分布列綜合歸類（解析版）

11設(shè)X為一個非負(fù)隨機變量，其數(shù)學(xué)期望為E(X(，則對任意ε>0，均有P(X≥ε(≤,望間的關(guān)系.設(shè)X的分布列為P(X=xi(=pi,i=1,2,?,n,其中pi=1，則對任意ε>0，Pxipi≤xipi=其中符號表示對所有滿足xi≥ε的指標(biāo)i所對應(yīng)的Ai求和．的影響，與之前的n無關(guān).ii，所以，P(B2(=P(A1B2(+P(B1B2(=P(A1(P(B2|A1(+P(B1(P(B2|B1(P(Ai+1(=P(AiAi+1(+P(BiAi+1(=P(Ai(P(Ai+1|Ai(+P(Bi(P(Ai+1|Bi(，即pi+1=0.6pi+(1-0.8(×(1-pi(=0.4pi+0.2，構(gòu)造等比數(shù)列{pi+λ{，設(shè)pi+1+λ=(pi+λ(，解得則pi+1-(pi-1-所以當(dāng)n∈N*時，E(Y(=p1+p2+?+pn= A.p1=B.P(X1=2(=C.數(shù)列是等比數(shù)列D.Xn的數(shù)學(xué)期望E(Xn(=122利用pn=pn-1+P(Xn-1=0(+P(Xn-1=2(推出pn-=-pn-1-，可判斷C；利用P(Xn=0(=P(Xn=2(=可判斷D.當(dāng)n≥2(n∈N?(時，pn=pn-1+P(Xn-1=0(+P(Xn-1=2(=pn-1+[P(Xn-1=0(+P(Xn-1=2([=pn-1+(1-pn-1(=-pn-1+整理得pn-=-pn-1-，p1-=-=-，P(Xn=1(=pn，P(Xn=0(=×pn-1+P(Xn-1=0(=pn-1+P(Xn-1=0(，P(Xn=2(=×pn-1+P(Xn-1=2(=pn-1+P(Xn-1=2(，因P(X1=0(=P(X1=2(，所以P(Xn=0(=P(Xn=2(=，E(Xn(=0×P(Xn=0(+1×P(Xn=故D正確， n的數(shù)學(xué)期望E(Xn(為定值.2=n=-×(-n-1+n+pn-1=(2qn-1+pn-1-1(，結(jié)合2q1+p1-1=0，由此可得qn=、分布列以及數(shù)33n由題意知p1==，q1==，所以p2=p1+q1+(1-p1-q1(=.n=pn-1+qn-1+(1-pn-1-qn-1(=-pn-1+，所以pn-=-pn-1-.所以pn-=-×(-n-1，pn=-×(-n-1+.n=pn-1+qn-1=pn-1+qn-1①,1-qn-pn=pn-1+(1-qn-1-pn-1(=pn-1+(1-qn-1-pn-1(②.所以①-②,得2qn+pn-1=(2qn-1+pn-1-1(.又因為2q1+p1-1=0，所以2qn+pn-1=0．所以qn=.所以Xn的概率分布列為：Xn012p1-pn-pn1-pn2所以Xn的數(shù)學(xué)期望E(Xn(為定值1. 44=,a1=,b2=,a2=(3)E(Xn(=+n-1n+1=1-bn-an，an+1=bn+an，2=b1+b1+a1=b1+a1=×+×=，a2=b1+a1=×+ -an-bn，nnnn55nnnnn-bn，n+1=bn+bn+an+1-bn-an=1-bn-an.an+1=bn+an，n+1+bn+1-=2n+n(+1-n-n-=n+n-=1n+bn-2an+bn-2an+bn-6，+b1-=，n+bn-=n-1，所以2an+bn=+n-1，E(Xn(=0×(1-an-bn(+1×bn+2an=bn+2an=+n-1.ξ01P1-pPEξ=p，Dξ=p(1-p(.66則Pn+1=0.6Pn+0.2(1-Pn)=0.4Pn+0.2，i-1，由題意得甲第i次投籃次數(shù)Yi服從兩點分布，且P(Yi=1)=1-P(Yi=0)=Pi，i-1+， 圖.77人的概率.