🔙 Quay lại trang tải sách pdf ebook Giáo trình xác suất - thống kê và ứng dụng Ebooks Nhóm Zalo TRl/CfNG DAI HOC CONG NGHIEP TP.HCM KHOA KHOA HOC CO BAN TS. N g u y en P h u V in h (C hu b ie n ) T a p T h e G iao V ien To T o an K h o a K H C B G I A O T R ' l N H X A C S U A T ■ T H O N G K E V A U N G D U N G ■ DUNG CHO BAC DAI HOC V A CAO DANG t=x . L_ = J -;= t U A NHA XUAT BAN THONG KE - 2008 e 2 d t LCfl NOIDAU Xac su§'t thong k§ (XSTK) dtidc bien soan dtia tren de ctidng mon hoc cung ten do Bo Giao Due va Dao Tao qui dinh bat buoc cho sinh vien cac khoi nganh cong nghe va khoi nganh kinh te cua Trtidng Dai hoc Cong nghiep Tp.HCM. Rieng sinh vien khoi kinh te co the bo qua cac phan co lien quan den tich phan boi. Trong giao trinh hau het cac dinh ly deu dtidc chting minh mot cach chi tiet, tham chi co nhieu chting minh muon hieu, doi hoi ngtidi doc phai co trinh do toan nhat djnh, nhat la chticfng vectd nglu nhien, nhiftig chu trtidng khong bat buoc sinh vien phai hieu ttfcfng t$n cac chting minh, ma chi can nhd dinh ly va van dung chung vao bai t|p va thtic te. Song song do giao trinh van luon coi trong phan cong nghe thtic hanh tmh toan, nen sinh vien se bat gap nhting ham thong ke trong excel, cung nhti cac ham trong calculator 570MS diTdc long ghep trong phan tmh toan, de sinh vien co da dang cong cu tinh toan khac nhau. Co sau bang tra so lieu cac phan pho'i thong dung dtidc lap trinh so' bang MALAB, d dau m5i bang co vi du cach tra cu the tting tham so" cho tting bang so li6u. Tat ca bai tap, d£*u co lcTi giai trong sach Ngan hang c§u hoi XSTK vdi cung tac gia va da xuSt ban nam 2006. Sach duoc viet boi Tap the giao vien To toan thuoc Khoa Cd ban trifdng Dai hoc Cong nghiep Tp.HCM. Tap the tac gia cung xin bay to long cam chan thanh den Ban Giam Hieu trticfng Dai hoc Cong Nghiep Tp.HCM, dac biet Tign ST. Hieu Trtiofng Ta Xuan T l, Ngai da dong vien giup d3 de giao trinh nay kip ra mat phuc vu sinh vien va thay giao. Mac du cac tac gia da rat co gang nhOhg chac han khong tranh khoi thieu sot, bdi mot le "khuyet tat nglu nhien ton tai mot cach tat nhien trong bat ky mot ca'u true vat the nao", do la "tat nhien tan ltic, ngau nhien l6i lam" va chung ta dang di tim qui luat tCf trtic quan sinh dong trong khong gian lion don ngau nhien den qui luat tat nhien trong tti duy ttiu ttidng, tim cai tat dinh trong khong gian bat dinh. Day cung chmh la vi nhan sinh, la muc dlch cua mon hoc nay. Nhting y kien dong gop ve nOi dung va hinh thtic cua tap sach se dticJc tiep thu vdi sti tran trong va long biet cm sau sac, xin gdi ve TS.Nguyln Phu Vinh, Khoa Cd Ban, Trtidng Dai hoc Cong nghiep Tp.HCM. Tp.HCM, ngay 01 thang 04 nam 2007 Thay mat cac tac gia . Tien ST: Nguy§n Phu Vinh LCfl NOI DAU Xac suat thong ke (XSTK) difdc bien soan di/a tren de ctidng mon hoc cung ten do Bo Giao Due va Dao Tao qui dinh bat buoc cho sinh vien cac khoi nganh c6ng nghe va khoi nganh kinh te cua Trtidng Dai hoc Cong nghiep Tp.HCM. Rieng sinh vien khoi kinh te co the b6 qua cac phan co lien quan den tich phan boi. Trong giao trinh hau het cac dinh ly deu dtidc chting minh mot cach chi tiet, tham chi co nhieu chting minh muon hieu, doi hoi ngtidi doc phai co trinh do toan nhat dinh, nhat la chtidng vectd ngSu nhien, nhtftig chu trtidng khong bat buoc sinh vien ph£i hieu ttidng t3n cac chting minh, ma chi can nhd dinh ly va van dung chung vao bai tfp va thtic te. Song song do giao trinh vSn luon coi trong phan cong nghe thtic hanh tmh toan, nen sinh vien se bat gap nhting ham thong ke trong excel, cung nhti cac ham trong calculator 570MS dtidc long ghep trong phan tmh toan, de sinh vien co da dang cong cu tmh toan khac nhau. Co sau bang tra so' lieu cac phan phoi thong dung dtidc lap trinh so bang MALAB, d dau m5i bang co vi du cach tra cu the tting tham so cho tting bang so lieu. Tat ca bai tap, deu co lefi giai trong sach Ng§n hang c§u hoi XSTK vdi cung tac gia va da xuSt ban nam 2006. Sach dugc viet boi Tap the giao vien To toan thu0c Khoa Cd ban trticJng Dai hoc Cong nghiep Tp.HCM. Tap the tac gia cung xin bay to long cam chan thanh den Ban Giam Hieu trticfng Dai hoc Cong Nghiep Tp.HCM, dac biet Tien ST. Hieu Trtftfng Ta Xu3n T l, Ngai da dong vien giup d3 de giao trinh nay kip ra mat phuc vu sinh vien va thay giao. Mac du cac tac gia da rat co gang nhtihg chac han khong tranh khoi thieu sot, bdi mot le "khuye't tat ng§u nhien ton tai mot cach tat nhien trong bat ky mot cau true vat the nao", do la "ta't nhien tan ltic, ngau nhien 16i lam" va chung ta dang di tim qui luat tti trtic quan sinh dong trong khong gian hon don ngSu nhien den qui M t tat nhien trong tti duy ttiu ttidng, tim cai tat djnh trong khong gian bat dinh. Day cung chinh la vj nhan sinh, la muc dich cua mon hoc nay. Nhting y kien dong gop ve nfli dung va hinh thtic cua tap sach se dticJc tiep thu vdi sti tran trong va long biet On sau sac, xin gefi ve TS.Nguyen Phu Vinh, Khoa Cd Ban, Trtidng Dai hoc Cong nghiep Tp.HCM. Tp.HCM, ngay 01 thang 04 nam 2007 Thay mat cac tac gia . Tien ST: Nguyln Phu Vinh LCfl n 6 i d a u Xac sua't thong k§ (XSTK) dufdc bien soan dtia tren de ctidng mon hoc cung ten do Bo Giao Due va Dao Tao qui djnh bat buoc cho sinh vien cac khoi nganh c6ng nghe va khoi nganh kinh te cua Trtidng Dai hoc Cong nghiep Tp.HCM. Rieng sinh vien khoi kinh te co the bo qua cac phan co lien quan den tich phan b0i. Trong giao trinh hau het cac djnh ly deu dtidc chting minh mot cach chi tiet, tham chi co nhieu chting minh muon hieu, doi hoi ngtfdi doc phai co trinh do toan nhat djnh, nhat la chtfdng vectd ngau nhien, nhtfng chu trtidng khong bat buoc sinh vien ph&i hieu ttidng tan cac chting minh, ma chi can nhd dinh ly va van dung chung vao bai t$p va thtic te. Song song do giao trinh van luon coi trong phan cong nghe thtic hanh tmh toan, nen sinh vien se bat gap nhting ham thong ke trong excel, cung nhti cac ham trong calculator 570MS dUdc long ghep trong phan tmh toan, de sinh vien co da dang cong cu tinh toan khac nhau. Co sau bang tra so' lieu cac phan phoi thong dung dtidc lap trinh so bang MALAB, d dau mdi bang co vi du cach tra cu the tting tham so cho tting bang so' lieu. Tat ca bai tap, d£»u co lcTi giai trong sach Ng$n hang cSu hoi XSTK vtfi cung tac gia va da xuat ban nam 2006. Sach duoc vie't bdi Taip the giao vien To toan thuoc Khoa Cd ban trtidng Dai hoc Cong nghiep Tp.HCM. Tap the tac gia cung xin bay to long cam chan thanh den Ban Giam Hieu trtidng Dai hoc Cong Nghiep Tp.HCM, dac biet Tien ST. Hieu Trtidng Ta Xuan Te, Ngai da dong vien giup dd de giao trinh nay kip ra mat phuc vu sinh vien va thay giao. Mac du cac tac gia da rat co gang nhiftig chac han khong tranh khoi thie'u sot, bdi mot le "khuye't tat ngiu nhien ton tai mOt cach tat nhien trong bat ky mot ca'u true vat the nao", do la "ta't nhien tan ltic, ngau nhien 16i lam" va chung ta dang di tim qui luat tti triTc quan sinh dong trong khong gian hon don ng2u nhien den qui luat ta't nhien trong tti duy ttiu ttidng, tim cai ta't djnh trong khong gian ba't djnh. Day cung chinh la vj nhan sinh, la muc dich cua mon hoc nay. Nhting y kie'n dong gop ve noi dung va hinh thtic cua tap sach se dtidc tiep thu vdi sti tran trong va long biet dn sau sac, xin gdi ve TS.Nguyin Phu Vinh, Khoa Cd Ban, Trtidng Dai hoc Cong nghiep Tp.HCM. Tp.HCM, ngay 01 thang 04 nam 2007 Thay mat cac tac gia . Tien ST: Nguyen Phu Vinh Left noi d^u Muc luc.... MUCLUC ...3 ....4 PHAN I: LY THUYET XAC SUAT CHl/CfNGl: XAC SUAT BIEN CO VA CONG THLfC XAC SUAT.......................................................................5 CHl/CfNG 2: BIEN NGAU NHIEN (BNN).......................... 