Công thức xác suất toàn phần và công thức Bayes (trung bình số 2)

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":"8429","level":2,"total":10,"point":10,"point_extra":0},"segment":[{"id":"6266","post_id":"8429","mon_id":"1159285","chapter_id":"1159383","question":"H\u1ed9p th\u1ee9 nh\u1ea5t có 3 viên bi xanh và 6 viên vi \u0111\u1ecf. H\u1ed9p th\u1ee9 hai có 3 viên vi xanh và 7 viên bi \u0111\u1ecf. Các viên bi có cùng kích th\u01b0\u1edbc và kh\u1ed1i l\u01b0\u1ee3ng. L\u1ea5y ng\u1eabu nhiên m\u1ed9t viên bi t\u1eeb h\u1ed9p th\u1ee9 nh\u1ea5t chuy\u1ec3n sang h\u1ed9p th\u1ee9 hai. Sau \u0111ó l\u1ea1i l\u1ea5y ng\u1eabu nhiên \u0111\u1ed3ng th\u1eddi hai viên t\u1eeb h\u1ed9p th\u1ee9 hai, bi\u1ebft hai bi l\u1ea5y ra t\u1eeb h\u1ed9p th\u1ee9 hai là bi màu \u0111\u1ecf, tính xác su\u1ea5t viên bi l\u1ea5y ra t\u1eeb h\u1ed9p th\u1ee9 nh\u1ea5t c\u0169ng là bi màu \u0111\u1ecf.<\/p>","options":["A.<\/strong> $\\dfrac{8}{15}$<\/span>","B.<\/strong> $\\dfrac{7}{15}$<\/span>","C.<\/strong> $\\dfrac{8}{11}$<\/span>","D.<\/strong> $\\dfrac{7}{13}$<\/span>"],"correct":"3","level":"2","hint":"","answer":"Ch\u1ecdn C.<\/strong> $\\dfrac{8}{11}$<\/span>.<\/span><\/p>G\u1ecdi $A_1$<\/span>: “L\u1ea5y ra m\u1ed9t bi màu xanh \u1edf h\u1ed9p th\u1ee9 nh\u1ea5t”. Và $A_2$<\/span>: “L\u1ea5y ra m\u1ed9t bi màu \u0111\u1ecf \u1edf h\u1ed9p th\u1ee9 nh\u1ea5t”. Nên $A_1$<\/span>, $A_2$<\/span> là h\u1ec7 bi\u1ebfn c\u1ed1 \u0111\u1ea7y \u0111\u1ee7. G\u1ecdi B : “Hai bi l\u1ea5y ra t\u1eeb h\u1ed9p th\u1ee9 hai là màu \u0111\u1ecf”. Ta có: $P(A_1)=\\dfrac{C^1_3}{C^1_9}=\\dfrac{1}{3}$<\/span>, $P(A_2)=\\dfrac{C_6^1}{C_9^1}=\\dfrac{2}{3}$<\/span>.<\/p>N\u1ebfu l\u1ea5y \u0111\u01b0\u1ee3c bi xanh t\u1eeb h\u1ed9p th\u1ee9 nh\u1ea5t b\u1ecf vào h\u1ed9p th\u1ee9 hai thì $P(B|A_1)=\\dfrac{C^2_7}{C^2_{11}}=\\dfrac{21}{55}$<\/span>.<\/p>N\u1ebfu l\u1ea5y \u0111\u01b0\u1ee3c bi \u0111\u1ecf t\u1eeb h\u1ed9p th\u1ee9 nh\u1ea5t b\u1ecf vào h\u1ed9p th\u1ee9 hai thì $P(B|A_2)=\\dfrac{C^2_8}{C^2_{11}}=\\dfrac{28}{55}$<\/span>. Áp d\u1ee5ng công th\u1ee9c xác su\u1ea5t toàn ph\u1ea7n <\/p>$P(B)=P(A_1).P(B|A_1)+P(A_2).P(B|A_2)=\\dfrac{1}{3}.\\dfrac{21}{55}+\\dfrac{2}{3}.\\dfrac{28}{55}=\\dfrac{7}{15}$<\/span>.<\/p>Xác su\u1ea5t viên bi l\u1ea5y ra t\u1eeb h\u1ed9p th\u1ee9 nh\u1ea5t màu \u0111\u1ecf, bi\u1ebft hai bi l\u1ea5y ra t\u1eeb h\u1ed9p th\u1ee9 hai màu \u0111\u1ecf, ta áp d\u1ee5ng công th\u1ee9c Bayes<\/p>$P(A_2|B)=\\dfrac{P(A_2).P(B|A_2)}{P(B)}=\\dfrac{\\dfrac{2}{3}.\\dfrac{28}{55}}{\\dfrac{7}{15}}=\\dfrac{8}{11}$<\/span>.<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2025-02-07 09:47:04","option_type":"math","len":0},{"id":"6268","post_id":"8429","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>.