解得m=0.012.ξ0123P 8 88表.992442k=1(3)Dξ1<Dξ3<Dξ2所以P(A(估計為=；99P(X=2(=P(ABC-+AB-C+A-BC(=P(A(P(B(P(C-(+P(A(P(B-(P(C(+P(A-(P(B(P(C(=；P(X=3(=P(ABC(=P(A(P(B(P(C(=，則X的分布列為：P P (3)Dξ1<Dξ3<Dξ2 =；==；==；所以Dξ1<Dξ3<Dξ2. 布列.列.X01P 37 47P(Y=0)==，P(Y=1)==.Y01P 7 67若在一次實驗中事件發(fā)生的概率為p(0<p<1(，則在n次獨立重復(fù)實驗中，在第k次首次發(fā)生的概率為p(k(=(1-p(k-1p，k=1,2,?,Eξ=。?,k，其中k為M與n的較小者，P(ξ=m(=服從參數(shù)為N,M,n的超幾何分<23.4≥23.4P(K2≥k0(則P(X=0)==，P(X=1)==，P(X=2)==，所以X的分布列為：X012P <23.4≥23.466K2===6.4>3.841，(3)s<s<sP(X=0(=C=5P(X=1(=CC=15P所以X的分布列為：X012P x1==0.4,x2==0.3,x3==0.2,x4==0.1，=0.25，所以s==.y1=80=8,y2=80=4,y3=80=4,y4=y1+y2+y3+y41=所以s==.對應(yīng)的平均數(shù)為x1+x2+x3+x4+y1+y2+y3+y4=所以s= =(0.4-0.25(2+(0.3-0.25(2+(0.2-0.25(2+(0.1-0.25(2+-2+-2+-2+-2 =8 所以s<s<s. P(K2≥k0(2P(X=0(=CC=4=1P(X=1(=CC=12=3可得X的分布列為X012P 5 35 5 *3次.E(X(；k則P(X=0(==，P(X=1(==，P(X=2(==，P(X=3(==.X的分布列為X0123P1 數(shù)學(xué)期望E(X(=0×kP(A1(=1-=，P(A2(=++=，P(A3(=+若在一次實驗中事件發(fā)生的概率為p(0<p<1(，則在n次獨立重復(fù)實驗中恰好發(fā)生k次概率p(ξ=k(=Cpk(1-p(n-k(k=0,1,2,?,n(，稱ξ服從參數(shù)為n,p的二項分布，記作ξ~B(n,p(，Eξ=np，Di=所以X～B(4,0.25(，所以P(X=0(=(，P(X=1(=C，P(X=4(=C×4=，所以X的分布列為：X01234P 所以E(X(=4×=1；所以P(N|M(= ≤x2≤?≤x20，經(jīng)計算2=s=443.6.故X的分布列為：X01234P 則E(X)=4×=. (2)首先算出P(B(進一步結(jié)合二項分布的概2+C22=，P(X=4(=2×2=，所以X的分布列為X01234P 數(shù)為X，X=k(k=0,1,?,n)的概率記為P(X=k(，則n為何值時，P(X=6(的值最大？則X～B(n,，可得P(X=6(=C6(1-n-6=C6n-6，令>1，解得n<7，可知當(dāng)n≤6，可得an+1>a令<1，解得n>7，可知當(dāng)n≥8，可得an+1>an；數(shù)，且σ>0，-∞<μ<+∞)。(3)ξ~N(μ,σ2(，則ξ在(μ-σ,μ+σ(，(μ-2σ,μ+2σ(，(μ-3σ,μ+3σ(上取值的概率分別為68.3%，，則P(μ-σ<Z<μ+σ)≈0.6827，P(μ-2σ<Z<μ+2σ)≈0.9545，P(μ-3σ<Z<μ=15，則X的分布列為：X012P5 2=362故P(Z≥91)=P(Z≥μ+2σ)=[1-P(μ-2σ<Z<μ+2σ)]≈0.02275， 附：若X~N(μ,σ2((σ>0)，則P(μ-σ<X<μ+σ(≈0.683，P(μ-2σ<X<μ+2σ(≈0.954，P(μ-3σ<X<μ+3σ(≈0.997.