Công thức xác suất toàn phần và công thức Bayes (nâng cao)

Home Lớp 12 Toán lớp 12 - Sách Kết nối tri thức Bài 19. Công thức xác suất toàn phần và công thức BayesBài tập nâng cao

{"common":{"save":0,"post_id":"8428","level":3,"total":10,"point":10,"point_extra":0},"segment":[{"id":"6254","post_id":"8428","mon_id":"1159285","chapter_id":"1159383","question":"M\u1ed9t thùng có các h\u1ed9p lo\u1ea1i I và lo\u1ea1i II, trong \u0111ó có 2 h\u1ed9p lo\u1ea1i I, m\u1ed7i h\u1ed9p có 13 s\u1ea3n ph\u1ea9m t\u1ed1t và 2 ph\u1ebf ph\u1ea9m và có 3 h\u1ed9p lo\u1ea1i II, m\u1ed7i h\u1ed9p có 6 s\u1ea3n ph\u1ea9m t\u1ed1t và 4 ph\u1ebf ph\u1ea9m. Ch\u1ecdn ng\u1eabu nhiên trong thùng m\u1ed9t h\u1ed9p và t\u1eeb h\u1ed9p \u0111ó l\u1ea5y ra hai s\u1ea3n ph\u1ea9m \u0111\u1ec3 ki\u1ec3m tra, gi\u1ea3 s\u1eed hai s\u1ea3n ph\u1ea9m \u0111ó \u0111\u1ec1u t\u1ed1t thì xác su\u1ea5t \u0111\u1ec3 hai s\u1ea3n ph\u1ea9m \u0111ó thu\u1ed9c h\u1ed9p lo\u1ea1i I b\u1eb1ng:<\/p>","options":["A.<\/strong> $\\dfrac{9}{16}$<\/span>","B.<\/strong> $\\dfrac{6}{13}$<\/span>","C.<\/strong> $\\dfrac{4}{13}$<\/span>","D.<\/strong> $\\dfrac{52}{87}$<\/span>"],"correct":"4","level":"3","hint":"","answer":"Ch\u1ecdn D.<\/strong> $\\dfrac{52}{87}$<\/span>.<\/span><\/p>G\u1ecdi A : “Ch\u1ecdn \u0111\u01b0\u1ee3c trong thùng m\u1ed9t h\u1ed9p lo\u1ea1i I”. Và B : “Ch\u1ecdn \u0111\u01b0\u1ee3c trong thùng m\u1ed9t h\u1ed9p lo\u1ea1i II”. Xác su\u1ea5t ch\u1ecdn h\u1ed9p lo\u1ea1i I là $P(A)=\\dfrac{2}{5}$<\/span> và xác su\u1ea5t ch\u1ecdn h\u1ed9p lo\u1ea1i II là $P(B)=\\dfrac{3}{5}$<\/span>.<\/p>G\u1ecdi C là bi\u1ebfn c\u1ed1 “C\u1ea3 2 s\u1ea3n ph\u1ea9m l\u1ea5y ra \u0111\u1ec1u t\u1ed1t”. Xác su\u1ea5t l\u1ea5y \u0111\u01b0\u1ee3c 2 s\u1ea3n ph\u1ea9m t\u1ed1t t\u1eeb h\u1ed9p lo\u1ea1i I là $P(C|A)=\\dfrac{C^2_{13}}{C^2_{15}}=\\dfrac{26}{35}$<\/span>.<\/p>Xác su\u1ea5t l\u1ea5y \u0111\u01b0\u1ee3c 2 s\u1ea3n ph\u1ea9m t\u1ed1t t\u1eeb h\u1ed9p lo\u1ea1i II là $P(C|B)=\\dfrac{C^2_6}{C^2_{10}}=\\dfrac{1}{3}$<\/span>.<\/p>V\u1eady xác su\u1ea5t hai s\u1ea3n ph\u1ea9m l\u1ea5y ra t\u1eeb m\u1ed9t h\u1ed9p trong thùng \u0111\u1ec1u t\u1ed1t là <\/p>$P(C)=P(C|A).P(A)+P(C|B).P(B)=\\dfrac{26}{35}.\\dfrac{2}{5}+\\dfrac{1}{3}.\\dfrac{3}{5}=\\dfrac{87}{175}$<\/span>.<\/p>Áp d\u1ee5ng công th\u1ee9c Bayes, xác su\u1ea5t l\u1ea5y ra hai s\u1ea3n ph\u1ea9m \u0111\u1ec1u t\u1ed1t thu\u1ed9c h\u1ed9p lo\u1ea1i I là<\/p>$P(A|C)=\\dfrac{P(C|A).P(A)}{P(C)}=\\dfrac{\\dfrac{26}{35}.\\dfrac{2}{5}}{\\dfrac{87}{175}}=\\dfrac{52}{87}$<\/span>.<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2025-02-07 08:38:12","option_type":"math","len":0},{"id":"6256","post_id":"8428","mon_id":"1159285","chapter_id":"1159383","question":"Gi\u1ea3 s\u1eed 5% email c\u1ee7a b\u1ea1n nh\u1eadn \u0111\u01b0\u1ee3c là email rác. B\u1ea1n s\u1eed d\u1ee5ng m\u1ed9t h\u1ec7 th\u1ed1ng l\u1ecdc email rác mà kh\u1ea3 n\u0103ng l\u1ecdc \u0111úng email rác c\u1ee7a h\u1ec7 th\u1ed1ng này là 95% và có 10% nh\u1eefng email không ph\u1ea3i là email rác nh\u01b0ng v\u1eabn b\u1ecb l\u1ecdc. Gi\u1ea3 s\u1eed ch\u1ecdn m\u1ed9t email, tính xác su\u1ea5t \u0111\u1ec3 email ch\u1ecdn \u0111\u01b0\u1ee3c là m\u1ed9t email b\u1ecb h\u1ec7 th\u1ed1ng l\u1ecdc b\u1ea5t k\u1ec3 có là rác hay không.