高等数学概率论与数理统计《初等概率论》-概率计算经典例题.pdf

《高等数学概率论与数理统计《初等概率论》-概率计算经典例题.pdf》由会员分享，可在线阅读，更多相关《高等数学概率论与数理统计《初等概率论》-概率计算经典例题.pdf（16页珍藏版）》请在淘文阁 - 分享文档赚钱的网站上搜索。

1、5?V15?V14 466-VO,VO,1 VO,1.1?1.(Chung-Erd os?)?Ak,1 6 k 6 n?KP ni=1Ak!?nPi=1P(Ak)?2nPi=1nPj=1P(Ai Aj).(Trans.Amer.Math.Soc.72(1952),p.179-186.)1.2?.2.k?7a?%2Ib?.?zI?Vp=1/2,zI?a1?1?Vq(a).).AL1IBkLk?7k?1.dKq(0)=1,q(a+b)=0,q(k)=P(Bk)=P(A)P(Bk|A)+P(Ac)P(Bk|Ac)=12P(Bk+1)+12P(Bk1)=12q(k+1)+q(k 1).uk2q(k)=

2、q(k+1)+q(k 1)l?q(k+1)q(k)=q(k)q(k 1)=.=q(1)q(0)=q(1)1.dq(n)1=nq(1)1.?n=a+b,?0 1=(a+b)q(1)1,i.e.,q(1)1=1a+b.?q(a)=1+aq(1)1=1 aa+b=ba+b.dV?7ak K?%b?1?V?.XJe?(b )71.15?V146-VO,1.3K1.3.1K3.n?N(N n)fz?V?zzNBe?V(1)A=/,?nk0(2)B=/Tknk0(3)C=/,km(m 6 n)0(4)D=/kkm(m 6 n)0(5)E=/?k30.).(1)?oNn,?k|n!,?P(A)=n!Nn.(

3、2).lNf?n|?Nn?2d(1)?P(B)=?Nn?n!Nn.(3).ln?m,?k?nm?U?nmN 1?P(C)=?nm?(N 1)nmNn.(4).d(3)?P(D)=?N1?nm?(N 1)n1Nn.(5).EBp?P(E)=1?Nn?n!Nn.4.n?N(N 6 n)z?V?zzNBe?V(1)A=/z0(2)B=/Tkmf?0(m N)(3)C=/,?kk0.225?V6).(1).?Ai=/1if?0 i=1,2,.,N P(A)=P(Ac1 Ac2 AcN)=1 P?Ni=1Ai?.5?P?Ni=1Ai?=NXi=1P(Ai)X16ij6NP(Ai Aj)+X16ijk6N

4、P(Ai Aj Ak).+(1)NP?Ni=1Ai?,?P(Ai)=?N 1N?n,i=1,.,N,P(Ai Aj)=P(Ai)P(Aj|Ai)=?N 1N?n?N 2N 1?n=?N 2N?n,i 6=j=1,2,.,N,.P?Ni=1Ai?=?N NN?n=0.P?Ni=1Ai?=NXi=1(1)i1?Ni?N iN?n.?P(A)=1 P?Ni=1Ai?=NXi=0(1)i?Ni?N iN?n.(2).NTkm?k?Nm?U.,mf?VP(A1 A2 Am Acm+1 AcN)=P?mi=1Ai?P?Ni=m+1Aci?mi=1Ai?P?mi=1Ai?=?N mN?n,P?Ni=m+1A

5、ci?mi=1Ai?=NmXi=0(1)i?N mi?N m iN m?n?P(B)=?Nm?P(A1 A2 Am Acm+1 AcN)=?Nm?N mN?n NmXi=0(1)i?N mi?N m iN m?n335?V146-VO,=?Nm?NmXi=0(1)i?N mi?N m iN?n.(3).”?ckk?dK?N kzfk?P(C)=P(Ac1 Ac2 Ack(Ak+1+Ack+1)(AN+AcN)=P(Ac1 Ac2 Ack)=kXi=0(1)i?ki?N iN?n.5.n?N,?k.?z?V?zzNBgk(1 6 k 6 n)?VP(k).).?1 6 k n ck 1gv 1k