所以P(X≥72(≈=0.1585.P(Y=2(=×(1-2=；(1-=；P(Y=8(=(1-×2=；P(Y=10(=×2=. 附：若X~N(μ,σ2(，則P(μ-σ<X≤μ+σ(=0.6827，P(μ-2σ<X≤μ+2σ(=0.9545，P(μ-3σ<X≤μ+3σ(=0.9973.∴P(ξ=0(=P(ξ=4(=4=，P(ξ=1(=C1-3=，P(ξ=2(=C2(1-2=，P(ξ=3(=C3(1-=，ξ01234P 38 4 41∴P(Y≥250(=P(Y≥231+19(=×(1-P(μ-σ<Y≤μ+σ((連續(xù)型隨機變量X，定義其累積分布函數(shù)為F(x)=P(X≤x)．已知某系統(tǒng)由一個電源和并聯(lián)的A，B，C間工作相互獨立.t,t,1>t2>0，證明：P(T>t1|T>t2)=P(T>t1-t2)；t<0t≥0.【分析】(1)根據(jù)正態(tài)分布的對稱性即可結(jié)合F(x)=P(X≤x)的定義求解，【詳解】(1)由題設(shè)得P(38<X<42)=0.6827，P(36<X<4所以F(44)-F(38)=P(X≤44)-P(X≤38)=P(40≤X≤44)+P(38≤X≤40)PT>t1|T>t2)=======4t2-t1，P(T>t1-t2)=1-P(T≤t1-t2)=1-G(t1-t2)=4t-t,所以P(T>t1|T>t2)=P(T>t1-t2).(ⅱ)由(ⅰ)得P(T>n+1|T>n)=P(T>1)=1-P(T≤1)=1-G(1)=，所以第n+1天元件B，C正常工作的概率均為.束.∴進行3局比賽決出冠亞軍的概率為P=+=故X的分布列為：X12P 23 3 X234P 9 49 49期望EX=所以結(jié)束時恰好打了6局的概率為P=P1+P2=+=.則P(X=2(=2=，P(X=3(=3+C2×=，P(X=4(=C×3×+C×2=，所以X的分布列如下：X234P 9 49 49 P(X=4)=4+4=；P(X=5)=C4×+C4×=； 隨機變量X的分布列為：X4567P 8 45 P(X=1(=1-=，P(X=2(=×(1-=，P(X=3(=×=，P123X 25 相互獨立.P(A(=α=，P(B(=β=，P(C(=γ=，所以P(N)=P(ABAA)+P(BAAA)+P(ACCA)+P(CACA)+P(CCAA)4+32=.P(X=2)=α2+β2，所以X的分布列為X245P2+β22+β2)所以X的期望E(X)=2(α2+β2)+8αβ(α2+β2)+20α2β2學(xué)員贏得比賽的概率與比賽一開始甲學(xué)員贏得比賽的概率相同.=P(A)P(A)+P(A)P(B)P(M)+P(B)P(A)P(M)=α2+αβP(M)+βαP(M)=α2+2αβP(M)(α+β)2-2αβα2+2αβ+β2-2αβα2+β2(α+β)2-2αβα2+2αβ+β2-2αβα2+β2. ×=，P(X3=3(=1--= X3123P 3 7所以=k(1≤k≤n-1,k∈N*(時，P(=k(=((k-1×;當(dāng)=n時P(=n(=((n-1，故YYn123?n-1nP 3 ?n-2n-1+nn-1(n∈N*，n≥2).E(Yn+1(-E(Yn(=nn-1×+(n+1(n-nn-1=n>0，故E(Yn(單調(diào)遞增；由上得E(Y2(=，故E(Yn(=E(Y2(+[E(Y3(-E(Y2([+[E(Y4(-E(Y3([+?+[E(Yn(-E(Yn-1([，∴E(Yn(=+2+3+?+n-1=+=3-2×n-1<3，故E(Y2(<E(Y3(<E(Y4(<E(Y5(<?<E(Yn(<3. 