh x\u00e1c su\u1ea5t l\u00f4 \u0111\u1ec1.<\/p>\n
N\u1ebfu anh em c\u00f3 \u0111i\u1ec1u th\u1eafc m\u1eafc, h\u00e3y chat v\u1edbi ch\u00fang t\u00f4i. Ch\u00fang ta s\u1ebd c\u00f9ng th\u1ea3o lu\u1eadn th\u00eam v\u1ec1 v\u1ea5n \u0111\u1ec1 n\u00e0y. Ch\u00fac anh em lu\u00f4n may m\u1eafn.<\/p>\n<\/div>\n<\/div>\n<\/article>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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