6、g?VP(k)=?N 1N?k11N.?k=nP(n)=?N 1N?n1?N 1N+1N?=?N 1N?n1.?P(k)=1N?N1N?k1,k=1,2,.,n 1,?N1N?n1,k=n.1.45K6.?1,-E?zg?Vp,(0 p l?Vp(0 p?Vq=1 p,?m3?:/Ln m?:0?V.).?nP(A)=0,?n duLn m?:?3nLm?n/2?P(A)=?nn/2?pn/2qn/2d?P(A)=?nn/2?pn/2qn/2,n,0,n.8.(k9?i=1K)?:3?80,1,2,3,n+mim un mm l!l!?VOp,q,rvp+q+r=1?u0:m+n:2?0:?V

7、P0(n).).?P0(j):uj?0:?Vj=0,1,2,3,n+m.w,P0(0)=1,P0(n+m)=0,d?V?P0(j)=pP0(j+1)+rP0(j)+qP0(j 1),1 6 j 6 n+m 1r=1 q p,?P0(j)=pP0(j+1)+(1 p q)P0(j)+qP0(j 1)l?kP0(j+1)P0(j)=qpP0(j)P0(j 1)=(qp)jP0(1)P0(0)=(qp)jC,C=P0(1)1555?V146-VO,uP0(j)=P0(j 1)+C(qp)j11=P0(n+m)P0(0)=m+nXj=1P0(j)P0(j 1)=Cm+nXj=1(qp)j1=C1 (q

8、/p)m+n1 (q/p),?p 6=qC=1 (q/p)1 (q/p)m+n.?P0(j)=C(qp)j1+P0(j 1)=C(qp)j1+(qp)j2+qp+1+P0(0)=1 1 (q/p)1 (q/p)m+n1 (q/p)j1 (q/p)=(q/p)j(q/p)m+n1 (q/p)m+n.P0(n)=(q/p)n1 (q/p)n1 (q/p)m+n?p=q dP0(j)P0(j 1)=P0(1)P0(0)=C?P0(m+n)P0(0)=m+nXj=1P0(j)P0(j 1)=m+nXj=1C=C(m+n)=1=C(m+n)665?V6l?C=1m+nqP0(j)=C+P0(j 1)=C

9、j+P0(0)=1+Cj=m+n jm+nu?VP0(n)=(q/p)n1(q/p)m1(q/p)m+n,p 6=qmm+n,p=q.dd?:m un:?n+m:?V:Pm+n(n)=1 P0(n)=1(q/p)n1(q/p)m+n,p 6=q1 P0(n)=nm+n,p=q.1.6m5K9.!?1,m?zm?Vp?Vq=1 p,m?1o3e?m5Ke?V(1)2n+1n+1(2)XknX(3)3mckn3mckn(4)Xk2X(5)XknX(6)nm(7)XknX?XJyn 1n 1K?Xm(8)XknX?XJyn 1n 1Km-#m)w,(1)(3)?n,mn+1?A(2)(3)?n,mn

10、?A?(3)(4)w,Xm7L?1?B=/0;Ai=/3cmTi0,i=0,1,2,d?V?P(B)=2Xi=0P(B|Ai)=0+?21?pq P(B)+?22?p1 1775?V146-VO,=2pqP(B)+p2,q=1 p.l?kP(B)=p21 2pq(4)?,)?An=”31ngm?”Kn 31n1n2n2/2 P(A2)=p2,P(A4)=2pqp2=2p3q,P(A6)=p2(2pq)2.?kP(A2m)=p2(2pq)m1,m=1,2duA2mp?VP(Xm=1A2m)=Xm=1P(A2m)=p21 2pq.(5)?b?p 6=q?P(j)?n j?e?Vj=0,1,2n,w