27 CHUCfNG 3: CAC DINH LY G I0I HAN..............................93 CHl/CfNG 4: VECTCf NGAU NHIEN (VTNN)..«................ 113 PHAN II: LY THUYET THONG KE CHl/CfNG 5: LY THUYET M AU......................................... 198 CHl/CfNG 6: l/CfC LI/0NG DAC TRl/NG DAM DONG... 217 \CHl/CfNG 7: KIEM DINH GIA THIET THONG K E ___ 241 CHl/CfNG 8: LY THUYET Tl/CfNG QUAN......................... 271 BANG SO A, GIA TRI HAM MAT D O ............................... 298 BANG SO B, TICH PHAN LAPLACE................................ 299 B ANG SO C, PHAN PHOI STUDENT.... ...........................300 BANG SO D, PHAN PHOI CHI BINH PHl/CfNG.............302 BANG SO E, PHAN PHOI FISHER - SNEDECORE........306 BANG SO F, PHAN PHOI CHUAN DCfN GIAN N(0,1) ... 314 TAI LIEU THAM KHAO ....................................................... 316 Chufdng 1: Xdc suaft cua bien co v£ c ic cone thtfc xac suat. 5 PHAN I: LY THUYET XAC SUAT Chtfctag 1 Xac suat cua bien cd" va cac cong thufc xac suat I. Xac sufi't cua bien co' 1.1. Dinh nghTa co dien v i xac suSft Trong xac su§t, ta hieu phep thu* nhu* la viSc thifc hien mot nh6m dieu kien xac djnh (chSng han lam thi nghiem) de nghlen clTu mot doi tifdng hay mpt hien tutfng nao do. Ket qua cua — DheD thCraol la bien_co. --------------- Vi du: Cac vi du ve phep thu* - Tha hon bi tir do cao lm , bien co: "hon bi rcfi xuong". - Gieo 1 dong xu, bien co: "dong xu sap", "dong xu ngfra". - Gieo 1 con xuc x3c, bien co: "xuat hien mat 3 cham*. - Kiem tra 1 SV ve mon XSTK, bien co:’ "dat", "khong dat". 1.2. Cac loai bien co 0B i|i^c6 c'hSc chin (ky hi§u Q) la bien co nhat djnh xay ra khittuJeirffn^phep thu*. @Biifr^c6 khong the (ky hieu 0) la bien co nhat djnh khoncfxay ra RTfi thut hien phep thir. Q Bien co tonq: C la bi£?n co tong cua hai bien co A, B, ky hieu la C=AuB, neu C xay ra khi va chi khi it nhaft m6t trong hai bien co A hay B xay ra. (p)Bien co tich: C la bien co tich cua hai bien co A, B, ky hieu la C=AnB=A.B=AB, neu C xay.ra khi va chi khi ca hai bien co A hay B deu xay ra. Vf du: -Bien co "hon bi rcfi xuong" la Q. -Bien co "xuat hien dong thcli mSt s£p va ngfra" khi gieo dong xu la 0. 8 Chitong 1: X£c suat cua bien co va c£c cong thtfc x£c su£t. « f o - ( c |c J + c |c J ) Cach 2: Ta dung BC doi l|p : P(c) = f16 c 10 • Han che: Djnh nghTa theo quan dielm co dien con nhtfng han ch§ sau: + s6 lifting cSc bcsc la hull han. + tmh chat dong kh& nSng kh6ng phai bao gid cung xac djnh dutfc. Clilnh v) the ta co mot djnh nghla khac nhir sau: I I I . Djnh nghla xac suat theo thong ke: Neu I3p lai n lln phep thu*, trong do co m I3n xu§t hien bien co A thi ty s6: f„(A) = — dutfc goi la tBn suSt cua ft bien c6 A. Vdi n du I6n thi tl so nay co gi6i han being so p nao do, di/dc goi la xac su§t cua A. P(A)= Um f„ (A ) = p n—»oo Ta se th§y y nghla cua dinh nghla n&y qua djnh ly Bernuilli d chutfng 3 ro rang hdn. - Uu di£m: Kh6ng gian bcsc Q g$m v6 han bcsc mS kh6ng din doi hoi tmh dong kha nSng. - Nhu’dc diem: ©di h6i phai lap lai nhieu Ian phep thCr. Trong thi/c te, nhiSu bai toSn kh6ng cho phep th u t hien do dieu kien va kinh phi lam phep thir. V i.d n li Trong th6ng ke dSn s6, ngi/di ta da tong ket dutfc xac suSt em be ra ddi la trai hay gai x§p xi being 0,5. b Vf du2: Xac suat dugc m ^t sap khi tung dong xu la —.2 IV. Djnh nghla x£c suat b ln g hinh hoc Chitting 1: X£c suat cua bien co va c ic c6ng thtic xac suat. 9 Cho m i4n f t . Goi do do cua ft la do dai, dien tich, t h i tich (ting vdri ft la diffrng cong, m ien phang, khoi). Goi A la bien co la (diem, dtf&ng cong, m ien phang, kho'i) M e S d f t . Ta c6 - P(A) = ^ . d6 do f t Vf du: Tim x^c suat cua diem M rcri vao hinh tron noi tiep hinh vuong canh 2 cm. Giai: _ dodoS _ ;rx l 2 _ /r do do ft ~ 4 ” 4 Ngoai ra mot djnh nghla tong quat nhat cua xac suat la theo ti§n de do do, do nha toan hoc Nga Kolmogorop nam 1933, dieu nay da day XSTK len mot tarn cao m6i, nhu*ng cut ky phut tap, ta khong xet trong giao trinh nay. V. Tmh chat va y nghla cua xac suat: 5.1. Tmh chat cua x£c s u it i. P(0)=O ii. P(Q)=1 iii. 0 | = 30-10 = 20, |(Z )n r)\C | = 25-10 = 15, |(Z> n C) \ rj = 30 -10 = 20, | C\(D uT)! = 60-(20+ 10+ 20) = 10, |Z )\(C u 7 ’)| = 75-(20 + 10 + 15) = 30, |r\(C u Z ))| = 7 0 -(2 0 + 10+ 15) = 25, vay |(7, uCuZ>)| = (20 + 10 + 30 + 20 + 10 + 15 + 25) = 130, Vay tong so'phan tti = |ru C u £ > | + 50 = 130+50=180 . 130 13 P(TuCuD) =130 + 50 18 Cach khac: 12 Chtfcfng 1: Xac suat cua bien co va c£c c6ng thtic x£c suat. P{T'u CkjD) = P (T ) + P[C )+ P{ D )-P {T r \ C ) - P [T r \ D ) - P [ C n D ) , ^ . 60 75 70 30 25 30 10 130 _ + P ( T u C(j D) =---- + -----+ --------------------------- + -----= ----- » v ’ 180 180 180 180 180 180 180 180 Vi du3: L6p hoc day bon mon: co 55 em gioi dai so' (A=Algebra), co 65 em gi6i sinh hoc (B=Biology), co 65 em gioi may tmh (C=Computer), co 75 em gi6i thiet ke do hoa (D=Designer), co 20 em gioi A&B, co 30 em gi6i A&C, co 30 em gioi A&D, co 35 em gioi B&C, co 45 em gi6i B&D, co 50 em gi6i C&D, co 10 em gi6i A&B&C, co 15 em gi6i A&B&D, co 25 em gioi A&C&D, co 30 em gioi B&C&D, co 10 em gi6i ci bo'n A&B&C&D va co 10 em khong gi6i mon nao. Gi6i mot mon thi dUcfc thifdng. Goi ten NN mot em, tinh XS de em do diftfc thifcfng. Giai: Vi Ar^BnC A n B n C n D va \Ar\B r\C\ = \ A n B r^C r\D\ = \Q nen ta co hinh ve difdi day. Bang phtfdng phap loai truf dan ta se co cau true phan tuf nhif hinh ve tren. Tong so phan tuf = \A u B u C u D\ +15 = = |Z)| + 5 + 5 + 10 + 5 + 5 + 5 +15= 75+35+15 =125 Va theo cong thtic xac suat co dien ta co ket qua: P ( A u B u C u D ) = Z)|+ 5 + 5 +10+ 5 + 5 + 5 75 + 35 110 125 125 125 Chiftftig 1: XAc suaft cua bien co v i cic cong thufc xac suat. 13 Cach khdc: P ( A v B v C u D ) = P{A) + P(B) + P(C) + P(D) - P(A n B) - P(A n C) -P( A n D ) - P(B n C )- P(B n D ) - P{C n D ) + + P ( A n B n C ) + P ( A n B n D ) + P ( A n C n D ) + P ( B n C n D ) - -P (A nB nC nD ) 55 65 65 75 20 30 40 35 45 50 ~ 125 125+ 125 125 125 125 125 125 125 125+ 10 15 25 30 10 110 H-------1------H------ + ------------ —----- S 125 125 125 125 125 125 ]6L2l Xac suS't co dieu kien, cong thtfc nhan xac suat. 6.2.1. Xac suat co dieu kien - Dinh nahia: Cho 2 bien co A va B. Xac su§t co dieu kien cua^A vdi dieu kien B, ky hi§u P(A/B), la xac suat cua A du'dc tmh sau khi B da xay ra. - Cong thtic tm h: P(Al B) = => P(AB) = P(B)P(A / B) = P(A)P(B / A) "(•o) 14 Chifdng 1: Xac suat cua bien co v i cdc cone thufc xac suat. Vj d u l: Mot cong ty can tuyen 2 nhan vien. Co 6 ngi/di nop trong do co 4 ntf va 2 nam. Kha nang dutfc tuyen cua moi ngutfi la nhif nhau. Tinh xac suat de a/ ca 2 ngutfi nu' dutfc chon, biet ring co It nhat 1 ngutfi ntt da dutfc chon. b/ Neu Hoa la mot trong 4 ntf, tinh xac suat de Hoa dutfc chon, suy ra xac suat de Hoa dutfc chon vdi dieu ki e n r3ng co it nhat 1 ngutfi nu' da dutfc chon. Giai: a/ Ta dung xac suat co dieu kien: C: BC co 2 nur difdc chon. D: BC co it nhat mot nff dtfdc chon. b/ Xac suat de Hoa difcfc chon la: ~ = — , do do xac suat de Cl 15 5 Hoa difcfc chon vdi dieu kien da co mot ntf dtfdc chon la: — = — 14 1 4 15 Vi du2: Mot cong ty can tuyen 2 nhan vien. Co 6 ngu'di nop dcfn, trong do co 4 nu1 va 2 nam. Tuyen NN lien tiep 2 ngi/di. Tinh xac suat de ca 2 ngutfi nur dutfc chon. Giai Ta dung xac suat co dieu kien: P(Ai A2) = P(A l ) p [ Ay /A^j 4 3 = 2 = C^ 6 5 ~ 5 c l ChiftJnp 1: Xdc suat cua bien co" va cac cong thtfc xac suat. 15 V« du3: Mot xi/dng phim, tuyen 100 ngifcfi, tiong do co 40 ngUfri la nff, co 10 ngirdi la d vi tri quan ly, trong do co 5 ngifcfi vijfa la ntf vCfa la quan ly. Goi ngSu nhien mot ngifcfi. Tinh XS de: a/ la nu\ b/ la ngtfdi khong phai d vi tri quan ly. c/ la quan ly nhtfhg vdi dieu kien lai la nff. Giai: 40 a/B: BC la ntf: P{B) = 100 -X 100-10 90 b/ Q: BC la vi tri quan ly: p [q ) = 100 100 c/ Ta dung xac suat co dieu kien:P(/l//?). 