<\/p>Xác su\u1ea5t \u0111\u1ec3 ng\u01b0\u1eddi \u0111ó m\u1eafc b\u1ec7nh khi ch\u01b0a ki\u1ec3m tra: $P(A)=0,01$<\/span>.<\/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: $P(B|A)=0,99$<\/span>.<\/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>Áp d\u1ee5ng công th\u1ee9c tính xác su\u1ea5t toàn ph\u1ea7n, xác su\u1ea5t \u0111\u1ec3 k\u1ebft qu\u1ea3 ki\u1ec3m tra c\u1ee7a ng\u01b0\u1eddi \u0111ó d\u01b0\u01a1ng tính là<\/p>$P(B)=P(A).P(B|A)+P(\\overline{A}).P(B|\\overline{A})=0,01.0,99+0,99.0,01=0,0198$<\/span>.<\/p>Áp d\u1ee5ng công th\u1ee9c Bayes, ta có $P(A|B)=\\dfrac{P(A).P(B|A)}{P(B)}=\\dfrac{0,01.0,99}{0,0198}=0,5$<\/span>.<\/p>Xác su\u1ea5t \u0111\u1ec3 ng\u01b0\u1eddi \u0111ó m\u1eafc b\u1ec7nh n\u1ebfu k\u1ebft qu\u1ea3 ki\u1ec3m tra ng\u01b0\u1eddi \u0111ó d\u01b0\u01a1ng tính là 0,5.<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2025-02-07 09:58:36","option_type":"txt","len":0},{"id":"6269","post_id":"8429","mon_id":"1159285","chapter_id":"1159383","question":"Trong m\u1ed9t tr\u01b0\u1eddng h\u1ecdc, t\u1ec9 l\u1ec7 h\u1ecdc sinh n\u1eef là 52% . T\u1ec9 l\u1ec7 h\u1ecdc sinh n\u1eef và t\u1ec9 l\u1ec7 h\u1ecdc sinh nam tham gia câu l\u1ea1c b\u1ed9 ngh\u1ec7 thu\u1eadt l\u1ea7n l\u01b0\u1ee3t là 18% và 15% . G\u1eb7p ng\u1eabu nhiên m\u1ed9t h\u1ecdc sinh c\u1ee7a tr\u01b0\u1eddng. Bi\u1ebft h\u1ecdc sinh \u0111ó có tham gia câu l\u1ea1c b\u1ed9 ngh\u1ec7 thu\u1eadt. Tính xác su\u1ea5t h\u1ecdc sinh \u0111ó là nam<\/p>","options":["A.<\/strong> $\\dfrac{207}{1230}$<\/span>","B.<\/strong> $\\dfrac{207}{1250}$<\/span>","C.<\/strong> $\\dfrac{10}{27}$<\/span>","D.<\/strong> $\\dfrac{10}{23}$<\/span>"],"correct":"4","level":"2","hint":"","answer":"Ch\u1ecdn D.<\/strong> $\\dfrac{10}{23}$<\/span>.<\/span><\/p>G\u1ecdi $A_1$<\/span>, $A_2$<\/span> l\u1ea7n l\u01b0\u1ee3t là các bi\u1ebfn c\u1ed1 g\u1eb7p \u0111\u01b0\u1ee3c m\u1ed9t h\u1ecdc sinh n\u1eef, m\u1ed9t h\u1ecdc sinh nam. Nên $A_1$<\/span>, $A_2$<\/span> là h\u1ec7 bi\u1ebfn c\u1ed1 \u0111\u1ea7y \u0111\u1ee7.<\/p>G\u1ecdi B “H\u1ecdc sinh \u0111ó tham gia câu l\u1ea1c b\u1ed9 ngh\u1ec7 thu\u1eadt”.<\/p>$P(A_1)=0,52$<\/span>, $P(A_2)=1-0,52=0,48$<\/span>.<\/p>$P(B|A_1)=0,18$<\/span>, $P(B|A_2)=0,15$<\/span>.<\/p>Áp d\u1ee5ng công th\u1ee9c xác su\u1ea5t toàn ph\u1ea7n<\/p>$P(B)=P(A_1).P(B|A_1)+P(A_2).P(B|A_2)$<\/span> $=0,52.0,18+0,48.0,15=\\dfrac{207}{1250}=0,1656$<\/span>.<\/p>Xác su\u1ea5t \u0111\u1ec3 h\u1ecdc sinh \u0111ó là nam, bi\u1ebft h\u1ecdc sinh \u0111ó tham gia câu l\u1ea1c b\u1ed9 ngh\u1ec7 thu\u1eadt, ta áp d\u1ee5ng công th\u1ee9c Bayes<\/p>$P(A_2|B)=\\dfrac{P(A_2).P(B|A_2)}{P(B)}=\\dfrac{0,48.0,15}{0,1656}=\\dfrac{10}{23}$<\/span>.<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2025-02-08 07:39:54","option_type":"math","len":0}]}