所以P=P1+P2=160+40則P(X=2(=2=，P(X=3(=C+C3=+=，P(X=4(=C×2×=，所以隨機變量X的分布列為：X234P 49 3 29 金的占比.P(Y=4)=4×2=P(Y=5)=C5×2=P(Y=6)=C6×2=P(Y=7)=C7×2=Y4567P 8 4 5P(X=3)=(1-p)3;P(X=4)=C(1-p)3?p設(shè)乙贏得全部獎金為事件A，則P(A)=P(X=3)+P(X=4)=(1-p)3(1+3p)max=f==0.0272<0.05 =.所以隨機變量X的分布列為：X2345P 49 則P(A(=,P(B(=,P(C(=，P(ξ=1(=P(AB)=P(A)P(B)=×=，=P(A)P(C)+P(A)P(B)P(C)+P(A)P(C)P(B)=×+ ξ123P 3 38 決策.比賽四局結(jié)束，所以P(X=4(=1-P(X=2(-P(X=3(=.所以X的分布列為X234P 49 3 29事件D.=.=.所以P(D(=P(AD(+P(BD(+P(CD(=P(D∣A(P(A(+P(D∣B(P(B(+P(D∣C(P(C(==P(D∣B(>P(D∣C(， 所以三人總積分X的分布列為X468P555為p3.故P(A(=p3(1-p1(+p3p1(1-p2(p3+(1-p3(p2(1-p1(p3；同理可得P(B(=(1-p1(p3+(1-p1((1-p3(p2(1-p1(+p1(1-p2(p3(1-p1(；P(C(=p2(1-p1(p3+(1-p2(p3(1-p1(=p3(1-p1(；顯然P(B(-P(C(=(1-p1((1-p3(p2(1-p1(+p1(1-p2(p3(1-p1(>0，故P(B(>P(C(，P(A(-P(B(=[p3p1(1-p2(p3-p1(1-p2(p3(1-p1([+[(1-p3(p2(1-p1(p3-(1-p1((1-p3(p2(1-p1([=(p1+p3-1(p1(1-p2(p3+(p1+p3-1((1-p3(p2(1-p1(=(p1+p3-1([p1(1-p2(p3+(1-p3(p2(1-p1([，由于p1+p3>1，故P(A(-P(B(=(p1+p3-1([p1(1-p2(p3+(1-p3(p2(1-p1([>0，所以P(A(>P(B(； 概率為：P1=1-1-=；(1-1-1-=，所以丙獲得冠軍的概率P=+=.所以甲獲得冠軍的概率P=+=. iiiD=B1B2A3A4A5+B1C2A3A4A5+A1A2B3B4A5+A1A2B3C4A5+A1C2C3A4A5+A1C2B3A4A5,所以P(D(=+++++=.；(2)證明見解析，Pn=-n-1+(3)E(Y)=1-(-n+n-1n-1+i-1+ -qi((sinqi+1-sinqi(<2=n=+-n-1(3)證明見解析2=3=p2+(1-p2(=；n=pn-1+(1-pn-1(，∴pn=-pn-1+，即pn-=-pn-1-，顯然qn>qn+1，則h(qn(>h(qn+1(，∴qn-sinqn>qn+1-sinqn+1，則=qn-qn+1>sinqn-sinqn+1，即(qn+1-qn((sinqn+1-sinqn(=(qn-qn+1((sinqn-sinqn+1(<， P(X=1(==；P(X=2(==；P(X=3(==.所以隨機變量X的分布列為：X123P3 351則有P1=0,P2==,P3==.An+1=A?An+1+An?An+1所以Pn+1=P(A?An+1+An?An+1(=P(A?An+1(+P(An?An+1(=P(A(?P(An+1∣A(+P(An(?P(An+1∣An(=(1-Pn(?+Pn?0=(1-Pn(.即Pn+1=-Pn+.