6.2.2. Cong thufc nhan: P{AnB) 5/100 1 P(B) 40/100 8 Bien co doc Up : 2 bien co A va B goi la doc lip neu P(A/B) = P(A) (hoac P(B/A)=P(B)), tift la si/xay ra hay khong cua bien co nay khong anh hi/dng den kha nang xay ra cua bien co kia. Chu y: ♦ Bien co A, Bdoc lap -> A, B doc lap. ♦ A va B la 2 bien co doc lap neu B co xay ra hay khong cung khong anh hiforng den kha nang xay ra A. Ta co P(AB) = P(A).P(B) ♦ Vdi A, B,C khong doc lap thi : P(AB) = P(A n 5 ) = P(B)P(A / B) P(ABC) = P(A)P(B / A)P(C / AB) ♦ Md rong: ^ P{A\A2A$....An) = P(A X )'P(AZ / a x).p (a z / AxA1).P(Aa I AxA2/^):.. p ( A / A. A- A~ A A 1 ■ ’r 16 Chifdng 1: X£c suat cua bien co \ k c£c c6ne thufc xac su a t Vf d u l: 'Itu JA. Ba vien dan doc lap bin vao 1 bia Xac suat trung dlch cua vien thu* nhit, vien thu* hai, vien thuf ba tUdng CiTng la 0,4; 0,5; 0,7. Tim xac suat de: a/ Co dung 1 vi§n trung dlch. b/ Co It n h it 1 vien trung dlch. Giai: a/ A i: BC vien thu* nhSt trung dlch. A2: BC vien thu* hai trung dlch. A3: BC vi§n thu* ba trung dlch. A: BC co dung 1 vien trung dlch A — A j A j A j u Aj A 2 A j u A x A j A^ P(A ) = p (a 1 A ^ A 3) + p ( a1A1 ^ ) + p (a i A^A3) b/ A: BC co It nhat 1 vien trung dlch. P { T ) = p ( A ^ A ^ A i ) = > P ( A ) = i ~ p ( T ) U Vj du2: TCf 16 san pham co 20 san pham trong do c6 5 san pham xau, liy lien tiep 2 sin pham (khong hoan lai). Tinh xac suat de ca 2 san phcfm deu la san ph3m xiu. Giai: 5 4 . n / j . 4 5 Cf P(Al ) = ^ - , P \ * 2 / = -!= > P(A\ o^[2) = — — = —f v w 2 0 I / A lJ 1 9 V l 2 ) 192Q c 2q (lay cung mot luc) ■ V j du3: Trong tren, ISy lien tiep 3 san phSm (khong hoan lai). Tinh xac suat de c i 3 san pham deu xau. Giai: Tuong tu bai tren:P(A ) = A ± A = i i _ (i£y cftng mot luc) ■ Vi du4: Mot to co 4 nam va 3 nuf. Chon lien tiep 2 ngtfcri. Tim xac suat de: yhyjggg i: Xac suat cua bien co va cic c6ng thufc xac suat. 17 a/ Ca 2 l i nuf. b/ Co 1 nam v& 1 nur. rfT Q iai: B at SW Ai: “ chon dUtfc nuf d lan thuf i”. Bi:“chon diXofc nam d lan thuf i”. aJ Goi A: “ chon difcfc 2 nuf”. Ta co A = A,A2 => Pi A) = P(AlA1) = P(A,)P(A2/A,) = 1 4 = \ ■ " lb 7 b/ Goi B:“chon difcrc m ot nam va mot nuf”. Ta c6 P(B) = P(AxB2 U A2Bl ) xk P(AxB2) + P(A2Bx) = P(AX )P(B2 / Ax ) +P(BX )P(A2 / Bx) 3 4 4 3 4 = —x - + - x '- = — 7 6 7 6 7 Cach khdc: Dung x£c suat c8 dien (lay cung mot luc): c \c \ 4 C72 7 ' Nhan xet: Dufa v&o vl du 2,3,4 ta nhan xet rang viec xac suat lay lien tiep lan lufoft n vat, m6i lan mot vat va khong hoan lai, th i ttfcrng difomg vdi xac suat lay cung luc n vat. \ , ' V II. Cdng thufc xac suat d iy du, cong thufc Bayes Ta xem hinh duoi day: Tap Q dirge phan cat thanh n tap rdi nhau, ta. ' goi: • He day du modi: Ai, A2, A3, A4 ,.....An rdi nhau c Q , thoa: DAI HOC TRAi NGUYEN TRDUG TAM HOC Lltl) 18 Chifdng 1: Xdc suat cua bien co v i c ic cone thtfc xac suat, i=n Al 'uA2 u... theo cong thtic nhan ta co: P(B) = P (A ,)p [ B /Ay p { A 1)p[B/Ay . . . _ +P[A„)p{B/A'' 7.2. Cdng thufc Bayes (gia thiet, aia sufl: Xac suat de xuat hien Ak vtfi gia suf da xuat hien B la: , / ) p { M bA ) ' B J P{B) That vay: P \^k B Vf dul: p ^ i % J r ( A k B) P(B) P(B) Co 2 lo san pham. Lo 1 co 20 san pham, trong do co 15 san pham tot. Lo 2 co 20 san pham, trong do co 10 san pham tot. Lay ngau nhien 1 lo va tu* lo do chon ngiu nhien 1 san pham. a/ Tim xac suat de san phSm lay ra la san pham tot. b/ Gia sir san pham lay ra la san pham tot. Tinh xac suat de san pham do thuoc lo thu* nhat, lo thu* hai. Giai: b/ PP(A)P (h )P \Ay P(A) Vi du2: Co 3 hop thuoc. Hop 1 co 5 ong tot va 2 ong xau. H6p 2 co 4 ong tot va 1 ong xau. Hop 3 co 3 ong tot. Lay ngau nhien 1 hop va tu* hop do rut ngau nhien 2 ong thuoc. 20 Chifdng 1: Xac suat ctia bien co v i c£c c6ne thtfc xac suat. a/ Tim xac suat de di/dc 1 ong thuoc tot va 1 ong thuoc xau. b/ Gia su* khi rut 2 ong thuoc, ta thay co 2 ong thuoc tot. Tim xac suat de cac ong do d hop 2. Giai: a / P ( » i ) = i = P (H j) = P ( //3 ) , ! ; 1 c \c \ 1 c \c \ 1 0 = —X , + -X . -I— X—r 3 C72 3 c | 3 c | b/ B: BC difcfc 2 ong thuoc tot. 1 c f 1 c l 1 c f = — X — H— X — H— X —~ 3 Cj 3 cj 3 c | ' 2/ | = B P(B) Vi d M? : Mot nha may san xuat bong den co 2 phan xi/dng 1 va 2. Biet ring phan xuting 1 san xuat gap 4 lan phan xuting 2, ty le bong den hu* cua phan xi/dng 1 la 10%, phan xu’dng 2 la 20%. Mua ngau nhien 1 bong den cua nha may. a/ T]m xac suat de bong den n&y hu*. b/ Gia sCf mua phai bong hu*. Tim xac suat de bong den nay thuoc ph§n xu’dng 1, ph5n xutfng 2. Giai: a / p ( * i ) = ; - ^ ( ^ 1 = 7 P(A) = P ( X t ) p ( f x J +P ( X 2) p [ f x J =4 10 1 20 5 100 5 100 Chifdng 1: X£c suat cua bien co v i c£c cdng thtic xac suat. 21 . . r < . x i) p [ y y . \ ,v/N b/ P f v i - l p fV V -— . P(/4) ’ I , P(/l) Vf du4: Co 3 hop gio'ng nhau: hop I chtia 20 bi tr^ng; hop II chtia 10 bi tra n g v& 10 bi xanh; hop III chtia 20 bi xanh. Chon nglu nhien m ot hop v& tir d6 gia sti boc ngau nhien ra dticrc 1 bi trang. Tim xac suat de vien bi d6 1& cua hop I. Giai: Goi Ak: “ chon h6p thti k” (k = 1; 2; 3). Suy ra he {Ak) day du vk xung khac. Goi B: “ boc dtfdc bi tran g ”. P W B j . , j . X P(Aj)P(BIAj) 3 j 1 BAI T^P4 1/ Mot thu quy co chum chia khoa gom 9 chiec chia gio'ng het nhau, trong do 2 chia co the md difdc tu. Anh ta thuf ngau nhien tting chia (chia khong trung difdc bd ra ngoai). Tim XS de anh ta md difdc tu dung vao lan thti 3. 2/ Hai Sv chdi mot tro chdi nhif sau: Ca hai luan phien lay moi lan mot bi tti mot hop difng 2 bi trang va 4 bi den (bi difcfc rut ra khong tra lai vao hop). Ngifdi nao lay ra difdc bi trang tnfdc thi thang cuoc. Tinh XS thang cuoc cua ngifdi lay trifdc. 3/ Hai Sv chdi mot tro chdi nhif sau: Ca hai luan phien lay mdi lan mot bi t£f mot hop difng 2 bi trang va 4 bl den (bi difdc rut ra khong trd lai vao hop). Ngifcfi nao lay ra difdc 2 bi den trifdc thi thang cuoc. Tinh XS thang cu6c cua ngifcfi lay tnfdc. 4/ Mot xifdng phim, tuyen 100 ngifcfi, trong dd cd 40 ngifdi la nff, cd :i 5 ngifdi vifa la nff vffa la quan ly. 22 Chtfdng 1: Xac suat cua bien co v i cac cone thtfc xac su a t goi nglu nhien m6t ngifdi. Tinh XS de: a/ la nff. b/ la ngifdi khong phai d vi tri quin ly. c/ la qudn ly nhifng vdi dieu kien lai la nff. 5/ Co 4 hop difng bi, difdc danh so tff 1 den 4. Hop 1 co 5 X, 3D, 2V. Hop 2 co 6 X, 8D, 6V. Hop 3 co 4 X, 7D, 4V. Hop 4 co 8 X, 5D, 7 V. Ngoai ra con co mot hop nho difng 4 chff so' 0,1,2,2. Lay NN hai chff so tff hop nh6, roi lay NN 1 bi tff h6p bi co danh so bang tong hai chff so vffa lay difdc. a/ Tim XS de difdc bi do. b/ Biet rang lay difdc bi xanh, tim XS de bi do difdc lay tff hop 1. 