<\/p>","options":["A.<\/strong> 0,7956","B.<\/strong> 0,7965","C.<\/strong> 0,9756","D.<\/strong> 0,1425"],"correct":"4","level":"3","hint":"","answer":"Ch\u1ecdn D.<\/strong> 0,1425.<\/span><\/p>G\u1ecdi A : “Email \u0111\u01b0\u1ee3c ch\u1ecdn là email rác”. Và B : “Email \u0111\u01b0\u1ee3c ch\u1ecdn là email b\u1ecb h\u1ec7 th\u1ed1ng l\u1ecdc”. Vì 5% email nh\u1eadn \u0111\u01b0\u1ee3c là rác nên xác su\u1ea5t nh\u1eadn \u0111\u01b0\u1ee3c m\u1ed9t email rác là $P(A)=5\\%=0,05$<\/span>. Xác su\u1ea5t b\u1ecb l\u1ecdc c\u1ee7a email rác là $P(B|A)=95\\%=0,95$<\/span>. Xác su\u1ea5t email nh\u1eadn \u0111\u01b0\u1ee3c không ph\u1ea3i rác là $P(\\overline{A})=1-P(A)=1-0,05=0,95$<\/span>. Xác su\u1ea5t b\u1ecb l\u1ecdc c\u1ee7a email không ph\u1ea3i rác là $P(B|\\overline{A})=0,1$<\/span>. V\u1eady xác su\u1ea5t ch\u1ecdn m\u1ed9t email b\u1ecb l\u1ecdc b\u1ea5t k\u1ec3 là rác hay không là $P(B)=P(B|A).P(A)+P(B|\\overline{A}).P(\\overline{A)}=0,95.0,05+0,1.0,95=0,1425$<\/span>.<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2025-02-07 08:45:29","option_type":"txt","len":0},{"id":"6258","post_id":"8428","mon_id":"1159285","chapter_id":"1159383","question":"Gi\u1ea3 s\u1eed 5% email c\u1ee7a b\u1ea1n nh\u1eadn \u0111\u01b0\u1ee3c là email rác. B\u1ea1n s\u1eed d\u1ee5ng m\u1ed9t h\u1ec7 th\u1ed1ng l\u1ecdc email rác mà kh\u1ea3 n\u0103ng l\u1ecdc \u0111úng email rác c\u1ee7a h\u1ec7 th\u1ed1ng này là 95% và có 10% nh\u1eefng email không ph\u1ea3i là email rác nh\u01b0ng v\u1eabn b\u1ecb l\u1ecdc. Gi\u1ea3 s\u1eed ch\u1ecdn m\u1ed9t email, tính xác su\u1ea5t \u0111\u1ec3 email ch\u1ecdn \u0111\u01b0\u1ee3c là m\u1ed9t email rác b\u1ecb h\u1ec7 th\u1ed1ng l\u1ecdc.<\/p>","options":["A.<\/strong> $\\dfrac{9}{16}$<\/span>","B.<\/strong> $\\dfrac{6}{13}$<\/span>","C.<\/strong> $\\dfrac{4}{13}$<\/span>","D.<\/strong> $\\dfrac{1}{3}$<\/span>"],"correct":"4","level":"3","hint":"","answer":"Ch\u1ecdn D.<\/strong> $\\dfrac{1}{3}$<\/span>.<\/span><\/p>G\u1ecdi A : “Email nh\u1eadn \u0111\u01b0\u1ee3c là email rác”. Và B : “Email \u0111\u01b0\u1ee3c ch\u1ecdn là email b\u1ecb h\u1ec7 th\u1ed1ng l\u1ecdc b\u1ea5t k\u1ec3 có là rác hay không”. Vì 5% email nh\u1eadn \u0111\u01b0\u1ee3c là rác nên xác su\u1ea5t nh\u1eadn \u0111\u01b0\u1ee3c m\u1ed9t email rác là $P(A)=5\\%=0,05$<\/span>. Xác su\u1ea5t b\u1ecb l\u1ecdc c\u1ee7a email rác là $P(B|A)=95\\%=0,95$<\/span>. Xác su\u1ea5t email nh\u1eadn \u0111\u01b0\u1ee3c không ph\u1ea3i rác là $P(\\overline{A})=1-P(A)=1-0,05=0,95$<\/span>. Xác su\u1ea5t b\u1ecb l\u1ecdc c\u1ee7a email không ph\u1ea3i rác là $P(B|\\overline{A})=0,1$<\/span>. V\u1eady xác su\u1ea5t ch\u1ecdn m\u1ed9t email b\u1ecb l\u1ecdc b\u1ea5t k\u1ec3 là rác hay không là $P(B)=P(B|A).P(A)+P(B|\\overline{A}).P(\\overline{A)}=0,95.0,05+0,1.0,95=0,1425$<\/span>. Xác su\u1ea5t ch\u1ecdn m\u1ed9t email trong s\u1ed1 nh\u1eefng email b\u1ecb l\u1ecdc th\u1ef1c s\u1ef1 là email rác là Công th\u1ee9c Bayes: $P(A|B)=\\dfrac{P(A).P(B|A)}{P(B)}=\\dfrac{0,05.0,95}{0,1425}=\\dfrac{1}{3}$<\/span>.<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2025-02-07 08:52:18","option_type":"math","len":0}]}