Công thức xác suất toàn phần và công thức Bayes (trung bình 1)

Home Lớp 12 Toán lớp 12 - Sách Kết nối tri thức Bài 19. Công thức xác suất toàn phần và công thức BayesBài tập trung bình

{"common":{"save":0,"post_id":"8427","level":2,"total":10,"point":10,"point_extra":0},"segment":[{"id":"6260","post_id":"8427","mon_id":"1159285","chapter_id":"1159383","question":"M\u1ed9t tr\u1ea1m ch\u1ec9 phát hai tín hi\u1ec7u A và B v\u1edbi xác su\u1ea5t t\u01b0\u01a1ng \u1ee9ng 0,85 và 0,15. Do có nhi\u1ec5u trên \u0111\u01b0\u1eddng truy\u1ec1n nên $\\dfrac{1}{7}$<\/span> tín hi\u1ec7u A b\u1ecb méo và thu \u0111\u01b0\u1ee3c nh\u01b0 tín hi\u1ec7u B còn $\\dfrac{1}{8}$<\/span> tín hi\u1ec7u B b\u1ecb méo và thu \u0111\u01b0\u1ee3c nh\u01b0 A. Gi\u1ea3 s\u1eed \u0111ã thu \u0111\u01b0\u1ee3c tín hi\u1ec7u A.  Xác su\u1ea5t thu \u0111\u01b0\u1ee3c \u0111úng tín hi\u1ec7u lúc phát là:<\/p>","options":["A.<\/strong> $\\dfrac{272}{1120}$<\/span>","B.<\/strong> $\\dfrac{373}{279}$<\/span>","C.<\/strong> $\\dfrac{173}{279}$<\/span>","D.<\/strong> $\\dfrac{272}{279}$<\/span>"],"correct":"4","level":"2","hint":"","answer":"Ch\u1ecdn D.<\/strong> $\\dfrac{272}{279}$<\/span><\/span><\/p>G\u1ecdi A là bi\u1ebfn c\u1ed1 “Phát tín hi\u1ec7u A”, <\/p>B là bi\u1ebfn c\u1ed1 “Phát tín hi\u1ec7u B”, <\/p>$T_A$<\/span> là bi\u1ebfn c\u1ed1 “Thu \u0111\u01b0\u1ee3c tín hi\u1ec7u A”, <\/p>$T_B$<\/span> là bi\u1ebfn c\u1ed1 “Thu \u0111\u01b0\u1ee3c tín hi\u1ec7u B”.<\/p>Theo \u0111\u1ec1 bài, ta có: P(A) = 0,85; P(B) = 0,15; $P(T_B|A)$<\/span> = $\\dfrac{1}{7}$<\/span>; $P(T_A|B)$<\/span> = $\\dfrac{1}{8}$<\/span>.<\/p>Suy ra $P(T_A|A)$<\/span> = 1 − $\\dfrac{1}{7}$<\/span> = $\\dfrac{6}{7}$<\/span>.<\/p>Ta có: $P(T_A)$<\/span> = $P(A).P(T_A|A)+P(B).P(T_A|B)$<\/span><\/p>                      = 0,85. $\\dfrac{6}{7}$<\/span> + 0,15. $\\dfrac{1}{8}$<\/span> = $\\dfrac{837}{1120}$<\/span>.<\/p>Theo công th\u1ee9c Bayes, ta có: $P(A|T_A)=\\dfrac{P(A).P(T_A|A)}{P(T_A)}=\\dfrac{0,85.\\dfrac{6}{7}}{\\dfrac{837}{1120}}=\\dfrac{272}{279}$<\/span>.<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2025-02-07 09:05:59","option_type":"math","len":0},{"id":"6263","post_id":"8427","mon_id":"1159285","chapter_id":"1159383","question":"M\u1ed9t c\u0103n b\u1ec7nh có 1% dân s\u1ed1 m\u1eafc ph\u1ea3i. M\u1ed9t ph\u01b0\u01a1ng pháp chu\u1ea9n \u0111oán \u0111\u01b0\u1ee3c phát tri\u1ec3n có t\u1ef7 l\u1ec7 chính xác là 99%. V\u1edbi nh\u1eefng ng\u01b0\u1eddi b\u1ecb b\u1ec7nh, ph\u01b0\u01a1ng pháp này s\u1ebd \u0111\u01b0a ra k\u1ebft qu\u1ea3 d\u01b0\u01a1ng tính 99% s\u1ed1 tr\u01b0\u1eddng h\u1ee3p. V\u1edbi ng\u01b0\u1eddi không m\u1eafc b\u1ec7nh, ph\u01b0\u01a1ng pháp này c\u0169ng chu\u1ea9n \u0111oán \u0111úng 99 trong 100 tr\u01b0\u1eddng h\u1ee3p. N\u1ebfu m\u1ed9t ng\u01b0\u1eddi ki\u1ec3m tra và k\u1ebft qu\u1ea3 là d\u01b0\u01a1ng tính (b\u1ecb b\u1ec7nh), xác su\u1ea5t \u0111\u1ec3 ng\u01b0\u1eddi \u0111ó th\u1ef1c s\u1ef1 b\u1ecb b\u1ec7nh là bao nhiêu?