6/ Mot ngifcfi co ba ngifcfi tinh, viet 3 la thif cho 3 ngifdi tinh, co 3 phong bi. Tinh xac suat de: a/ Ghep dung hoan toan. b/ Ghep sai hoan toan. c/ Co it nhat ghep mot cai dung. 7/ Tung 3 con xuc sac, X bien cd tong. Tim luat phan phoi cua X. 8/ Mot ngifcfi co n ngifcfi tinh, viet n la thif cho n ngifcfi tinh, co n phong bi. Tinh xac suat de co it nhat ghep mot cai dung. 9/ Tren ban co san 5 dong xu (3N,2S), tung them 3 dong xu nffa, sau do khoanh lay NN 4 dong xu. Tinh: a/ XS de chi co 3 N. b/ Lap phan phoi XS cua X la bien cd cac dong xu ngifa (N). 10/ Mot binh chffa 10 bi, va co 4 bi d6. Lay NN lan I ra 2 bi de tren ban, sau do lay lan II ra 1 bi nffa. a/ Tinh XS de lan hai lay ra difdc 1 bi do. ChiftJng 1: X£c suat cua bien co v i cac cong thtfc xac suat. 23 b/ Tinh XS de tdng hai lan lay ra chi difdc 2 bi do. 11/ Trong nhom gom 10 Sv di thi co 3 Sv chuan bi tot, 4 Sv chuan bi kha, 2 Sv chuan bi trung binh va mot chuan bi kem. Trong cac phieu thi co 20 cau h6i. Sv chuan bi tot co the tra lcfi difdc ca 20 cau, chuan bi kha tra lcfi difdc 16 cau, chuan bi trung binh tra ldi difdc 10 cau, Cdn Sv kem co the tra ldi 5 cau. Mot Sv difdc goi NN tra ldi difdc 3 cau hoi tuy y. Tmh Xs de Sv do: a/ difdc chuan bi tot. b/ difdc chuan bi kem. 12/ Mot binh chtfa 10 bi, va co 4 bi do. Lay NN lan I ra 1 bi de tren ban, sau do lay lan II ra 2 bi nffa. a/ Tinh XS de lan II lay ra chi difdc 1 bi do. ' b/ Tinh XS de tdng hai lan lay ra chi difdc 2 bi do. c/ Chdi tro chdi nhif sau: Neu tong hai lan lay ra t£f m6t den hai bi do, toi difdc 4$, con lai toi phai mat 7$, toi co ldi khong? 13/ Hai binh m6i binh chtfa 10 bi, va m6i binh co 4 bi do. Lay tuf binh I ra 2 bi bo vao binh 2, sau do lay tif binh II ra 3 bi (xet lay cung luc). X la so bi do difdc lay tif binh II. a/ Tim phan phoi XS cua X. b/ Gia suf X=2 bi do, thi bien co' X phu thuoc nhieu nhat vao bien co nao d lan lay thtf nhat. 14/ Hai binh m6i binh chtfa 8 bi, va m6i binh co 3 bi do. Lay tif binh I ra 3 bi bo vao binh II, sau do lay tif binh II ra 4 bi (xet lav cung luc). X la so" bi do difdc lay tif binh 2. a/ Tim phan phoi XS cua X. b/ Gia suf X=3 bi do, thi bien co X phu thuoc nhieu nhat vao bien co nao d lan lay thi? nhat. c/ Tim phan phoi XS cua X, neu tif binh II lay ra 4 bi, xet lav co hoan lai. 24 Chitong 1: Xdc suat cua bien cd va cdc cone thufc xac suat. 15/ Lan I rut 2 la bdi trong b6 52 Id de trSn ban. Lan II rut them 2 Id nffa de tren ban. Sau do khoanh NN 2 la. X la so la cd co trong 2 la khoanh sau cung. a/ Tim phan phoi XS cud X. b/ Tinh XS trong 2 la do chi c6 1 con cd. 16/ Lan I rut 3 la bai gdm 2 cd, 1 ro trong b§ 52 Id bai de tren ban. Lan II rut them 2 la nffa de tren ban. Sau do khoanh NN 2 la. X la so 1 d cd co trong 2 la khoanh sau cung. a/ Tim phan pho'i XS cua X. b/ Tinh XS trong 2 la do chi co 1 con cd. 17/ Lan I rut 2 la bai trong b6 52 Id bdi de tren ban. Lan II rut them 1 la nffa de tren ban. a/ Tinh XS trong 3 la do la ba con cd. b/ Tinh XS rut lan II la do la m6t con cd. c/ Co mot tro chdi nhif sau: Gid suf rut lan n, la do la m6t con cd. Vdi bien cd lan I rut difdc 2 la cd thi toi difdc 5$, chi difdc mot la cd toi difdc 2$, con lai (khong co Id cd nao ) toi mat 1$. Vay toi co ldi trong tro chdi nay khong? 18/ Lan I rut 1 la bai trong bo 52 la bai de tren ban. Lan II rut them 1 la nffa de tren ban. a/ Tinh XS de la thtf hai la con cd. b/ Vdi hai la bai tren ban. Goi Ai la bien co" co i la cd, i=0,l,2. Chffng minh he nay la day du. c/ Mot tro chdi dat ra la: Neu cd hai deu khong co la cd, thi toi phai mat 10$, Neu chi co mot la cd, thi toi difdc 15$, cdn lai toi mat 2$. Vay trung binh mot lan chdi toi ldi, 16 bao nhieu. 19/ Lan I rut 2 la bai trong bo 52 la bai de tren ban. Lan n rut them 1 la nffa de tren ban. a/ Tinh XS de la thtf hai la con cd. b/ Tinh XS 3 la bai tren ban la 3 la cd. Chitons 1: X£c suat cua bien co va c£c cong thtic xac suat. 25 20/ C6 2 cay sung cung ban vao m6t bia, XS sung I ban trung bia la 70%, XS sung II ban trung bia Id 80%. a/ Sau khi ban hai phat. Tinh XS de sung II ban trung bia. b/ Sau khi ban hai phat. Tinh XS de c6 m6t sung ban trung bia. c/ Sau khi ban hai phat, ngifcfi ta thay rang co 1 vidn dan trung bia, Tinh XS de vien dan trung la sung II. d/ Sau khi ban hai phat, ngifdi ta thay rang c6 1 vien dan trung bia, Tinh XS de vien dan trung la sung I. 21/ Tai hoi chd co 3 loai cuTa hang. Cifit hang I phuc vu cho nhffng ngifcfi may man, bin hang co ty le phe pham la 1%. Cifa hang II phuc vu cho nhffng ngifdi binh thifdng, tan hang co ty le phe" pham la 5%. Cifd hang ID phuc vu cho nhffng ngifdi rui ro, ban hang co ty le phe pham la 10%. Mot ngifdi vao hoi chd phai gieo 2 dong xu. Ngifdi do la may man neu ca hai deu sap, la rui ro neu ca hai deu ngufa. Con lai la binh thifdng. a/ Tinh XS de mot ngifdi vao hoi chd mua phai hang phe' pham. b/ Neu phdi mua phai hang phe pham. Thi theo ban ngifdi do may man hay rui ro, hay binh thifcfrtg?. c/ Gia suf ngifdi do vao hoi chd mua phai hang phe pham the thi trung binh ngifdi do phdi tra gia bao nhieu cho mon hang phe' pham do, biet rang gia phe' pham tai cffa hang I, n, HI tifcfng tfng la: 9$, 6$, 4$. 22/ A.nh A clidi tro chdi nhif sau: tung 2 con xuc sac, X bien co' tong. a/ X < 4, A mat 6 dong, b/ X > 9, A mat 5 dong, c/ Con lai, A difdc 8 dong. H6i A cd nen chdi khong? 23/ Tung 3 con xuc sac tinh tdng ba mat, A chdi tro chdi: tong < 7, A difdc 8$, tdng > 15, A dUdc 9$, con lai A mat 3$. Tro chdi co ldi cho A? 26 Chtfdng 1: Xac suat cua bien co va cic c6ng thufc xac suat. 24/ Mot binh chtfa 10 bi, va co 4 bi do, 3 bi vang. Lay NN lan I ra 2 bi de tren ban, sau do lay lan II ra 2 bi nffa de tren ban. a/ Tinh XS de lan II lay ra chi difdc 2 bi do. b/ Tinh XS de tdng bi hai lan lay ra chi co 2 bi d6 va 1 bi vang. c/ Cff hai lan nit bi nhif the, toi chdi tro chdi nhif sau: Neu tdng hai lan lay ra: • Co tff mot den hai bi do va khong co bi vang nao: toi difdc 4$. • Co tff hai bi vang trd len, toi mat 7$. • Ngoai hai trifdng hdp tren, con lai toi difdc 2$. Toi co ldi khong trong tro chdi nay? Chiftfng 2: Bien ngau nhien (BNN). 27 Chtftfng 2 BIEN NGAU NHIEN (BNN) I. Bien ngiu nhien Dinh nghla: Bien nglu nhien (bnn) (con goi la dai lifdng nglu nhien). / . / Anh xa X tuf khong gian mau Q vao tdp so thifc R: x-.n - * r w e X,n> X ( w ) e R diftfc goi la bnn. Thifdng difcfc ky hieu la: X,Y,Z.... • Bien ngau nhien rcfi rac: Neu X (ft) la hffu han hay vo han dem difcfc. • Bien nglu nhien lien tuc: Neu X(Q) la mot khoang hay mot so khoang hay toan bo R. Vi dul: Ban lien tiep n vien dan doc lap vao bia. Goi X la so" vien dan trung dich => X = {0,1,2, , la bnn rcfi rac. Vi du2: Tuoi tho cua mot thiet bi ta bat dau khao sat la bnn: Cl = {/: 0 < / < °°} = [0, °o( => X(t ) = t , la bnn lien tuc. U. Luat phan phoi xac suat cua bien ngiu nhien 2.1 Bang ph£n phoi xac suat cua bnn rcfi rac Ta co bang phan phoi sau: X Xl X2 Xi . xn P(X) Pi P2 Pj • Pn Vi d u l: Tung 2 con sue xac, X bnn la tong 2m at G iai 28 Chitong 2: Bien nglu nhien (BNN). Vi du2: Trong 10 san p h lm c6 6 chlnh pham , lay ngiu nhien khong ho&n lai ra 2 s£n ph lm , lap bang p h an phffi x&c suat cua so chinh pham dugc l£y ra. Gi&i X la so chinh pham dugc lay ra, ta c6: X = 0,1,2 * P(o)2 = 15 ’ II K I06 ■p(#)=^ ' p (1)=^ ’ ,( 2 ) ‘ b X 0 1 2 P(X) 2 8 5 15 15 15 C h u v: Trong hai vi du tre n ta luon c6: £/*• =1 III. Ham ph5n phoi (hpp) Dinh nghla: Ham phan phoi (hpp). Fx (jc) = / ;,(jc) = jP(A: [l< A :< 31= F (3)-F (l) = l - ^ = ^ 42 42 42 37 ^2 0 v& J f(x)dx = 1. a Gio'ng n h a rcfi rac, bien nn lien tuc cung difcfc dinh nghia: < — >F(x) = £ P(xj 32 Chtfdng 2: Bien nglu nhien (BNN). 