Xác suất có điều kiện - trung bình số 2

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

{"common":{"save":0,"post_id":"8343","level":2,"total":10,"point":10,"point_extra":0},"segment":[{"id":"6223","post_id":"8343","mon_id":"1159285","chapter_id":"1159383","question":"Cho m\u1ed9t h\u1ed9p kín có 6 th\u1ebb ATM c\u1ee7a BIDV và 4 th\u1ebb ATM c\u1ee7a Vietcombank. L\u1ea5y ng\u1eabu nhiên l\u1ea7n l\u01b0\u1ee3t 2 th\u1ebb (l\u1ea5y không hoàn l\u1ea1i). Tìm xác su\u1ea5t \u0111\u1ec3 l\u1ea7n th\u1ee9 hai l\u1ea5y \u0111\u01b0\u1ee3c th\u1ebb ATM c\u1ee7a Vietcombank n\u1ebfu bi\u1ebft l\u1ea7n th\u1ee9 nh\u1ea5t \u0111ã l\u1ea5y \u0111\u01b0\u1ee3c th\u1ebb ATM c\u1ee7a BIDV.<\/p>","options":["A.<\/strong> $\\dfrac{5}{9}$<\/span>","B.<\/strong> $\\dfrac{2}{3}$<\/span>","C.<\/strong> $\\dfrac{7}{9}$<\/span>","D.<\/strong> $\\dfrac{4}{9}$<\/span>"],"correct":"4","level":"2","hint":"","answer":"Ch\u1ecdn D.<\/strong> $\\dfrac{4}{9}$<\/span>.<\/span><\/p>G\u1ecdi A là bi\u1ebfn c\u1ed1 “L\u1ea7n th\u1ee9 hai l\u1ea5y \u0111\u01b0\u1ee3c th\u1ebb ATM Vietcombank“, B là bi\u1ebfn c\u1ed1 “L\u1ea7n th\u1ee9 nh\u1ea5t l\u1ea5y \u0111\u01b0\u1ee3c th\u1ebb ATM c\u1ee7a BIDV".<\/p>Ta c\u1ea7n tìm $P(A|B)$<\/span>.<\/p>Sau khi l\u1ea5y l\u1ea7n th\u1ee9 nh\u1ea5t (bi\u1ebfn c\u1ed1 B \u0111ã x\u1ea3y ra) trong h\u1ed9p còn l\u1ea1i 9 th\u1ebb (trong \u0111ó 4 th\u1ebb Vietcombank) nên $P(A|B)$<\/span> = $\\dfrac{4}{9}$<\/span>.<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2025-01-19 16:34:45","option_type":"math","len":0},{"id":"6226","post_id":"8343","mon_id":"1159285","chapter_id":"1159383","question":"M\u1ed9t bình \u0111\u1ef1ng 9 viên bi xanh và 7 viên bi \u0111\u1ecf. L\u1ea7n l\u01b0\u1ee3t l\u1ea5y ng\u1eabu nhiên ra 2 bi, m\u1ed7i l\u1ea7n l\u1ea5y 1 bi không hoàn l\u1ea1i. Tính xác su\u1ea5t \u0111\u1ec3 bi th\u1ee9 2 màu xanh n\u1ebfu bi\u1ebft bi th\u1ee9 nh\u1ea5t màu \u0111\u1ecf.<\/p>","options":["A.<\/strong> $\\dfrac{3}{5}$<\/span>","B.<\/strong> $\\dfrac{9}{16}$<\/span>","C.<\/strong> $\\dfrac{7}{16}$<\/span>","D.<\/strong> $\\dfrac{21}{80}$<\/span>"],"correct":"1","level":"2","hint":"","answer":"Ch\u1ecdn A.<\/strong> $\\dfrac{3}{5}$<\/span>.<\/span><\/p>G\u1ecdi A là bi\u1ebfn c\u1ed1 “L\u1ea7n th\u1ee9 nh\u1ea5t l\u1ea5y \u0111\u01b0\u1ee3c bi màu \u0111\u1ecf”. G\u1ecdi B là bi\u1ebfn c\u1ed1 “L\u1ea7n th\u1ee9 hai l\u1ea5y \u0111\u01b0\u1ee3c bi màu xanh”. Ta c\u1ea7n tìm P(B | A). Không gian m\u1eabu $n(\\Omega)=16.15$<\/span> cách ch\u1ecdn.