06:19:40 L THUYTXC SUT V THNG K TON HC Bi Hong Hi, Phan Vn Tn H KHTN 06:19:40 Chng 3. I LNG NGU NHIN V HM PHN B 3.1 Khi nim v i lng ngu nhin Ktqungunhincaphpthcthctrngnh tnh bi s kin ngu nhin o M t bng li: A={ ng tin nhn mt sp } ctrngnhlngchoktqungunhinca php th ngi ta dng khi nim i lng ngu nhin Cc nh ngha: o Mtilngnhn ccgi trca nvixcsuttngng no gi l i lng ngu nhin o ilngngunhinlilngmkhitinhnhmtlot phpthtrongcngmtiukinnhnhaucthmiln nhncgitrnyhocgitrkhchontonkhngbit trc c 06:19:40 Chng 3. I LNG NGU NHIN V HM PHN B 3.1 Khi nim v i lng ngu nhin Cch gi:o Nhiukhiilngngunhincncgilbinngu nhin Hai cch gi tng ng nhau K hiu:o Thngthngccilngngunhin(hayccbinngu nhin)ckhiubngccchciLatinhinhoa:X,Y, Z,, hoc cc k t Hylp:, q, ,, o Cc gi tr c th ca i lng ngu nhin (cc gi tr m i lng ngu nhin c th nhn) c k hiu bng cc ch ci Latinh in thng tng ng: x, y, z, 06:19:40 Chng 3. I LNG NGU NHIN V HM PHN B 3.1 Khi nim v i lng ngu nhin Phn loi:Cn c vo tp gi tr c th ca i lng ngu nhin ngi ta phn bit hai loi i lng ngu nhin o i lng ngu nhin ri rc: Tp hp cc gi tr c th c ca n l hu hn hoc v hn m c Vd:GiXlilngngunhinchsimnhnckhigieo mt con xc xc. Vy X={1,2,3,4,5,6} hay x1=1, x2=2,, x6=6 o ilngngunhinlintc:Tphpccgitrcthca n lp y mt khong no y ca trc s hoc c trc s, tc n l tp hp v hn v khng m c Vd:GiYlilngngunhinchnhitkhngkh(oC)o c H Ni. Vy Y={y, ye[-10; 50]} 06:19:41 Chng 3. I LNG NGU NHIN V HM PHN B 3.2 i lng ngu nhin ri rc Xt i lng ngu nhin ri rc X m cc gi tr c th can l tp {x1, x2,, xn,} vi P(X=xi) = pi, i=1,2, o m t bin ngu nhin ri rc X ta s dng bng phn b xc sut sau o Trong Epi = 1, pi > 0 i=1,2, Vd1:Gieongthihaingtinginghtnhau.GiXlbin ngu nhin ch s ln xut hin mt sp. Hy lp bng phn b ca X o Gii: S ln xut hin mt sp ch c th l 0, 1 hoc 2, do X={0,1,2} o Gi Ai l ng tin th i xut hin mt sp (i=1,2), P(Ai)=0.5 oS kin X=0: X=1: hoc S kin X=2:o V cc Ai c lp nhau: P(X=0)=0.5x0.5, P(X=1)=2x(0.5x0.5), P(X=2)=0.25Xx1 x2 ...xi xn... Pp1 p2...pi pn... 2 1A A2 1A A2 1A A2 1A AX012 P0.25 0.50.25 06:19:41 Chng 3. I LNG NGU NHIN V HM PHN B 3.2 i lng ngu nhin ri rc V d 2: Mt x th c 3 vin n. Anh ta bn tng pht cho ti khi hoc trng mc tiu hoc ht c 3 vin th thi. Hy lp bng phn b xc sut ca s n chi ph, bit xc sut trng ch mi pht l 0.8 o Gii: Gi X l bin ngu nhin ch s n chi ph. Vy X={ 1, 2, 3 } o S kin X = 1: Bn pht th nht trng ch (do khng bn tip na), P(X=1) = p1= 0.8 o S kin X = 2: Bn pht th nht trt v pht th hai trng,P(X=2)= p2 = (1-0.8)0.8 = 0.16o S kin X = 3: Bn pht th nht v th hai u trt (do cn bn pht th ba), P(X=3) = p3= (1-0,8)2 = 0,04X123 P0.8 0.160.04 06:19:41 Chng 3. I LNG NGU NHIN V HM PHN B 3.2 i lng ngu nhin ri rc V d 3: Tin hnh n php th c lp, xc sut xut hin s kin A mi php th khng i bng p. Gi X l bin ngu nhin ch s ln xut hin s kin A trong n php th. Hy lp bng phn b xc sut ca X. o Gii: Ta c X={ 0, 1, 2, 3,, n } o Xc sut ca s kin X=k (0 s k s n) c tnh theo cng thc Bernoulli o T X01kn P n k p p C k P k X P pk n k kn n k,..., 1 , 0 , ) 1 ( ) ( ) ( = = = = =0 0 0 nnq p C1 1 1 nnq p Cn n n nnq p C(q = 1-p) n h thc nh thc Newton k n k knq p C== +nkk n k knnb a C b a0) (ta c ng thc 1 ) (0 0= + = = ==nnkk n k knnkkq p q p C p06:19:41 Chng 3. I LNG NGU NHIN V HM PHN B 3.3 i lng ngu nhin lin tc Xt i lng ngu nhin X m cc gi tr c th ca n lp y mt khong hoc c trc s. Khi X l i lng ngu nhin lin tc o mtbinngu nhinlintcX ta s dng khi nimhmmtxc sut (Probability density function) o Hm f(x) c gi l hm mt xc sut ca bin ngu nhin lin tc X nu n tha mn hai iu kin sau: o Khi , xc sut X nhn gi tr trong khong (a,b) c xc nh bi } + = + e >1 ) ( ) 2) , ( , 0 ) ( ) 1dx x fx x f}= < x) = 1F(x) o Hm TDIST(x, n, Tails) c tham s Tails=1 hoc 2 Nu Tails=2, kt qu tr v l xc sut P(|X|>x), x>0 Nu Tails=1, kt qu tr v l xc sut P(X>x), x>0 V hm mt i xng qua trc tung suy ra nhnh x 0A