Bài tập ôn cuối chương VIII

Home Lớp 11 Toán lớp 11 - Sách kết nối tri thức Bài tập ôn cuối chương VIIIBài tập nâng cao

{"common":{"save":0,"post_id":"7155","level":3,"total":10,"point":10,"point_extra":0},"segment":[{"id":"5811","post_id":"7155","mon_id":"1158924","chapter_id":"1159182","question":"Gieo hai con xúc x\u1eafc I và II cân \u0111\u1ed1i, \u0111\u1ed3ng ch\u1ea5t m\u1ed9t cách \u0111\u1ed9c l\u1eadp. Ta có bi\u1ebfn c\u1ed1 A : “Có ít nh\u1ea5t m\u1ed9t con xúc s\u1eafc xu\u1ea5t hi\u1ec7n m\u1eb7t 6 ch\u1ea5m”. Lúc này giá tr\u1ecb c\u1ee7a P(A) là<\/p>","options":["A.<\/strong> $\\dfrac{25}{36}$<\/span>","B.<\/strong> $\\dfrac{11}{36}$<\/span>","C.<\/strong> $\\dfrac{1}{36}$<\/span>","D.<\/strong> $\\dfrac{15}{36}$<\/span>"],"correct":"2","level":"3","hint":"","answer":"Ch\u1ecdn B.<\/strong> $\\dfrac{11}{36}$<\/span><\/span><\/p>G\u1ecdi $A_1$<\/span> là bi\u1ebfn c\u1ed1 : “Con xúc x\u1eafc I xu\u1ea5t hi\u1ec7n m\u1eb7t 6 ch\u1ea5m”.<\/p>       $A_2$<\/span> là bi\u1ebfn c\u1ed1 : “Con xúc x\u1eafc II xu\u1ea5t hi\u1ec7n m\u1eb7t 6 ch\u1ea5m”.<\/p>Suy ra $A_1$<\/span> và $A_2$<\/span> là hai bi\u1ebfn c\u1ed1 \u0111\u1ed9c l\u1eadp và ta có $P(A_1)=\\dfrac{1}{6}$<\/span>, $P(A_2)=\\dfrac{1}{6}$<\/span>.<\/p>Bi\u1ebfn c\u1ed1 $\\overline{A}$<\/span>: "Hai con xúc x\u1eafc I và II không xu\u1ea5t hi\u1ec7n m\u1eb7t 6 ch\u1ea5m” là bi\u1ebfn c\u1ed1 \u0111\u1ed1i c\u1ee7a bi\u1ebfn c\u1ed1 A: “Có ít nh\u1ea5t m\u1ed9t con xúc x\u1eafc xu\u1ea5t hi\u1ec7n m\u1eb7t 6 ch\u1ea5m”.<\/p>Ta có: $\\overline{A}=\\overline{A_1} \\space \\overline{A_2}$<\/span>, suy ra $P(\\overline{A})=P(\\overline{A_1}).P(\\overline{A_2})=\\bigg(1-\\dfrac{1}{6}\\bigg)\\bigg(1-\\dfrac{1}{6}\\bigg)=\\dfrac{25}{36}$<\/span><\/p>V\u1eady $P(A) = 1-P(\\overline{A})=1-\\dfrac{25}{36}=\\dfrac{11}{36}$<\/span>.<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2024-05-27 04:57:57","option_type":"math","len":0},{"id":"5813","post_id":"7155","mon_id":"1158924","chapter_id":"1159182","question":"Gieo m\u1ed9t \u0111\u1ed3ng xu liên ti\u1ebfp hai l\u1ea7n. G\u1ecdi A là bi\u1ebfn c\u1ed1: “L\u1ea7n \u0111\u1ea7u xu\u1ea5t hi\u1ec7n m\u1eb7t s\u1ea5p”; B là bi\u1ebfn c\u1ed1: “Hai l\u1ea7n gieo xu\u1ea5t hi\u1ec7n m\u1eb7t gi\u1ed1ng nhau. Khi \u0111ó bi\u1ebfn c\u1ed1 A∪B \u0111\u01b0\u1ee3c mô t\u1ea3:<\/p>","options":["A.<\/strong> A∪B = {SS}","B.<\/strong> A∪B = {SS; SN}","C.<\/strong> A∪B = {SS; SN; NN}","D.<\/strong> A∪B = {SS; SN; NN; NS}"],"correct":"3","level":"3","hint":"","answer":"Ch\u1ecdn C.<\/strong> A∪B = {SS; SN; NN}<\/span><\/p>Ta có: A = {SS<\/span>; SN} và B = {SS<\/span>; NN}<\/p>V\u1eady A∪B = {SS; SN; NN}<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2024-05-27 05:02:46","option_type":"txt","len":2},{"id":"5815","post_id":"7155","mon_id":"1158924","chapter_id":"1159182","question":"M\u1ed9t h\u1ed9p ch\u1ee9a 10 chi\u1ebfc th\u1ebb cùng lo\u1ea1i, trên m\u1ed7i th\u1ebb \u0111\u01b0\u1ee3c \u0111ánh s\u1ed1 t\u1eeb 1 \u0111\u1ebfn 10. Hai th\u1ebb khác nhau thì \u0111\u01b0\u1ee3c ghi 2 s\u1ed1 khác nhau. Rút ng\u1eabu nhiên 1 chi\u1ebfc th\u1ebb t\u1eeb h\u1ed9p. Xét các bi\u1ebfn c\u1ed1 A: “S\u1ed1 xu\u1ea5t hi\u1ec7n trên th\u1ebb rút ra là s\u1ed1 ch\u1eb5n”; B: “S\u1ed1 xu\u1ea5t hi\u1ec7n trên th\u1ebb rút ra nh\u1ecf h\u01a1n 7”. Khi \u0111ó bi\u1ebfn c\u1ed1 A∪B \u0111\u01b0\u1ee3c mô t\u1ea3:<\/p>","options":["A.<\/strong> A∪B = {2; 4; 6}","B.<\/strong> A∪B = {2; 4; 6; 8; 10}","C.<\/strong> A∪B = {2; 4; 6; 8}","D.<\/strong> A∪B = {1; 2; 3; 4; 5; 6; 8; 10}"],"correct":"4","level":"3","hint":"","answer":"Ch\u1ecdn <\/span>D.<\/strong> A∪B = {1; 2; 3; 4; 5; 6; 8; 10}<\/span><\/p>Ta có: A = {2; 4; 6<\/span>; 8; 10} và B = {1; 2<\/span>; 3; 4<\/span>; 5; 6<\/span>}<\/p>V\u1eady A∪B = {1; 2; 3; 4; 5; 6; 8; 10}<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2024-05-27 05:08:43","option_type":"txt","len":2}]}