Xác suất có điều kiện

Home Lớp 12 Toán lớp 12 - Sách Kết nối tri thức Bài 18. Xác suất có điều kiệnBài tập nâng cao

{"common":{"save":0,"post_id":"8344","level":3,"total":10,"point":10,"point_extra":0},"segment":[{"id":"6213","post_id":"8344","mon_id":"1159285","chapter_id":"1159383","question":"L\u1edbp Toán S\u01b0 Ph\u1ea1m có 95 Sinh viên, trong \u0111ó có 40 nam và 55 n\u1eef. Trong k\u1ef3 thi môn Xác su\u1ea5t th\u1ed1ng kê có 23 sinh viên \u0111\u1ea1t \u0111i\u1ec3m gi\u1ecfi (trong \u0111ó có 12 nam và 11 n\u1eef). G\u1ecdi tên ng\u1eabu nhiên m\u1ed9t sinh viên trong danh sách l\u1edbp. Tìm xác su\u1ea5t g\u1ecdi \u0111\u01b0\u1ee3c sinh viên \u0111\u1ea1t \u0111i\u1ec3m gi\u1ecfi môn Xác su\u1ea5t th\u1ed1ng kê, bi\u1ebft sinh viên \u0111ó là n\u1eef?<\/p>","options":["A.<\/strong> $\\dfrac{1}{5}$<\/span>","B.<\/strong> $\\dfrac{11}{23}$<\/span>","C.<\/strong> $\\dfrac{12}{23}$<\/span>","D.<\/strong> $\\dfrac{11}{19}$<\/span>"],"correct":"1","level":"3","hint":"","answer":"Ch\u1ecdn A.<\/strong> $\\dfrac{1}{5}$<\/span>.<\/span><\/p>G\u1ecdi A là bi\u1ebfn c\u1ed1 “G\u1ecdi \u0111\u01b0\u1ee3c sinh viên n\u1eef” và B là bi\u1ebfn c\u1ed1 “G\u1ecdi \u0111\u01b0\u1ee3c sinh viên \u0111\u1ea1t \u0111i\u1ec3m gi\u1ecfi môn Xác su\u1ea5t th\u1ed1ng kê”.<\/p>Ta \u0111i tính P(B | A).<\/p>Ta có $P(A)=\\dfrac{55}{95}$<\/span>, $P(AB)=\\dfrac{11}{95}$<\/span>.<\/p>Do \u0111ó $P(B|A)=\\dfrac{P(AB)}{P(A)}=\\dfrac{11}{95}:\\dfrac{55}{95}=\\dfrac{1}{5}$<\/span>.<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2025-01-19 15:39:15","option_type":"math","len":0},{"id":"6214","post_id":"8344","mon_id":"1159285","chapter_id":"1159383","question":"M\u1ed9t nhóm h\u1ecdc sinh có 30 h\u1ecdc sinh, trong \u0111ó có 16 em h\u1ecdc khá môn Toán, 25 em h\u1ecdc khá môn Hóa h\u1ecdc, 12 em h\u1ecdc khá c\u1ea3 hai môn Toán và Hóa h\u1ecdc. Ch\u1ecdn ng\u1eabu nhiên m\u1ed9t h\u1ecdc sinh trong s\u1ed1 \u0111ó. Tính xác su\u1ea5t \u0111\u1ec3 h\u1ecdc sinh \u0111ó h\u1ecdc khá môn Toán bi\u1ebft h\u1ecdc sinh \u0111ó h\u1ecdc khá môn Hóa h\u1ecdc.<\/p>","options":["A.<\/strong> 0,53","B.<\/strong> 0,75","C.<\/strong> 0,48","D.<\/strong> 0,84"],"correct":"3","level":"3","hint":"","answer":"Ch\u1ecdn C.<\/strong> 0,48.<\/span><\/p>G\u1ecdi bi\u1ebfn c\u1ed1 A: “H\u1ecdc sinh \u0111ó h\u1ecdc khá môn Toán” và bi\u1ebfn c\u1ed1 B: “H\u1ecdc sinh \u0111ó h\u1ecdc khá môn Hóa h\u1ecdc”.<\/p>Ta c\u1ea7n tính P(A | B).<\/p>Vì $n(AB)=12$<\/span> và $n(B)=25$<\/span> nên $P(A|B)=\\dfrac{n(AB)}{n(B)}=\\dfrac{12}{25}=0,48$<\/span>.<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2025-01-19 15:57:26","option_type":"txt","len":0},{"id":"6218","post_id":"8344","mon_id":"1159285","chapter_id":"1159383","question":"Gi\u1ea3 s\u1eed trong m\u1ed9t nhóm ng\u01b0\u1eddi có 91% ng\u01b0\u1eddi là không nhi\u1ec5m b\u1ec7nh. \u0110\u1ec3 phát hi\u1ec7n ra ng\u01b0\u1eddi nhi\u1ec5m b\u1ec7nh, ng\u01b0\u1eddi ta ti\u1ebfn hành xét nghi\u1ec7m t\u1ea5t c\u1ea3 m\u1ecdi ng\u01b0\u1eddi c\u1ee7a nhóm \u0111ó. Bi\u1ebft \u0111\u1ed1i v\u1edbi ng\u01b0\u1eddi nhi\u1ec5m b\u1ec7nh thì xác su\u1ea5t xét nghi\u1ec7m có k\u1ebft qu\u1ea3 d\u01b0\u01a1ng tính là 85%, nh\u01b0ng \u0111\u1ed1i v\u1edbi ng\u01b0\u1eddi không nhi\u1ec5m b\u1ec7nh thì xác su\u1ea5t xét nghi\u1ec7m có ph\u1ea3n \u1ee9ng d\u01b0\u01a1ng tính là 7%. Tính xác su\u1ea5t \u0111\u1ec3 ng\u01b0\u1eddi \u0111\u01b0\u1ee3c ch\u1ecdn ra không nhi\u1ec5m b\u1ec7nh và không có ph\u1ea3n \u1ee9ng d\u01b0\u01a1ng tính.