Thực hành tính xác suất theo định nghĩa cổ điển

Home Lớp 10 Toán lớp 10 - Sách Kết nối tri thức Bài 27. Thực hành tính xác suất theo định nghĩa cổ điểnBài tập nâng cao

{"common":{"save":0,"post_id":"6494","level":3,"total":10,"point":10,"point_extra":0},"segment":[{"id":"6870","mon_id":"1158767","chapter_id":"1158895","post_id":"6494","question":"M\u1ed9t \u0111\u1ed9i g\u1ed3m 5 n\u1eef và 8 nam. L\u1eadp m\u1ed9t \u0111oàn \u0111\u1ea1i bi\u1ec3u g\u1ed3m 4 ng\u01b0\u1eddi \u0111\u1ec3 tham d\u1ef1 h\u1ed9i ngh\u1ecb, tính xác su\u1ea5t \u0111\u1ec3 trong 4 ng\u01b0\u1eddi \u0111\u01b0\u1ee3c ch\u1ecdn có ít nh\u1ea5t 3 nam<\/span>.<\/p>","options":["A. $\\dfrac{70}{143}$<\/span> <\/strong>","B. $\\dfrac{73}{143}$<\/span><\/strong>","C. $\\dfrac{56}{143}$<\/span><\/strong>","D. $\\dfrac{87}{143}$<\/span><\/strong>"],"correct":"1","level":"3","hint":"","answer":"Không gian m\u1eabu là ch\u1ecdn tùy ý 4 ng\u01b0\u1eddi t\u1eeb 13 ng\u01b0\u1eddi. Suy ra s\u1ed1 ph\u1ea7n t\u1eed c\u1ee7a không gian m\u1eabu là $n(\\Omega)=C_{13}^4=715$<\/span>.<\/p>G\u1ecdi là bi\u1ebfn c\u1ed1 4 ng\u01b0\u1eddi \u0111\u01b0\u1ee3c ch\u1ecdn có ít nh\u1ea5t 3 nam. Ta có hai tr\u01b0\u1eddng h\u1ee3p thu\u1eadn l\u1ee3i cho bi\u1ebfn c\u1ed1 nh\u01b0 sau: \u25cf TH1: Ch\u1ecdn 3 nam và 1 n\u1eef, có $C_8^3.C_5^1$<\/span> cách. \u25cf TH2: Ch\u1ecdn c\u1ea3 4 nam, có $C_8^4$<\/span> cách. Suy ra s\u1ed1 ph\u1ea7n t\u1eed c\u1ee7a bi\u1ebfn c\u1ed1 A là n(A) = $C_8^3.C_5^1$<\/span> + $C_8^4$<\/span> = 350.<\/p>V\u1eady $P(A)=\\dfrac{n(A)}{n(\\Omega)}=\\dfrac{350}{715}=\\dfrac{70}{143}$<\/span><\/p>Ch\u1ecdn A. $\\dfrac{70}{143}$<\/span> <\/strong><\/span><\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2023-05-30 04:00:25","option_type":"math","len":0},{"id":"6872","mon_id":"1158767","chapter_id":"1158895","post_id":"6494","question":"M\u1ed9t nhóm g\u1ed3m 8 n\u1eef và 7 nam. Ch\u1ecdn ng\u1eabu nhiên 5 b\u1ea1n \u0111i tham gia v\u0103n ngh\u1ec7 c\u1ee7a \u0110oàn tr\u01b0\u1eddng. Tính xác su\u1ea5t \u0111\u1ec3 trong 5 b\u1ea1n \u0111\u01b0\u1ee3c ch\u1ecdn có c\u1ea3 n\u1eef l\u1eabn nam mà n\u1eef nhi\u1ec1u h\u01a1n nam<\/span> là:<\/p>","options":["A. $\\dfrac{60}{143}$<\/span> <\/strong>","B. $\\dfrac{238}{429}$<\/span><\/strong>","C. $\\dfrac{210}{429}$<\/span><\/strong>","D. $\\dfrac{82}{143}$<\/span><\/strong>"],"correct":"2","level":"3","hint":"","answer":"Không gian m\u1eabu là ch\u1ecdn tùy ý 5 ng\u01b0\u1eddi t\u1eeb 15 ng\u01b0\u1eddi. Suy ra s\u1ed1 ph\u1ea7n t\u1eed c\u1ee7a không gian m\u1eabu là $n(\\Omega)=C_{15}^5=3003$<\/span>.