所以Pn+1-=-Pn-，且P1-=-. =A×3=.P(X=1(=3×3=，P(X=3(=A×3=，P(X=2(=1-P(X=1(-P(X=3(=，X的分布列如下：X123P 9 23 29所以，當(dāng)n≥4時，Pn=Pn-1+(1-Pn-1-3××=+Pn-1-所以Pn=+Pn-1- n-1. 本結(jié)果為陽性的概率是p(0<p<1).比較E(X)與E(Y)的大小.所以P(A(=Cp(1-p)2p=3p2(1-p)2.所以P(Y=1(=(1-p)20,P(Y=21(=1-(1-p)20，所以E(Y(=(1-p)20+21[1-(1-p)20[=21-20(1-p)20.所以E(Y(-E(X(=1-20(1-p)20，令1-20(1-p)20>0，解得p>1-20.所以當(dāng)p>1-20時，E(Y(>E(X(；當(dāng)p=1-20時，E(Y(=E(X(；當(dāng)0<p<1-20時，E(Y(<E(X(. (2)已知每個人患該疾病的概率為p(0<p<1(.(2)(i)E(X(=(1-p(n+(n+1([1-(1-p(n[；(ii)答案見解析②根據(jù)題意可得：P==；所以E(X(=(1-p(n+(n+1([1-(1-p(n[；方案二：檢查的次數(shù)期望為E(X(=(1-p(5+6[1-(1-p(5[，E(X(-5=[6-5(1-p(5[-5=1-5(1-p(5，記g(p(=1-5(1-p(5，因為0<1-p<1，所以g(p(單調(diào)遞當(dāng)1-<p<1時，g(p(>0， 1-=.P(X=1)=(、p)4=p2；P(X=5)=1-p2，X15Pp21-p2則E(X)=5-4p2，P(Y=2)=p2；P(Y=4)=Cp(1-p)=2p(1-p)；P(Y=6)=(1-p)2；Y246Pp2(1-p(2則E(Y)=2p2+4(2p-2p2)+6(1-p)2=6-4p，E(Y)-E(X)=6-4p-(5-4p2)=4p2-4p+1， *陽性的概率為p(0<p<1(.3=0.2048ξ16P5-0.85(=4.35.②當(dāng)采用混合檢驗的方案時E(ξ(=1×(1-p(k+(k+1([1-(1-p(k[=k+1-k(1-p(k，即k+1-k(1-p(k<k，化簡得0<p<1-k，所以當(dāng)P滿足0<p<1-k，用混合檢驗的方案能減少檢驗次數(shù). (1)若E(ξ1(=E(ξ2(，試求p關(guān)于k的函數(shù)關(guān)系式p=f(k(；1，E(ξ2(=k+1-k(1-p(k，根=E(ξ2(解得p=1-(即可得解；(ii)求出p，根據(jù)E(ξ1(>E(ξ2(得到lnk>k，再構(gòu)造函數(shù)f(x(=lnx-利用導(dǎo)數(shù)可求得結(jié)∴P(ξ2=1(=(1-p(k，P(ξ2=k+1(=1-(1-p(k.∴E(ξ2(=(1-p(k+(k+1([1-(1-p(k[=k+1-k(1-p(k.若E(ξ1(=E(ξ2(，則k=k+1-k(1-p(k，N?,且k≥2).-n-11=1k-1k②假設(shè)對任意的n=k時，xk=e3-+1+(e--e-+k+1-1=0，所以(e-k+1-1((e-+k+1+1(=0.∴xk+1=e或xk+1=-e-.kn-1所以==en{為等比數(shù)列.>E(ξ2(，3-xn(n+1n+1+1=nn+1，vn+1n+1的通項公式.n+1=p，vn+1=2pnqn，wn+1=q，=+(n-1(，其中q1=-==，+n，qn=，于是wn+1=q *-.(ii)依題可知，p=1-e-(ξ1(-E(ξ2(=k(1-p(k-1=ke--1>0，ke--1>0→lnk->0，構(gòu)造函數(shù)f(k(=lnk-(k≥2)，由；當(dāng)進行混合檢驗時，P(ξ2=1(=(1-p(k，P(ξ2=k+1(=1-(1-p(k則E(ξ2(=(1-p(k+(k+1([1-(1-p(k[=k+1-k(1-p(k“E(ξ1(=E(ξ2(，:k=k+1-k(1-p(k則(1-p(k=，即p=1-(-k-1②假設(shè)當(dāng)n=k(k≥1,k∈N*(時，xk=e3+....