<\/p>","options":["A.<\/strong> 0,4","B.<\/strong> 0,35","C.<\/strong> 0,5","D.<\/strong> 0,65"],"correct":"3","level":"2","hint":"","answer":"Ch\u1ecdn C.<\/strong> 0,5<\/span><\/p>G\u1ecdi A là bi\u1ebfn c\u1ed1 “Ng\u01b0\u1eddi \u0111ó m\u1eafc b\u1ec7nh”.<\/p>G\u1ecdi B là bi\u1ebfn c\u1ed1 “K\u1ebft qu\u1ea3 ki\u1ec3m tra ng\u01b0\u1eddi \u0111ó là d\u01b0\u01a1ng tính (b\u1ecb b\u1ec7nh)”<\/p>Ta c\u1ea7n tính $P(A|B)$<\/span>.<\/span><\/strong><\/p>V\u1edbi $P(A|B)=\\dfrac{P(A).P(B|A)}{P(A).P(B|A)+P(\\overline{A}).P(B|\\overline{A})}$<\/span><\/p>Xác su\u1ea5t \u0111\u1ec3 ng\u01b0\u1eddi \u0111ó m\u1eafc b\u1ec7nh khi ch\u01b0a ki\u1ec3m tra: P(A) = 1% = 0,01.<\/p>Do \u0111ó xác su\u1ea5t \u0111\u1ec3 ng\u01b0\u1eddi \u0111ó không m\u1eafc b\u1ec7nh khi ch\u01b0a ki\u1ec3m tra: $P(\\overline{A})=1-0,01=0,99$<\/span>.<\/p>Xác su\u1ea5t k\u1ebft qu\u1ea3 d\u01b0\u01a1ng tính n\u1ebfu ng\u01b0\u1eddi \u0111ó m\u1eafc b\u1ec7nh là: P(B|A) = 99% = 0,99.<\/p>Xác su\u1ea5t k\u1ebft qu\u1ea3 d\u01b0\u01a1ng tính n\u1ebfu ng\u01b0\u1eddi \u0111ó không m\u1eafc b\u1ec7nh là: $P(B|\\overline{A})=1-0,99=0,01$<\/span>.<\/p>Suy ra $P(A|B)=\\dfrac{P(A).P(B|A)}{P(A).P(B|A)+P(\\overline{A}).P(B|\\overline{A})}$<\/span> $=\\dfrac{0,01.0,99}{0,01.0,99+0,99.0,01}=0,5$<\/span><\/p>Xác su\u1ea5t \u0111\u1ec3 ng\u01b0\u1eddi \u0111ó th\u1ef1c s\u1ef1 m\u1eafc b\u1ec7nh n\u1ebfu k\u1ebft qu\u1ea3 ki\u1ec3m tra ng\u01b0\u1eddi \u0111ó là d\u01b0\u01a1ng tính b\u1eb1ng 0,5.<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2025-02-07 09:29:04","option_type":"txt","len":0},{"id":"6264","post_id":"8427","mon_id":"1159285","chapter_id":"1159383","question":"Cho hai bi\u1ebfn c\u1ed1 A, B tho\u1ea3 mãn P(A) = 0,4; P(B) = 0,3; P(A|B) = 0,25. Khi \u0111ó P(B|A) b\u1eb1ng<\/p>","options":["A.<\/strong> 0,1875","B.<\/strong> 0,48","C.<\/strong> 0,333","D.<\/strong> 0,95"],"correct":"1","level":"2","hint":"","answer":"Ch\u1ecdn A.<\/strong> 0,1875<\/span><\/p>Theo công th\u1ee9c Bayes có:<\/p>$P(B|A)=\\dfrac{P(B).P(A|B)}{P(A)}=\\dfrac{0,3.0,25}{0,4}=0,1875$<\/span>.<\/p>","type":"choose","extra_type":"shape2","user_id":"131","test":"0","date":"2025-02-07 09:31:03","option_type":"txt","len":0}]}