11、,P(0)=1,P(2n)=0,VP(n),l?d?V?P(j)=pP(j 1)+qP(j+1)dd?P(j+1)P(j)=pqP(j)P(j 1)4?P(j+1)P(j)=(pq)jP(1)P(0)=C(pq)j,C=P(1)P(0)l?P(j)=C(pq)j1+P(j 1)=C(pq)j1+(pq)j2+pq+1+P(0)=1+C1 (p/q)j1 (p/q)q1=P(2n)P(0)=2nXj=1P(j)P(j 1)=2nXj=1C(pq)j1=C1 (p/q)2n1 (p/q)885?V6l?C=1 (p/q)1 (p/q)2nP(j)=1 1 (p/q)1 (p/q)2n1 (p/q)

12、j1 (p/q)=(p/q)j(p/q)2n1 (p/q)2n?p=q yP(j)=1 j2n?P(j)=(p/q)j(p/q)2n1(p/q)2n,p 6=q,1 j2n,p=q,l?P(n)=pnpn+qn,p 6=q,12,p=q.(6)|)(5)?aq?P(n)=pn(qm pm)qm+n pn+m,p 6=q?p=qP(n)=1 nn+m=mn+m3)KI5.P(0)=1,P(m+n)=0.w,(4)(5)?n=2?A?(5)q(6)?m=n?A(7).U3yn1n1c U3yn1n1?OABw,kAB=,?VP(A+B)=P(A)+P(B),d?)2P(A)=2n2Xk=npn?k

13、 1n 1?qknqP(B)=?2n 2n 1?qn1pn1pmqm+pml?kP(A+B)=2n2Xk=npn?k 1n 1?qkn+?2n 2n 1?pn+m1qn1/(qm+pm)5?m=1(7)(2).(8).?A=/3yn 1c0B=/3yn 1n 1?0K?VP(A+B)=P(A)+P(B)?a=P(A)=2n2Xk=npn?k 1n 1?qkn995?V146-VO,b=?2n 2n 1?pn1qn1KP(A+B)=a+ba+b2a+b3a+=a1b,):?D=/0A=/3ck(n 6 k 6 2n2)n/B=/yn 1n 10,C=/3ck(n 6 k 6 2n 2)n0 w,

14、kP(A)=a,P(B)=b,d?V?P(D)=P(A)P(D|A)+P(B)P(D|B)+P(C)P(D|C)=P(A)+bP(D)+0 P(C)=a+bP(D).kP(D)=a1 b.1.7K10.(Polya?fK)?fCkbr?,?2c?(1)?1?g?V;(2)XY?1ng?n1n2?V(n1+n2=n);(3)y?g?Vbb+r?g?Vrb+r;(4)1mg1n(m m)K?Bj=/1jg?0 d(3)?P(B1Bn)=P(B1)P(Bn|B1)=bb+rP(Bn|B1)=bb+rb+c(b+c)+r=b(b+c)(b+r)(b+r+c).d?m=1n(n m)K?m=k 1n(n

15、 m)K ym=kn(n m)KP(BkBn)=P(B1)P(BkBn|B1)+P(Bc1)P(BkBn|Bc1),?P(BkBn|B1)?ulCkb+cr?1k 1gn 1g?V db?P(BkBn|B1)=b+cb+c+rb+2cb+2c+r.nP(BkBn|Bc1)=bb+c+rb+cb+2c+r.l?P(BkBn)=bb+rb+cb+c+rb+2cb+2c+r+rb+rbb+r+cb+cb+2c+r=b(b+c)(b+r)(b+r+c).11115?V146-VO,11.(Banach?K),mCk?zg?b?zkN1guy,Tkr?V(r=1,2,N)21g?(uy),eTkr?V(

16、r=1,2,N)(1)?k?kU5?L/0?mL/”0 Kuy mTkr?V-E?1N+1u)32N r+1g?V K?V=?2NrN?(12)N(12)Nr 12,?V?21?2N rN?(12)N+1(12)Nr=?2N rN?(12)2Nr(2)aqu(1)1Ng/0u)312N rg?V?21?2N rN?(12)N+112=?2N rN?(12)2Nr112.(?K)nO3?S?vkf3?V(n 1)Tkr?V).k5OG?ln?k?U?m /?G?ln?k?G?Akn/k+?,?l?n?G?(n 1)!?Ai=/1iO?0 i=1,2,nKvkf3?VP(ni=1Aci)=1 P(