+-2(k-1)-e3--1+....+=a1+a2+-2(k-1)-e3--1+....+(=e-+1-ek+1+e=0:xk+1=e或xk+1=-en-1n=e3對一切n∈N*都成立.即{xn{為等比數(shù)列.-E(ξ2(=k(1-p(k-1=ke--1>0:->ln→lnk->0令f(k(=lnk-(k≥2)，則f’(k(=-=∵f(9(=ln9-<0，f(8(=ln8->0則k的最大值為8. 是陽性還是陰性都是獨立的，且每份樣本是陽性結(jié)果的概率為p(0<p<1(N*(n≥2(，都有e-E(ξ2(.結(jié)合E(ξ1(=E(ξ2(，化簡即可關(guān)于k的函數(shù)關(guān)系式p=f(k(；得p=1-=1-，E(ξ1(>E(ξ2(，化簡可得lnk>k，構(gòu)造函數(shù)f(x(=lnx-x(x>0(，求得∴P(ξ2=1(=(1-p(k，P(ξ2=k+1(=1-(1-p(k.∴E(ξ2(=(1-p(k+(k+1([1-(1-p(k[=k+1-k(1-p(k.若E(ξ1(=E(ξ2(，則k=k+1-k(1-p(k，(1-p(k=，∴1-p=k，∴p=k.:p關(guān)于k的函數(shù)關(guān)系式為f(k(=1-(“x1=1，:下面證明對任意的正整數(shù)n，xn=e.k-1k-=，:e-:e-.x+1+(e--e-+k+1-1=0，(e-k+1-1((e-+k+1+1(=0.:xk+1=e或xk+1=-e-kn-1:xk+1=e3成立.:由①②可知，{xn{為等比數(shù)列，xn=e3.:k>k+1-k(1-p(k，得<(1-p(k=((k，:lnk>設(shè)f(x(=lnx-x(x>0(，f/(x(=，:當(dāng)x≥3時，f/(x(<0，即f(x(在[3,+∞(上單調(diào)減.:ln5<.:k的最大值為4. P(Xt+1|...,Xt-2,Xt-1,Xt(=P(Xt+1|Xt(.為A(A∈N*,A<B(，賭博過程如下圖的數(shù)軸所示.(1)請直接寫出P(0(與P(B(的數(shù)值.計含義.-P(n-1(=P(n+1(-P(n(，即可即P(n)=P(n-1)+P(n+1)，所以P(n(-P(n-1(=P(n+1(-P(n(，所以{P(n({是一個等差數(shù)列，設(shè)P(n(-P(n-1(=d，則P(n-1(-P(n-2(=d，?,P(1(-P(0(=d，累加得P(n(-P(0)=nd，故P(B(-P(0)=Bd，得d=-，-P(0(=nd得P(A(-P(0(=Ad,即P(A)=1-,當(dāng)B=200時，P(A(=50%， 4864P(x2≥m)m2==≈10.667>6.635P(X=0)=C04=；P(X=1)=C13=P(X=2)=C22=P(X=3)=C31=P(X=4)=C40=.X的分布列為：X01234P E(X)=4×= P(X=0(=CC7=136P(X=1(=CC7=51PX012P 3 E(X(.(a+b((c+d((a+c((b+d(2=n(ad-bc)2,n(a+b((c+d((a+c((b+d(αxα則χ2=≈9.524>3.841=x0.05，P(X=0(=3=,P(X=1(=C2=，P(X=2(=C2=,P(X=3(=C3=，所以隨機變量X的分布列為：X0123P ，則P(μ-σ<Y≤μ+σ(=0.6826，P(μ-2σ<Y≤μ+2σ(=0.9544)所以P(Y>90(=[1-P(80-10<Y≤80+10([