<\/p>","options":["A.<\/strong> 0,93","B.<\/strong> 0,8463","C.<\/strong> 0,0637","D.<\/strong> 0,7735"],"correct":"2","level":"3","hint":"","answer":"Ch\u1ecdn B.<\/strong> 0,8463.<\/span><\/p>G\u1ecdi A “Ng\u01b0\u1eddi \u0111\u01b0\u1ee3c ch\u1ecdn ra không nhi\u1ec5m b\u1ec7nh” và B “Ng\u01b0\u1eddi \u0111\u01b0\u1ee3c ch\u1ecdn ra có ph\u1ea3n \u1ee9ng d\u01b0\u01a1ng tính”.<\/p>Ta c\u1ea7n tính $P(A\\overline{B})$<\/span>.<\/p>Theo bài ta có: P(A) = 0,91 ; P(B | A) = 0,07 ; P(B | $\\overline{A}$<\/span>) = 0,85.<\/p>Ta l\u1ea1i có $P(\\overline{B}|A)$<\/span> = 1 – P(B | A) = 1 – 0,07 = 0,93.<\/p>Vì $P(\\overline{B}|A)=\\dfrac{P(A\\overline{B})}{P(A)}$<\/span> nên $P(A\\overline{B})$<\/span> = $P(A)$<\/span>.$P(\\overline{B}|A)$<\/span> = 0,91 . 0,93 = 0,8463.<\/p>Ta có s\u01a1 \u0111\u1ed3 hình cây nh\u01b0 sau:<\/p><svg 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<text fill=\"#000000\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"24\" id=\"svg_21\" stroke=\"#000\" stroke-width=\"0\" text-anchor=\"start\" x=\"325\" xml:space=\"preserve\" y=\"186\">_<\/text> <\/g> <text fill=\"#000000\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"24\" id=\"svg_23\" stroke=\"#000\" stroke-width=\"0\" text-anchor=\"start\" x=\"205\" xml:space=\"preserve\" y=\"212.25\">AB<\/text> <line fill=\"none\" id=\"svg_24\" stroke=\"#000\" stroke-linecap=\"undefined\" stroke-linejoin=\"undefined\" stroke-width=\"1.5\" x1=\"69\" x2=\"31\" y1=\"106.25\" y2=\"134.25\"><\/line> <line fill=\"none\" id=\"svg_25\" stroke=\"#000\" stroke-linecap=\"undefined\" stroke-linejoin=\"undefined\" stroke-width=\"1.5\" x1=\"70\" x2=\"119\" y1=\"107.25\" y2=\"127.25\"><\/line> <line fill=\"none\" id=\"svg_26\" stroke=\"#000\" stroke-linecap=\"undefined\" stroke-linejoin=\"undefined\" stroke-width=\"1.5\" x1=\"265\" x2=\"232\" y1=\"106.25\" y2=\"131.25\"><\/line> <line 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y=\"117.25\">0,85<\/text> <text fill=\"#000000\" fill-opacity=\"null\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"15\" id=\"svg_31\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" x=\"296\" xml:space=\"preserve\" y=\"116.25\">0,15<\/text> <text fill=\"#000000\" fill-opacity=\"null\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"15\" id=\"svg_32\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" x=\"86\" xml:space=\"preserve\" y=\"46.25\">0,91<\/text> <text fill=\"#000000\" fill-opacity=\"null\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"15\" id=\"svg_33\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" x=\"222\" xml:space=\"preserve\" y=\"40.25\">0,09<\/text> <text fill=\"#000000\" fill-opacity=\"null\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"24\" id=\"svg_34\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" 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