17、ni=1Ai)dV?kP(ni=1Ai)=nXi=1P(Ai)nX16ijP(AiAj)+(1)(n1)P(A1A2An)P(AiAj)L1iO?1jO?V?rOw?VP(ni=1Aci)=1+nXi=1(1)k2k(2n k 1)!(2n 1)!.12125?V6TkrO?k?nr?U urO?nrO?(Xcr?,n r?)?VP(A1A2ArAcrAcr+1Acn)=P(A1Ar)P(AcrAcr+1Acn|A1Ar)=2r(2n r 1)!(2n 1)!1+nrXi=1?n rk?(1)k2k(2n 2r k 1)!(2n 2r 1)!#.?V=TkrO?VP=?nr?2r(2n r 1

18、)!(2n 1)!1+nrXi=1?n rk?(1)k2k(2n 2r k 1)!(2n 2r 1)!#.13.(n)?3g?n 0,u+w+2v=1.?+N8/f0“+Nn?).dK/I0/10UeD?“P(SAA|FAA,MAA)=1,P(SAA|FAA,MAa)=1/2,P(SAA|FAa,MAA)=1/2,P(SAA|FAa,MAa)=1/4.pSL/f0 FL/I0 ML/10 du+N N?uk/?N uP(FAA,MAA)=P(FAA)P(MAA)=u2,P(FAA,MAa)=P(FAA)P(MAa)=2uv,P(FAa,MAA)=P(FAa)P(MAA)=2uv,P(FAa,M

19、Aa)=P(FAa)P(MAa)=4v2.u1:w1:2v1L/f0“+N?|?V?u1=P(SAA)=P(SAA|FAA,MAA)P(FAA,MAA)+P(SAA|FAA,MAa)P(FAA,MAa)+P(SAA|FAa,MAA)P(FAa,MAA)+P(SAA|FAa,MAa)P(FAa,MAa)=u2+uv+uv+v2=(u+v)2./?w1=P(Saa)=(w+v)2.|u1+w1=w+2v+w=1,14145?V6?2v1=P(SAa)=1 u1 w1=(u1+w1)2 u1 w1=2u1w1./f0“n?u1:w1:2u1w1l?/f0?“n?u2=(u1+u1w1)2=u1,w2

20、=(w1+u1w1)2=w1,2v2=2u2w2=2v1.+N l1?“m zU?D15.(Dx)aD,/?0?b?akaa?U b?3+NkAa?Np(kAA?N?1p,kaa?NK?“)(1).kxx*duaa?e Aa?Vf?D5?kxx*duaa?uI1kAaE=FAa,MAaL?I1kAaVPE(C)=P(C|E),KV=AA+Aa?PE(Aa|)=PE(Aa,)PE()=PE(Aa)PE(AA+Aa)=PE(Aa)PE(AA)+PE(Aa)=1/21/4+1/2=23,?kAA?V1-2/3=1/3.(2).?+N?cf(?“?duTfgc+N I1?f5P(FAA)=13,P(F

21、Aa)=23,P(MAA)=1 p,P(MAa)=p.2?V?P(SAA)=P(SAA|FAA,MAA)P(FAA,MAA)+P(SAA|FAA,MAa)P(FAA,MAa)+P(SAA|FAa,MAA)P(FAa,MAA)+P(SAA|FAa,MAa)P(FAa,MAa)15155?V146-VO,=13(1 p)+1213p+1223(1 p)+1423p=23p3.P(Saa)=P(Saa|FAa,MAa)P(FAa,MAa)=1423p=6p.P(SAa)=1 P(SAA)P(Saa)=13+pp.(3).(2)I1?cfAa?VP(SAa|c)=P(SAa)P(c)=P(SAa)P(SAa)+P(SAA)=(13+p6)/(13+p6)+(23p3)=2+p6 p.5?0 6 p 6 1,2+p6p23.1616