0 = J 0dt, x < 0 x 1 2x2 = \4tdt, 0/>}, 0 < p < 1 Doi vdi bnn lien tuc: trong hinh ve la do thi cua ham mat do, dien tich boi den bang p, tai hoanh do xp. Dac biet p = - difcfc goi la median ([Med], x y/ la diem trung vi cua 2 72 . X:hay noi cach khac: x y - med(X) <=> ' 2 -oo 2 Bo'i vdi bnn rdi rac th i trung vi la 1 gia tri cua X m a ta i d6 xac s u a t duCctc c h ia d e u tiicfns d o i h a i ben, nghla la med (X) = Xj F k - ] = P[X < x , ] < i < F [*(+1 ] = P[X < *1+1 ]. VI d u l: Cho X rdi rac co luat phan phoi X 1 2 3 4 5 P(X) 0,1 0,2 0,15 0,1 0,45 Chifdng 2: Bi£n ngau nhien (BNN). 35 Med\X\ = 4. Do tdng x&c suat or hai ben cua X=4 deu b^ng 0,45. VI du2: Cho X rdi rac c6 luat phan phoi X 0 1 2 3 P(X) 2 15 20 5 42 42 42 42 Med m = 2. vi F,x=2) - r | + i | < I , F(x=3)= + H > 1 4.2 Ky v o n g toan , m om ent cap k, phvfcfng sai. • Dinh nghTa: Ky vong thifcfng kl hieu: E(X) = M(X) = |i (X). E: Expectation, M: Mean, nen ky vong con dtfOc goi la gia tri trung binh. +Vori X rdi rac ta dinh nghla: X Xl X2 ... >Li ... Xn P (X ) Pi P2 Pi Pn E(X) = M (X ) = /i(X) = x l p} +x2p2 +... + x npn = X *//>(*/)• 1 = 1 -t-oo + Neu X lien tuc thi E(X) = M(X) = fii(X) = { xf(x)dx . —oo Vi du; Trong binh dung 10 qua cau giong nhau nhung khac trong liicrng gom 5 qua co trong ltftfng nang 1kg, 2 qua co trong liicfng 2kg va 3 qua co trong lirong 3 kg. Lay nglu nhien 1 qua, goi X la trong ltfcfng qua cau do. X co luat phan phoi: X 1kg 2kg 3kg P(X) 0,5 0,2 0,3 * 5 , 2 , 3 Ix5 + 2x2 + 3x3 Suy ra E(X) = l x ^ + 2 x ^ + 3 x ^ = -------- ^ --------- = i.Skg. (trong li/tfng trung binh cua 1 qua cau theo y nghla thong thtfcmg cua binh quan gia auyen trong dcri song thiTc te ). 36 Chitong 2: Bien ngau nhien (BNN). • Y n gh la ky vong: Ky vong \k gi& tri trung binh c6 trong so theo x&c suat cua dai lifcrng n g lu nhien X, 111 trung tam diem cua phan phoi mk cAc gik tri cu the cua X se tap trung quanh d6. • Moment cSfp k: Rdi rac: E(Xk ) = £ * ? /> ( * /) i=1 OO s Lien tuc: E[xk j = J xk f { x ) d x , neu tich phSn n&y hoi tu. ♦M om ent tam cap k: Rdi rac: E ([ X - EX]k ) = E ([X - /if ) = X [*i “ EX~f p(xi ) i- 1 oo Lien tuc: E ( X - EX)k = E ( X - n)k = \ (x-fii) f { x ) d x , —oo neu tich phan nay hoi tu. TrUdng hop rieng: phtfofng sa i la moment tam cap k= 2: ♦ Phtfcfng sai: ( thifcfng ki hidu: var(X ) = D(X) = a 2 ( X ) ) R « rac: £ ( [ * -E X ]1) = e ([X - r f ) = £ [*, - f l f p (* / ) 1=1 oo Lien tuc: var(X) = D(X) = a2 (X) = E ( X - E X ) 2 = J { x - f i ) 1 f ( x ) d x *Y n gh la phu’ofng sa i —oo Phuong sai \k sai so binh phtfotag trung binh cua dai ltfong n g lu nhien X so vdi trung tam diem ky vong. Phtfcrng sai dung de do mufc do phan ta n cua X quanh ky vong. ♦ Moment tam cap k tri tuyet: Chtfcfng 2: Bien nglu nhien fBNNV 37 E \ X - E X \ k = E \ X - ju\k = 1 /{x)dx, n£u tich phan nay hoi tu. C h u v; k=n Z c tx * ( - E X ) " - k k=0 C^Elxk).(-EX)n k k-0 = Z Ck ( -i ) n~k E [ x k \.(n)n~k , fi=EX k=o ♦ Do l£ c h tie u c h u a n : yja2 (X) = ^var(X ) = £ (r) = E(p[x]) = ChvCne m in h : i ] V[x]f(x)dx X = - o o 38 Chifdng 2: Bien ngau nhien (BNN). Ta chi can chting m inh cho BNN r&i rac, cdn BNN lien tuc ta thay dau tdng bang dau tich phan. 1/ Ta chting m inh tuan tif cac tieu tie t sau: • E (c) = c x l = c, c co the la BNN chi nhan gia tri c, vdi xac suat bang 1. • E{bX) = b E ( X ), beR E ( b X ) = Y . bxiPi = b 'L xiP i= bE(X ) 1 = 1 1=1 • E { X + Y) = E { X + Y ) E{X + Y) = 'Z'£(xj + yj )pij=Y*/ i=l 6/ Ta chting minh cac tieu tiet sau: • var (cX) = E [c2X 2) - E 2 (cX) = c2E (X 2) - c2E 2 (X ) = Chifdng 2: Bien ngau nhien (BNN). 39 - c 2 \^E^X2^ - E 1{X)^ = c1 var(AT) • var (X + F) = £-([* + F]2 ) - [ £ ( * ) + £’(y )].[£ (A ') + £ ( F ) ] = = £ ( X 2 + 2 A T+ Y2) - E 2 (X) - E 2 (Y) - 2E(X)E(Y) = =\e {x 2} - E 2 ( X]\ + \ e [y 2} - E 1 {Y) +2 E{ XY ) - 2 E{ X) E{ Y) = =0 = var(Ar) + var(F) Do X, Y la hai BNN doc lap nen E(XY) = E(X)E(Y) 7/ La he qua cua cac tinh cha't d tren. 8/ Ta co the suy ra cho trtfdng hop BNN rcJi rac, con BNN lien tuc ta chap nhan M Vi du 1: X 2 3 4 6 7 P(X) 0.1 0.2 0.3 0.2 0.2 Tinh: E(X), var(X ), cr(X), £ ( x 3 ), Med(X), P(\X - E X |< 2 Giai: E(X) = 2x0.1 + 3x0.2 + 4x0.3 + 6x0.2 + 7x0.2 = 4.6 E (X 2 ) = 22 x 0.1 + 32 x 0.2 + 42 x 0.3 + 62 x 0.2 + 72 x 0.2 = 24 var(X ) = e [^X2^- E 2 (A") = 24-4.62 =2.84 a (X ) = \/^84 = 1.685 E (X 3 ) = 23 x 0.1 + 33 x 0.2 + 43 x 0.3 + 63 x 0.2 + 73 x 0.2 = 137.2 Med(X)=4, vi F(4)= X P(*j <4)=0.3 < 0.5 Xj <4 va 0.5 < F(6)= X P(xj <6)=0.6 Xj <4 Tinh P(\X - E X \ < 2 ) = P(\X -4.6| < 2) = P(2.6 < X < 6.6)± ! i !>-• i£ 40 Chiftfng 2: Bien nglu nhien (BNN). = 0.2+ 0.3+ 0.2 = 0.7 Vi du2: BNN X co ham mat d6 xac suat: f ( x ) = X 1 < * < 3 a / Tim k. b/ Tim E(X) c/ Tim Med(X) 1 1 d/ Tim ky vong cua Y = X + —X x=°° x=3 0, con lai Giai: a/D ifa v&o J f(x)dx= J f[x)dx = l = > k = —. x=- x=3 x=l x=3 x=3 b/ ^(Al)= J xf(x)dx= | x — -tfcc == J :-----dx = -ln (3 ) x=l x=l 2x x = l2x 2 d Med(X)=x, theo dinh nghla ta co: t=X t t=x t=1 Ii 2t t=1 x=3 x=3. 3 jf-1 1 3 --------= — <=> JC = — 2 x 2 2 d / £ ( ? [ * ] ) = j 9, A mat 5 dong, c/ Con lai, A difdc 8 dong. Giai: Lap bang hai chieu va dem o tdng ta co the thay ngay phan phoi xac suat cua X nhif bang sau: Chifdng 2: Bi£n neau nhien (BNN). 41 Xt 2 3 4 5 6 7 8 9 10 11 12 P(X') 136236336436536636536436336236136 J’(Ii) = f,( I * / £ 4 ) = ! ^ = £ nVi) = P(4 < £ * / < 9) = i t ^ ±5 = f | W 3) = p (9 < £ X ,) = 4±it2±l = A| Ky vong cua Y: a = I r w > = (-5) £ + (+8) M + (-6) 1 | = a = , .95 1 = 1 Vay trung binh ma n6i thi chdi mot van A dtfdc ldi la 1.95$ V. M ot so lu a t p h an p h o i rcfi rac 5.1 P h a n p h o i cUlu tren X = {jfj,jc2,....,xn} BNN X goi 1& phan phoi diu tren X ={jcj,jc2,....,jcn) neu: P(xk ) = —» Vfc = l,n fl D i thay: E(X) = - . £ xt , var (X) = - ' £ x f - n i=l n i=l Dac biet neu X = {1,2, thi ' 1 " ^ -X*. E(X) /i2- ! 12 5.2 P h a n p h d i sie u b pi X e H (N ,N A,n) X6t tap co N phan tur, trong d6 c6 Na phan tuf co tinh chat A. Tif tap d6 lay ra n phan tuf. Goi X \k so phan tuf co tinh chat A thi X c6 phan phoi sieu boi. Ta ki hieu X e H (N ,N Airi), mot vki tai lieu con kl hieu X - H(N,NA,n). D inh n g h la Phan phoi sidu boi 111 phan phoi cua dai lifdng nglu nhien rdi rac tren tap X = {0; 1; 2 ; n} vdi x£c suat ttfdng ting Ik 42 Chiton g 2: Bien nglu nhien (BNN). /'•w k rk =p[x=k\= Za J L ± a. D inh lv: Cho X e H (N ,N A,n) thi: k-n k-nz-k r>n-k k=n L , , L , k - n K=n N - N . a/ V " 1 c * *=0 *=0 *=0 'yv N b/ £ (X ) = «p, trong do: P = ~ ^ n - n a N c/ var(X ) = «p^ Chting minh: N -n N - 1 k=nCkN C" * *=h a/ X — <*- = !<=> I C A=0 r "L /Vyv *=o \ N I t \ N A I t \ N ~ ^ A (1 + x) = (l + x) * (1 + X) r n-k _ /^n N - N . /=yv f* = yv. y c ‘ x1 I < ^ x ‘ = 1=0 I C* A=0 * ^ N - N A i= 0 * so sanh 2 he so cua x" d hai ve ta co: c yv = X = X c n a c n - n a = Z c n a c n - n a k + i=n k + i - n 4 A k - n - i A=0 k=n Z /n* f^n-k _ y ^ k n - k s r ^ k -n-k ^ n * n - n a ~ Z- c / N.^N-N. k-n '' A k=0 A *-k f-n-k h/P\X = k\= ‘Wa N~"a - r " C AI= H ( N ,N A,n,k) = yv A^/4 !(7V-yy/l)!(,V-/i)!/i! A!(A^/j( -A )!(«-/c)!