<\/p>G\u1ecdi A là bi\u1ebfn c\u1ed1: “5 b\u1ea1n \u0111\u01b0\u1ee3c ch\u1ecdn có c\u1ea3 n\u1eef l\u1eabn nam mà n\u1eef nhi\u1ec1u h\u01a1n nam “.<\/p>Ta có hai tr\u01b0\u1eddng h\u1ee3p thu\u1eadn l\u1ee3i cho bi\u1ebfn c\u1ed1 nh\u01b0 sau: \u25cf TH1: Ch\u1ecdn 1 nam và 4 n\u1eef, có $C_7^1.C_8^4$<\/span> cách. \u25cf TH2: Ch\u1ecdn 2 nam và 3 n\u1eef, có $C_7^2.C_8^3$<\/span> cách. Suy ra s\u1ed1 ph\u1ea7n t\u1eed c\u1ee7a bi\u1ebfn c\u1ed1 A là n(A) = $C_7^1.C_8^4$<\/span> + $C_7^2.C_8^3$<\/span> = 1666.<\/p>V\u1eady $P(A)=\\dfrac{n(A)}{n(\\Omega)}=\\dfrac{1666}{3003}=\\dfrac{238}{429}$<\/span><\/p>Ch\u1ecdn B. $\\dfrac{238}{429}$<\/span><\/strong><\/span><\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2023-05-30 04:10:19","option_type":"math","len":0},{"id":"6873","mon_id":"1158767","chapter_id":"1158895","post_id":"6494","question":"M\u1ed9t t\u1ed5 g\u1ed3m 8 n\u1eef và 7 nam, c\u1ea7n ch\u1ecdn m\u1ed9t nhóm 5 ng\u01b0\u1eddi \u0111i lao \u0111\u1ed9ng. Xác su\u1ea5t \u0111\u1ec3 ch\u1ecdn \u0111\u01b0\u1ee3c nhóm có \u0111úng 2 ng\u01b0\u1eddi na<\/span>m<\/span> là:<\/p>","options":["A. $\\dfrac{56}{143}$<\/span> <\/strong>","B. $\\dfrac{140}{429}$<\/span><\/strong>","C. $\\dfrac{28}{715}$<\/span><\/strong>","D. $\\dfrac{1}{143}$<\/span><\/strong>"],"correct":"1","level":"3","hint":"","answer":"Không gian m\u1eabu là ch\u1ecdn tùy ý 5 ng\u01b0\u1eddi t\u1eeb 15 ng\u01b0\u1eddi. Suy ra s\u1ed1 ph\u1ea7n t\u1eed c\u1ee7a không gian m\u1eabu là $n(\\Omega)=C_{15}^5=3003$<\/span>.<\/p>G\u1ecdi A là bi\u1ebfn c\u1ed1: “Ch\u1ecdn m\u1ed9t nhóm 5 ng\u01b0\u1eddi có \u0111úng 2 ng\u01b0\u1eddi nam”.<\/p>Ch\u1ecdn 2 nam và 3 n\u1eef, có $C_7^2.C_8^3$<\/span> cách.<\/p>Suy ra s\u1ed1 ph\u1ea7n t\u1eed c\u1ee7a bi\u1ebfn c\u1ed1 A là n(A) = $C_7^2.C_8^3$<\/span> = 1176.<\/p>V\u1eady $P(A)=\\dfrac{n(A)}{n(\\Omega)}=\\dfrac{1176}{3003}=\\dfrac{56}{143}$<\/span><\/p>Ch\u1ecdn A. $\\dfrac{56}{143}$<\/span> <\/strong><\/span><\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2023-05-30 04:15:04","option_type":"math","len":0}]}