<\/p>L\u1ea7n th\u1ee9 nh\u1ea5t l\u1ea5y 1 viên bi màu \u0111\u1ecf có 7 cách ch\u1ecdn, l\u1ea7n th\u1ee9 hai l\u1ea5y 1 viên bi trong 15 bi còn l\u1ea1i có 15 cách ch\u1ecdn, do \u0111ó $P(A)=\\dfrac{7.15}{16.15}=\\dfrac{7}{16}$<\/span>.<\/p>L\u1ea7n th\u1ee9 nh\u1ea5t l\u1ea5y 1 viên bi màu \u0111\u1ecf có 7 cách ch\u1ecdn, l\u1ea7n th\u1ee9 hai l\u1ea5y 1 viên bi màu xanh có 9 cách ch\u1ecdn, do \u0111ó $P(AB)=\\dfrac{7.9}{16.15}=\\dfrac{21}{80}$<\/span>.<\/p>V\u1eady xác su\u1ea5t \u0111\u1ec3 viên bi l\u1ea5y l\u1ea7n th\u1ee9 hai là màu xanh n\u1ebfu bi\u1ebft viên bi l\u1ea5y l\u1ea7n th\u1ee9 nh\u1ea5t là màu \u0111\u1ecf là $P(B|A)=\\dfrac{P(AB)}{P(A)}=\\dfrac{21}{80}:\\dfrac{7}{16}=\\dfrac{3}{5}$<\/span>.<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2025-01-19 16:42:49","option_type":"math","len":0},{"id":"6229","post_id":"8343","mon_id":"1159285","chapter_id":"1159383","question":"Trong h\u1ed9p có 20 n\u1eafp khoen bia Tiger, trong \u0111ó có 2 n\u1eafp ghi “Chúc m\u1eebng b\u1ea1n \u0111ã trúng th\u01b0\u1edfng xe Camry”. B\u1ea1n Minh Hi\u1ec1n \u0111\u01b0\u1ee3c ch\u1ecdn lên rút th\u0103m l\u1ea7n l\u01b0\u1ee3t hai n\u1eafp khoen, xác su\u1ea5t \u0111\u1ec3 c\u1ea3 hai n\u1eafp \u0111\u1ec1u trúng th\u01b0\u1edfng là:<\/p>","options":["A.<\/strong> $\\dfrac{1}{2}$<\/span>","B.<\/strong> $\\dfrac{1}{10}$<\/span>","C.<\/strong> $\\dfrac{1}{190}$<\/span>","D.<\/strong> $\\dfrac{1}{20}$<\/span>"],"correct":"3","level":"2","hint":"","answer":"Ch\u1ecdn C.<\/strong> $\\dfrac{1}{190}$<\/span>.<\/span><\/p>G\u1ecdi A là bi\u1ebfn c\u1ed1 “N\u1eafp khoen \u0111\u1ea7u trúng th\u01b0\u1edfng”.<\/p>G\u1ecdi B là bi\u1ebfn c\u1ed1 “N\u1eafp khoen th\u1ee9 hai trúng th\u01b0\u1edfng”.<\/p>Ta \u0111i tính P(B | A).<\/p>Khi b\u1ea1n rút th\u0103m l\u1ea7n \u0111\u1ea7u thì trong h\u1ed9p có 20 n\u1eafp trong \u0111ó có 2 n\u1eafp trúng do \u0111ó $P(A)=\\dfrac{2}{20}=\\dfrac{1}{10}$<\/span>.<\/p>Khi bi\u1ebfn c\u1ed1 A \u0111ã x\u1ea3y ra thì còn l\u1ea1i 19 n\u1eafp trong \u0111ó có 1 n\u1eafp trúng th\u01b0\u1edfng do \u0111ó $P(B|A)=\\dfrac{1}{19}$<\/span>.<\/p>Vì $P(B|A)=\\dfrac{P(AB)}{P(A)}$<\/span> nên $P(AB)=P(B|A).P(A)=\\dfrac{1}{19}.\\dfrac{1}{10}=\\dfrac{1}{190}$<\/span>.<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2025-01-19 16:56:00","option_type":"math","len":0}]}