(A r-iV y4 - /i + A)!jV! . r-n-k \ r - - V n a n ~ n a i cau a, nen: 2 , ---- *=0 c iV r " (*) Chtidng 2: Bien nglu nhien (BNN). 43 n n E(X) = X kJ>(X=k)= ' Z k J I { N , N A,n,k) = *=0 k=l n = X kJi(N,NA,n,k) (bo so' hang dau bang zero). k=1 Do tin h chat giai thtia N!= N (N-l) (N-2) (N-3)!, nen khi rut m ot s6' hang ra lam thtia so' chung thi phai btit di mot so hang tifong ting trong bieu thtic H(N;NA;n;k)tuy vao vi tri d tuf so' hay mlu so va ngi/Oc lai khi dufa vao cung the'. Vi the dd dang co: k.H (N, N a ,n,k) = ^ n.H (N - 1, N A - 1,n - 1,k - 1) dat j= k-1, va ket qua cua cau trtftic (**) ta co: E(X) = ^ - n ." f_ H ( N - l , N A - l , n - l , j ) = ^ n x l = np " j=0 N d hyZ k ( k - l ) H ( N , N A ,n,k) = h' Z k ( k - l ) H ( N , N A ,n,k) k=0 k=2 *=« Na {Na -1)nin-1) *=« X k(k-l)H(N,NA,n,k>= Ay * ----- 1 S H(N-%Na -2,n-2,k-2) k-2 k-1 Dat j=k-2, k=n j-n-2 X H(N-2,N a - 2, n -2 , k - 2 )= X H (N - 2, NA - 2, n - 2, j) k=2 j=0 = 1 *=« , . NA(NA -l)n(n-l) Vay: X * ( * “ l)H (N ,NA,n,k) = - AK ' V k = 2 N(N-l) kf k ( k - l ) H ( N , N A,„,k) = ^ k2H (N ,NA,«, *) - X” kH(N, N A,n,k) k - 2 k = 2 k = 2 k=n k' £ k ( k - \ ) H { N , N A,n,k) = M ( x 2y M ( X ) = M ( x i y ( n p )k=2 44 Chtfcftig 2: Bien n glu nhien (BNN). „ I Y 1\ N A (NA - \ ) n ( n - l ) NA I " ------- N fT T T ) + N — iiA^ [n N t-N a - n + jV] N(N-l) var (X) = E [ x 1^ - E i (X) =* A ^ [« A ^ -A ^ -w + Ar] TV(iV-l) JV#i Na „„NA[nNA - N A -n + N \ N 2 - N Na +nNA - n N N >-npN(N-l) it N N(N-l)----- f t = n/>NNr-n N - n ( N - l ) =nPq( N - l ) Vi du; TCr bo b&i 52 cay co 4 cay At, lay ra 3 cay. Tinh x£c suat de c6 2 cay At. G iai Goi X 111 so At trong 3 cay lay ra, X e H( 52,4,3). P\X = 2] = ^ 4 ^ - 0 , 0 1 . c k 5.3 P h an p h o i n h i thiJc X e B(n, p) h ay X ~ B(n, p) Day phep thuf Bernoulli la day n ph6p thif thoa 3 dieu kien i/ C£c phep thuf doc lap vdi nhau. ii/ Trong m6i phep thuf ta chi quan tam den mot bien co A. iii/ Trong m6i phep thuf xac suat thang lai luon la hang so P(A) = p , P ( A ) - l - p = q, (0< p £ ^ X 2 j = l) p 2 +E{ X) = n(n- l ) p 2 +np => var(X ) = E^X2 j - E 2 (X) = j^«(« -1) p 2 +np - n 2p 2 = = n p - n p 2 = n p ( l - p) = npq ■ 5.3.1. P h an p h o i B ern ou lli X e B(n = 1, p) = B(p) Day la trtfcrng hop rieng cua phan phoi nhi thufc vdi n = l, va sau nay ta cung co the xem day la trtffrng hop dac biet cua phan phoi sieu boi (xem phan sau: Dinh ly gicri han sieu boi ve nhi thufc), vdi n=l. Luat phan phoi cua no la: Chifdng 2: Bien ngau nhien (BNN). 47 X 0 1 p __ e_ Trong phan thong ke no dac tri/ng cho loai tap hop duoc xet ve m at dinh tinh (hoac ve chat) vdi ti le co tinh chat A ma ta N quan tam la p = —^ ~ . Vay co: E[ X) = p va var(.Y) = pq . V i d ill: Choi 10 van bau cua lien tiep, tim xac suat de co It nhat 1 van cufa cua th^ng. G iai Choi 10 v£n 111 10 phep thtf doc lap (n = 10). Goi A: “cufa cua thang” => P(A) = - .6 X: so' van cufa cua thang => X e B1#i P[X > 1] = 1 - P[X < 11 = 1 - P[X = 0] = 1-^10 v6y VI d u 2: Xac suat de 1 con ga de la 0,6. Trong chuong co 10 con. Tinh xac suat de trong 1 ngay co a/ 10 con ga de; b/ 8 con de; d ta t ca khong de. G iai X = {0;l;2;...;10},n = 10,p = 0,6 => X e B(10;0,6). P{X = k\ =Cf0(0,6)* (0,4)10_* . VI d u 3: Lay nglu nhien 9, lan (co hoan lai). m6 i lan mot vien bi tif hop gom 6 bi xanh, 9 bi do, 10 bi vang. a/ Tim xac suat de co 4 lan nhan difOc b jvang. b/ Neu biet rang co 4 lan nhan difdc bi vang, tim xac xuat de cac bi vane do nhan difdc 3 cac lan lav thuT chan. 48 Chifcftig 2: Bien ngau nhien (BNN). G iai a/ n = 9,p = — = 0.4 => P[X = 4] = C$(0.4)4 (0.6) 9 " 4 =0.25082 25 b/ Chi co m ot vi tri s&p xep 4 bi v&ng (V) thoa m an bai to&n la: V V V V V V V VV trong C9 vi tri co the co cua 4 bi vang, nen P = ~ ^ C9 5.4 P h an p h o i n h i thufc am X e NB(n,p) h ay X ~ NB(nyp) BNN X difdc goi la phan phoi nhi thufc am X e NB(n,p) neu n 6 c6 h&m x&c suat: P (X = x) = CrxZ \p rqx ~r , re N* = iV\{0}, x e {r,r+ l,r + 2,.....«>} D inh ly: Cho X e NB(n,p) ta co: X—°° X = o o a/ X /> (* = x ) = I C'x-\Pr1X~r = 1 . * 6 AT x -r x=r b/ E{ X) = -P d var(A') = ~ Chting minh a / V r ;r - \ „rax~r _ r v r r ~Xa x ~r - nr V 2 )**(* r + ^) „x-r L L x - \ P <1 -PL c x - i i1 ~ P L - - - - - - - - - 7— 77;- - - - - - - - <1 r 00 r ( 00 '\(r_1) vi ta da biet dao ham bac (r-1 ) cua da thtic '£ « * ] = £ f 1 'f ' ( r - 1 )! v*=o y \-l) (1 -q)' y=r- 1 a ' Va bieu thtic cuoi cung co difdc la do ta dat y=x-l. Do co (r-1) so hang trong tong sigma va no dung la dao ham bac (r-1 ), Chu’dng 2: Bien ngau nhien (BNN'). 49 (r-1) ;*-!) (x-2)...(x-r+ l -Dkr-i) x=r (r-1). M l (i-*)r nen cuoi cung ta co: Z r - r - l r' r-1 r x -r CX -\P 9 x=r ( r - l ) ' \ L qX J b/ £ ( * ) = I x C £ \/q x-r I x ( x - l ) . . ( x - r + l)qx~r x -r v *)’x=r Vi da co ket qud, dao ham bac r: I A 1 = I x(x-i)(x-2)....(x-r+l)qx~r = V*=l ) q x -r chu y: co (r+1 ) so' hang trong dau sigma. r! (»-*>r+1 E ( X ) =( r - 1 )! d M V>*=1 ) r! * ( * * ) = £ x 2C'xZ \p 'V* -' = r ^ - £ * * ( * - l) ..( * - r + l ) < r ' x=r vr )’x=r Ta da co ket qua: \ ( r ) I *qx V^=l Jn ( l - f ) r + 2 ( l - « ) r + 1 Pr+1 Pr+I x=r E i x 2) = 7- ^ - T l J ( r - 1 )!g (r + l)! r(r)! / + 2 Pr + 1 2 2 2 qr +qr + pr _ qr r ~ 2 _ _ 2 + _ 2 /> P P g r(r+ l) | r 2 _ 50 Chiftftig 2: Bien nglu nhien (BNN). var(A:) = £(A:2) - £ 2 (A') = V -+ T— 2 2 P P f \r \ P j = SL V Vi'dul: . . . . . Gieo ngau nhien lien tiep moi lan 2 con xuc sac ly tuong (deu dan, doi xung, d6 ng chit....) cho dSn khi gieo dugc 2 lin co tong hai mat xuc s5c la 4 thi dimg. a) Tim xac suat de so lan gieo nhieu nhat la 4. b) Tim gia tri trung binh cua so lan gieo do. Giai Lap bang hai chieu va dem o tdng ta co the thay ngay phan phoi xac suat cua X nhif bang sau: 2 3 4 5 6 7 8 9 10 11 12 P(X') l36236336436536636536436336236136 , ^ „ 3 1 11 a/ r =2, x= 4, p=— = — => a = — , 36 12 * 12 ✓ „ 2 -l 2 1 ^ . 1 2 1 1 11 33 ==> xac suat= Ca_ \p q = C\p q = 3 ------ = ----- 12 12 144 f 2 b/ E{ X) = — = — = 24, tuc trung binh phai gieo 24 lan moi ky vong p L 1 2 dugc 2 lan co tong mat la 4. ■ Vidu2: Gieo nglu nhien lien tiep m6 i lin 3 d6 ng tiSn ly tuong (deu dan, d6 xung, dong chit....) cho den khi gieo dugc hai mat sip thi dimg. a) Tim xac suat de so lan gieo nhieu nhit la 3. b) Tim gia tri trung binh cua so lin gieo do. Giai Xac suat dugc 2 mat sap khi gieo 3 d6 ng tidn la p - c \ (0,5)3 a) Chu y n§u gieo lin thu k dimg thi lan thu k gieo dugc 2 mat sip con (k - 1) lan dau la khong gieo dugc. Do do xac suit la p { l~ P) (p = ^ 3 j,k=3 the vao la ket qua. Chtfdng 2: Bien nglu nhien (BNN). 51 A -l b ) £ ( * ) = £ * p ( l - / ’)* ' = p I * ( 1 - p ) k=1 k=\ ' oo V = ~P = ~Pj P P (0.5)- Jfc=0 Chu v; Dung ra la gieo cho den khi gieo dugc it nhat hai mat sap thi dung. Tuc neu gieo dugc 3 mat sap cung dung. Va nhu vay thi r =1 la so lan toi thilu dat dieu kien la "xay ra it nhat 2 mat sap". p = cf(0,5)3,r=l, £ (X )= £ (X ) = i = ------ ------- P p = c | ( 0 , 5 ) 3 Vi du 3: Hai Sv choi mot tro chOi nhif sau: Ca hai luan phien lay m5i lan hai bi tuf mot hop difng 3 bi trang va 5 bi den (bi sau khi difdc rut ra difdc tra lai vao hop). Ngifdi nao lav ra trifdc cd difdc 2 bi trang thi thang cuoc, va tro chdi ket thuc. Tinh trung binh cd bao nhieu phien lay nhif the de ket thuc tro chdi. Giai: C2 3 p =—j = — , ta tinh ky vong so' lan lay (ke ca hai Sv) de tro chdi ket Cg 28 thuc: E (X ) = — = = 9.33, khoang 10 lan rut nhu the thi ky vong ket P 3 28 thuc tro chdi. 5.5 P h a n p h o i P o isso n (xac su a t h iem ) X e P(X) Phan phoi cua dai lifong nglu nhien rdi rac X nhan cac gia tri 0, 1, 2, ..., k , .........oo, vdi xac suat tuong ting la e pk =P[X=k 1 = ■-x j f k\ Do thi rdi rac :Vdi X = 2 v& 11 diem nhif hinh ve difdi day. e - l 2 * ____ pk =P[X=k] = — ^ - , k =0 , 1 0 ft • 52 Chifdng 2: Bien ngau nhien (BNN). k=oo a/ X pk = l k= 0 b/ £’(A') = A c/ var(AT) = A Chting minh: A=oo k=°° «/ E f t = S <*[*=*1 = I k\ * = 0z' * = 0 * = 0A -A v 1 + — + --- + ..+---- + . 1 ! 2 ! *! k=oo A ^ A __—A ^ < — e e =1 b/ £ ( * ) = £ * ft=0 “ A C .A . _3 it!V Ar=0 it! k=°°kAk- x *=~ Xk nhting ^ ——— chinh 1& dao ham theo X cua Y *=o -n k\- e ' " H • rk=oo n/ Vay * = 0 =(e* ) ,= eX- r t n E(X) = le~ XeX = A. k=0 K ' ) i * k -°° *.2p -A jk c/var(X)= = * = 0 it! Chtfdng 2: Bien ngiu nhien (BNN). 53 e XXk , 2 *= A Tuong dtfcrng or day hieu theo tuong duong phan phoi. Chufng m in h : Vi np —> A. nen suy ra p —> 0 c „ V ( i - , r * ■ ■ (- 1 )(- g - ( - * + 1 ) / ( 1 - J>y- * 8 1 n ( n - l ) ( n - 2 ) . . . ( n - k + l) k ------------ ~L -----------(*/») (!-/») k\ ta tim gidri h an cac thtia so: co dinh k, ta co: n \ ni - * 'nf k - \ \ 1 - —— v n )->1 [tip) —> A.k -kp 7k =1 - 1 ( 1 - p f * = ( ! - p Y p np~kp) = ( » - p ) 7 ( i - p ) 54 Chifdne 2: Bien nglu nhien (BNN), N A / _ k ~ e * A cuoi cung Ckpk ( 1 - p) D inh ly -A 2 k n-k e A k\ N. Khi N » n thi X e H (N ,N A,n)~ X e B(n,p), ~*P Nghia \k n co' dinh, N —> °° va —> /? * {0,1} Tuong duong d day hieu theo tuong duong phan phoi. C hting m inh: ,-ik s^n-k s^k (-n-k Dat n -a = n -Na, #i! 1 £!(/*-&)! c n ac n - n a _ V i V j _ r "C A)yv r ,/* t AJyv ta co ------j— — = Cw , n va k co' dinh. Khi N —»oo thi «! yv-: =n - n a va vi JV.N-> p nen yv^ -» oo nghia la N a A ^ a —1 )»—»(^V^4 -fc + 1 ) , la cac dai lifdng vo cung ldn (co 1 so hang), Wdngtti A^,(A^-l),...,(7V-- n + k + l) cung la cac da ltidng vo cung ldn (co n-k so' hang). Vay n-k f-k f-n-k. — n r n t IVx{ n a f Nl ) A < n a r k I N J [ n j. r'k k n-k ~^CnP <1 Y nghia cua dinh ly nay co the giai thich nhu sau: Chifdng 2: Bien nglu nhien (BNN). 55 Dieu kien de dp dung phan phoi nhi thufc Id cdc phep thuf phai doc lap. Neu n phan tuf lay ra tuf tap ltfn theo nguyen tac co h o a n lg.i, tufc Id lay m ot phan tuf ra xem xet, xorg tra lai ta p ldn roi mdi lay phan tuf khdc, thi cac lan lay la cjac phep thuf doc lap, co the tinh xac suat theo nhi thufc. NgUoc lai lay kh o n g hoan la i (hay lay n p h a n tv[ c u n g lu c ) thi cac lan lay xem nhif cac phep thuf phu thuoc va ta phai tinh xac suat theo sieu boi. Nhifng neu N kha ldn va n nho th i m6 i la n lay m ot phan tuT, tap lorn gan nhif khong doi, khong anh hifdrng gi dang ke den lan lay tiep theo nen vSn co the xem cac phep thuf ay la doc lap. Dac biet khi N — phan phoi sieu boi trung vdi phan phoi nhi thifc. Vi d u l M ot lo hang co 1 % phe pham. Tim xac suat de khi chon ra 500 san pham co a/ T at ca deu tot; b/ 1 phe pham. G iai X g Z?(500;0,01),X = np = 500x0,01 = 5. aJ P\X = 0J = e-s .5° 0 ! b/ P[X = \\ = * Vi d u 2 1 ! Ti le thuoc hong trong m ot lo thuoc lorn la p=0.05 (xac suat hiem). Lay NN 20 lo. Goi X la so lo hong nhan diforc. Tim ham m at do cua X va so sanh vdri gia tri xap xi bdi phan phoi Poisson. G iai X e Z?(20;0.05), nen co ham m at do: f(x) = C2 0 1 0*05^ (°-9 5)20_JC = 0. o aJ Ham gamma la xac dinh. b/ r c/ r ( i) = i d /r ( f l + l) = « r(a ), a > 0 = > r(« + l) = /i!, n<= N C h tfng m in h : a/ DI dang chting m inh tich phan suy rong tre n la hoi tu. b/ Xem lai phan to£n cao cap. oo c/ r ( l ) = Je~xdx = -e~* = 1 d /r(a + l)= fxae Xdx = - x ae 0 u dv - xOO OO + a \x a e xdx = 0+aT(a) = ar(a 0 o D in h lv Ham beta, a> 0, b>0. Chtfdng 2: Bien ngau nhien (BNN). 57 Be{a,b) = ^ ( l - x / A ^ dx la hoi tu dtfdi dau tich phan. ChuTng m inh: 00 1 f HT T an h d lai: x d vo ctfc (x>l) , f— dx = \\PK a < \ x cf zero (0 a>0 ( hien nhien ). Tai can 1: f ( l - x / * ^ b>0 ( hien nhien). ■ D in h ly Ham beta va ham gamma co quan he qua bieu thtic: Be(v„ v 2) = r ( v i) r ( v 2) r ( v i+ v 2) Ch-JTng m in h : Xem bai tap cua cung tac gia. VI. M p t so lu a t p h a n p h o i lie n tu c 6.1 P h a n p h o i d e u X e U[a,b], a b [,a,b\ b/ E { X) = ^ - , var(A') = ^ — 12 d F = — e l / [ 0 , 1 ] b -a Chtfng m inh: Xem nhu bai tap. 6.2 P h an p h o i m u XeE{X),X> 0. BNN X goi \k phan phoi mu XeE{X),X > 0 neu co ham mat do: f ( x) = \^'e ’ X , do thi du5i dai ra khi X T. [O, x < 0 Phan Phoi mu ap dung trong pham vi ly thuyet do tin cay. Neu so lan xuat hien cua mot bien co trong mot khoang thcJi gian co phan phoi Poisson, thi thdi gian giffa hai lan xuat hien bien co do co phan phoi mu. Chang han, thdi gian lam viec lien tuc cua mot thiet bi giffa 2 lan sura chffa, thdi gian chd cua mot khach hang de difdc phuc vu D inh ly Cho X e £ ( A ) , A > 0 thi: aJ Kiem chting day la ham m at do. Chifcfng 2: Bien nglu nhien (BNN). 59 b/ H&m phan phoi F ( j c ) = > * e > x > ® [0 , jc< 0 d E ( X ) = l , E (X * h j i - = d/ P han vi xp = —-ln (l-/> ), 0 < / 7 < l A. e / Y = A,Xg E[AL = l] Chufng m in h : x=°° x=°° a/ | f(x)dx = | A* *xdx = -e -Ax x=0 0 f=jr r=x = 1 b/ F (x )= J f(t)dt= J A.e~X‘dt = 1 -e -Ax t= 0 f=0 c / £ ( j v ) = i * ( * 2 ) = J r . ™ ( x ) = ~ e (x j = fAxe~Xxdx = j ue~u — = y J ue~udu = 0 u=Ax=0 * ^ 0 = r ( 2 ) _ i r ( i ) _ i A A A £:(Ar2)= jAx2e~Axdx= | A.[^~ 0 u=Ax=0 _ r ( 3) _ 2x i r ( i ) _ 2 ~ A2 A,2 A.2 i 2 1 1 => var (AT) = ——---- = — A. A. A,2 d/ Tun phan vi xp : F ( x p ) = p t s p = l - e ~ Xx o x = - i l n ( l - p ) e/ X e e ( A ) ^ > Y = A X e £ ( A = l) Fy(y) =P(¥c > 0 ,A > 0 Cu the A = 0.00125, E(A = 0.00125), hay tinh va noi y nghla: a/ P{X > 800) b/ P{llti -\,k> 0 XJ f/ Tim diem p h a n vi jc0 75 c u a X. Giai: a/ F (X > 800) = P(800 800 gicf la 36.78%. b/ P(720 < X < 880) = F (880) - F (720) = 1 - F (880) = = ^-0.00125x720 _ £-0.00125x880 = ^-0.9 ^-1.1 =Q 0?369 Y nghla la xac suat de tuoi tho cua thiet bi nam trong khoang [700 gid, 800 gid] la 7.36%. d E ( X ) = \ = 1 = 800 gicf = trung binh tuoi tho cua thiet bi. A 0.00125 d/ P(X < 1200) = F(0 < X < 1200) = F (1 2 0 0 )-F (0 ) = = F (1200) - 0 = F (1200) = 1 - e~15 =0.7768 Chtfdng 2: Bien ngau nhien (BNN). 61 e/ P f * > * x i ) " - ' r KA’J -A- = l - ( l - « - '* Jr)= « -'u = e •» = f/ Tim diem phan vi 7 5 cua X. m-k - » k e = e F (*0 .7 5 ) = #-75 « *0.75 = - I® (1 ■- 0-75) = 800x(-1.3862) = = 887.228 gid. Nghla la xac suat de thiet bi ngifng hoat dong trifdc 887 gid la 0.75, hay xac suat de tuoi tho cua thiet bi ldn hdn 887 gid la 0.25. ■ 6.3. P h a n p h o i gam m a X e r(ar,/?), a > 0,/? > 0 BNN X goi la phan phoi gamma X g r(a,/3), a > 0,/? > 0 neu ----- x>» co ham m at do: f(x) = r (a)pa 0 , jc<» Giif co' dinh p, do thi dudi dai ra khi a T C h n f: 62 Chifdne 2: Bien nglu nhien (BNN) 1/ Khi Ap dung vdi X e Y a = = 2^ <=> X e ^ (r) [ _£ ■ x '2 'e 2 , jc > 0 day la ham mat do: / ( x ) = 0 , x < 0 co phan phoi la (r) vdi r bac tu do. Co ting dung rat nhieu trong ty thuyet XSTK sau nay. 2/ Khi P =1, thi ta ki hieu dOn gian hdn: X e y(a),a>0. X e r(a r,/? = l),a r> 0 <=>Xe y ( a ) ya > 0 , cd ham mat do: —— x* 1 - x A >e , x > 0 / ( * ) = r(or) 0 , x < 0 D in h lv Cho X e r ( a , f } ) , a > 0 , f i >0 thi: a/ Kiem chting la ham mat do: b/ fi = E ( X ) = a0, a 2 =\ *r{X) = cfl2. Chufng m in h : That vay a/ ] f(x)dx = X) ---- l — xia- ' )e fl dx =-----?---- " f xia^ e ^dx x=0 t=0 r ( a ) f i a ii = — => Pdu = dx , P X - o ° j x ,a - ' ]e Pdx = “ j ( u p j a - lK - updu = p W “ = x=0 ii= 0 X = o o = /J(a)r( 0,p>0 BNN X goi la phan pho'i weibull X e W ( a ,fi) ,a > 0,0 > 0 neu a co ham m at do: / ( jc) = a P xP~le~ax , jc > 0 ,a r> 0 , p > 0 0, jc < 0 0.5 1 1.5 2 2.5 3 { Giff co' dinh a = 1 , cho P T, ta thay do thi co thang len, hep vao va dinh cao len khi P T. Cung gio'ng nhif ham mu, phan pho'i Weibull diTOc suf dung trong ly thuyet do tin cay, de kiem dinh tuoi tho suf dung, thdi gian phan huy, thdi gian phan ra cua cac nguyen to hoa hoc. D in h lv Cho X e W (or,/?), a> 0,/? > 0 thi: a/ Kiem chtfng la ham mat do: b/Co ham phan phoi: F (x ) = l - c -axP, x > 0 . 64 Chifcfag 2: Bien ngau nhien (BNNV l c/ fi = E( X) = av1+i 2f 2)-(V iYl2 r — r 1 + 4 I P) - e/ Diem phan vi xp cua X ditoc xac dinh: x p =ln ( l- p ) a 11 Y = a X P e E (A - 1), nghia la tpcm f(y) = - = e-y ke Chting m inh: a/ j f(x)dx= | a p xP~le~axP dx= J e~axP [a Px&~xdx\ jc=0 x=0 x=0 dat u = ax^ =>du = aPx^ Xdx, vay J f[x)dx = J e Udu=\ x=0 jc =0 t=x b /F (x ) = | / ( / ) * = J a p t P~xe~a(P dt t = 0 t=o u=ax& dat du = a P t^ xd t, F(jc)= J e Udu= 1 — e-axP x=<» « = 0 X = ° o c///= { x/(x)dx= j xapxP le axPdx= J a p x p e~[ x=0 x=0 dat u = a x P z=> du = a P x^ ~ xdx 1 x=0 U —oo « = oo —— tt=oo 1 1 = f x e~udu= f «=0 ii=o e Udu=a & J (u)pe Udu=a u=o ' i + i 'P jf=oo ^ JC=oo d / £ ( x 2 ) = J x 2f ( x ) d x = I a f i x 1x ^ - 'e - axl,Jx = jf= 0 jc= 0 dat u = axP => du = aPx^~ldx Chifdng 2: Bien neau nhien (BNN). 65 «=» £-(aT2 ) = J XLe~udu= J J {u)fie^du= H= 0 «f= 0 I ) __2 f. iV2 r - f. ^ =>0* = E [ x 2} - E 2 (X) = a P < +P ,r< + /> i e/ Diem phan vi: xp cua X difdc dinh nghla: -ax^ & l - e p = p P HP l - e ~ aXp =/><=> 1 - p = e~aXp <=>ln(l-/> ) = - a x P <=> <^xp =ln ( l- p ) a P 4/ Y = a X P e e{X = 1) <=> /( > ) = le~ Xy = e~y Chu y or > 0,/? > 0 va ham dong bien. FY{y)=P{Y 1 -PP , trong do thay x= —1 V 66 Chtidng 2: Bien nglu nhien (BNN). ddn gian cho a p , ta thay rang: ee > '~Al Y -b** =e-y. (/*-<) va x■P-i ya h iP _= 1 M S ' l i Vay f y { y ) - e y <=> Y = aX^ g e(A = l) ■ 6.5. P h a n p h o i C a u ch y to n g q u a t: X g C(a,0), a g R,0 > 0 BNN X goi \k phan phoi Cauchy X e C (a,0),a & R,0>O neu co ham m at do: 1 / ( * ) = , -oo < x < °°,a g R,0 > 0 7T0 1 +x-a^ 2 ’ e ) j 0 .6 - 0.5- a = 1,0 = 4 0.4- 1 a = 1,0 = 2 a = 1,0 = 1 03: 1 a = 1,0 = 0.5 0 2 -# Do thi vdi co dinh a =1 (chinh la vi tri hoanh do dinh), 9 =4, 2, 1, 0.5, ta thay do thi vut len tren va co hep lai khi 9 giam, Dinh lv : Gia suf X g C(a,0) , phan phoi Cauchy tdng quat thi: a/ Kiem chting la ham mat do. b/ Tim ham phan phoi cua X. Chifdng 2: Bien nglu nhien (BNN). 67 / Tim trung vi cua X ( medX= x <==> nedx 1 \ f(t)d t = F (medx) = - ) IJ Tim E(X), var(X). Chting minh: J Kiem chtfng la ham mat do. 1 7t0 1 +e , •»-. t- a d t, d a t----- = u 6 ' = J kO du / 2^ _U f I + «2) 1 + V / = — [arctanii] n n ~2 VI 0 ) J 1C ~~2= - n =1 n >/ Tim ham phan phoi cua X t — X X F(x)= j / « * = f / r —oo —oc2 'I ;n9 1 +I « J / , _ /-a d t , dat -----0 x - a ^U = — [arctan a] * (l + «2 ) *■ 1 ♦ ( x ~a = —arctan n e 4 x -ae - narctan ---- H--- — Q ) 2 _ HlcMA | medx c/ medX= * <==> J f(t)dt = F [medx] = - <==> / — \ x -a \ v ' narctan /■ (* )= - x - a \e 1 1 ^ h— = — <=> arctan 2 2 e= 0 <=>jc-a 0 <=>x = o 68 Chtfdng 2: Bien ngau nhien (BNN). d! E{X)= \ xf(x)dx= \ xdx r( \ 2 A x - - tc6 1 + x — \ V, e ) / x - a r( x 1 °°f _ 1 °°f {uO + a)du dat —— = «» £ ( * ) = - J —-------------------- 7 ^ - — J v i ] ^ = 0 (1+“2) J - — 1adu I 1 (u0+a)du _ 1 E(X)- * i ( n V ) " * “f 0 w \ X ——oo n Q 1 +U I VI « J J lam tifdng tif nhif tren ta se thay co so hang du = °°, nen var(X) khong ton tai. r U UU _ f “ u2 du 1 1 - —oo | l + W2 j -o c (1 + “2) Bay gid ta xet mot phan phoi curc ky quan trong trong ly thuyet XSTK. 6 .6 . Ph& n p h o i c h u a n : X e N 0,// = const. PhUn phoi cua dai lufdng nglu nhien lien tuc X nhan gia tri _ (x- m) 2 tre n R vdi h&m m at do phan pho'i: f(x) = — t = c (TyJlTC 2 a2 Chifdng 2: Bien nglu nhien (BNN). 69 hu y: J f(x )d x = —oo (x-M)2 -- [ — ] = e 2 dx = l o 4 in °° j lifa v&o | e~ * dx = >//r -4 0 .2 - O 1 5 yo y jr* V 4 8 10 ^dri // = 2, cr = 3 ta c6 d6 thi nhuftrdn, Tdng qudt ta c6 : Diem ctfc dai M = , u 2 = M+<7 rJ, = / i - a ,* /2xe J *.6.1. P h a n p h o i ch u an dcfn gian: n p h o i i vk 2 diem uon: x e A^(//,«72)= > r = ^ — e W(0 , 1 ) 3am m at do phan phoi: f{t) = —r= e 2 (tra bang A). yjllt Tich phan Laplace: P[|A '-//|<3 max f 2(x) = -----—= < -----^ = = m a xf\(x ) x=m a 2yl2jce a x yjlne x=n _ a Chu y d hinh ve tren: 4 do thi cung co n =1, nhiftig vdi ba phtfcfng sai: o \ =1, o \ =2, a \ =3, a \ =4, neu X*eN(n, o f ) thi ctfc dai cua cic. ham mat do co thuf tu la: maxf4(x) < maxf3(x) < maxf2(x) < maxfi(x) neu 0 < g\ « t\ « j\ < o\ D inh lv: Cho X e J V ^ u 2) thi: a/ Med(X) = M b/ f.i(X) = E(X) = /i d cr2 (X) = var(X) = a 2. Chufng minh: a/ H ien nhien. (x-n ) 2 OO OO - — ----------- b/ E(X)= f xf(x)dx= \ x — = e 2ffl d x , dat z = -x~ ^ _ v = e 2 Vay: var(A") = -z e j e 2 dz 6.6.4. C a c h tin h Cho X e D at r = ^ - ^ = > J e A r( 0 , 1 ) Phtfcrng ph&p tinh: Tinh P\xj < AT < x2 ]. D at a = = <7 O' => />[*, < X < x2 J = P[a < r < 0 \ = $*£) - p(ar). P[xx < X < x 2\ = llA g. G iai 0,5 ~36) _ ^ (_ o.91) a/ P[3,5 —— > 1.65 => .xq > 6.315 1,1 / 1,1 Trong Excel: NORMSinv( 0.5 +0.45) =1.65. Vi D u 3: Cho X e yv(4.5,l.l2), dung 570 MS, tinh: P(3.5< X < 6 ), P(l.5< AT < 3), P(-1.5 < X < 4). G iai 1/ Dung cong thtic: P ( a < X (&)-ham P(a) i y=t _y*L Trong do ham P(t): p (/) = - _ J e 2 dy . Theo cac bitfc: y=- 1/ Xoa so lieu thong ke trudrc do: |t], 1 MODE|, Scl=Stat Clear, MODE = SD Chiftftig 2: Bien n elu nhien (BNN). 75 2/ Vao 2 so lieu jcj =4.5+l.l = 5.6, x2 = 4.5-1.1-3.4 ♦ Bay gid tinh /*(3.5ZS7«|, [iP(|, U/VSl, [j], 0 0.18165, lu u vao o B, U a s |, | T s T o 1, @ • Goi A-B: \RCl\, 0 , 0 , F I 1 * 0 , 1 , @ , 0 0 .7 3 2 . So sanh vdi ket qua cua vi du trtfdc. ♦ Bay gid tinh /*(1.5L @ • Goi A-B: \RCL\, 0 , 0 , F l U cx!. 0 , 00.0 8 3 1 5 • Tuong tir tin h P(-1.5 < X < 4) P(1.5 < X < 3) = NormDist(3,4.5,1.1,1) - NormDist( 1.5,4.5,1.1,1) = = 8.31%. ■ Vi du 4: Xe TV j , tim moment tam bac k, va moment tam tri tuyet doi bac k. Giai: 76 Chifdne 2: Bien nglu nhien (BNNY {x-MV f ( x ) = — 7— p 1 (k-1 ) le, H — IV 2J w2 =<7 2 (2 - l)!! = c o I — J s o h a n g tic h , n e n Chifdng 2: Bien ngiu nhien (BNN). k-1 m — ±0*2 2k - th § t vay: k 13 ==> (k-1), (k-3)... chiln, k - 3 2 x-l=[k-2 ]--> co x= f A - l^ ^ . ----- so hang tich, nen 2 2 m- = J V ( 3 - l ) n = J V x 2 = 2 a > £ m5 = J p ( 5 - l ) ! l - J « * x 4 x 2 = S S J I V i d u 5 : X g A^S,*!2) . Neu P(X > 9) =0,2 tinh = 0.5-0.2 = 0.3 = > - = 0.85 =><7Z= 22.14

0 Do thi c6 cung |i =0 , duoi dai ra va dinh thap xuong khi tr2 T. 78 Chifcfag 2: Bien nglu nhien fBNNY v . D inh Iv: Cho X e log./v|//;c72 j thi: M+~ a/ Ky vong: E (X) = e b/ Phtfdng sai: var (X) = Chtfng m inh: Xem nhif bai tap VIII. Ham cua BNN. Trong thuc te, doi khi ta xet BNN phu thuoc vao 1 hay nhieu BNN khac da biet luat phan pho'i. Van de cua ta la dtfa vao luat phan pho'i (ham phan pho'i xac suat hay ham m at do xac suat) cua BNN nay de tim luat phan phoi cua ham bien do. 8.1. T rtid n g hcfp BNN rcfi ra c . VI D u B iet bang phan phoi xac suat cua X, Cho Y =