《概率论与数理统计》答案详解

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2、${K:1.Æ:A∪B=B∪A,AB=BA.2.(ÜÆ:(A∪B)∪C=A∪(B∪C)=A∪B∪C,(AB)C=A(BC)=ABC.3.©Æ:A(B∪C)=AB∪AC.4.éá¯$:A=A.5.éóÆ:A∪B=AB,AB=A∪B;A∪B∪C=ABC,ABC=A∪B∪C.1ݺ

3、^©¼ã5n)¯'X$9${K.éASK3,4,5,6.•n)ªÇVÇ¿Â,¿ÝºVÇ5:1.K5:0≤P(A)≤1.2.55:P(Ω)=1,P(φ)=0.3.5:P(A1+A2+A3+···)=P(A1)+P(A2)+P(A3)+···.4.IfA⊂B,thenP(A)≤P(

4、B).•ÝºVÇüO.:;VÇ.AÛVÇ..©Oݺùü.£ã±9éAVÇúª.1.;V.:(i).kmΩ={ω1,ω2,···,ωn}.(ii).P(ω1)=P(ω2)=···=P(ωn).¯AVÇ:P(A)=m=A¤¹:ê.nΩ¥:ê2.AÛV.:«ΩSÅÝ:,K:MáΩ,Ü©AVÇ:P(A)=A¡È.Ω¡È•ÄOên:1.{n:©õ«¹.2.¦{n:©õÚ½.•Äü

5、Ü:ü

6、Ü«O´ük^S(½ö?Ò),

7、ÜØÄ^S.1.Eü:1,2,3,···,n,ùnêi¥Ñkêiü¤,êi±E,

8、o
knk«{.2.ØEü:1,2,3,···,n,ùnêi¥kØÓêiü¤,o
kAk=n(n−1)···(n−k+1)«{.n3.

9、Ü:1,2,3,···,n,ùnêi¥Ñkêi,ØÄÙk^S,o
kCk«{.n•üêÚ

10、Üê:1.PüêAn=n!¡nü.5½0!=1.nkn!Ak2.

11、ÜêC==n.nk!(n−k)!k!~X:C3=10∗9∗8.103!3.Ck=Cn−k.nn~X:C7=C3.1010§1.2Y)1.2.1.(1)^1P1:,Ù§aí,KmΩ={1,2,3,4,5,6}.(2)^1,2,3PÑ?Ò1,2,3

12、n¥,Ù§aí,KmΩ={(1,2,3),(1,2,4),(1,2,5),(1,3,4),(1,3,5),(1,4,5),(2,3,4),(2,3,5),(2,4,5),(3,4,5)}.(3)3g¬ÜÑ,¤±3g,õ10g,u´mΩ={3,4,5,6,7,8,9,10}.(4)^1L¬,0Lg¬;(1,0)L1¬1g¬,daí.¦©(0,0),(0,1,0),(0,1,1,0),(0,1,1,1),(1,0,0),(1,0,1,0),m.(1,0,1,1),(1,1,0,0),(1,1,0,1),(1,1,1,0),(1,1,1,

13、1))1.2.2.^1L¬,0Lg¬.Km{ω1=(0,0,0,0),ω2=(0,0,0,1),ω3=(0,0,1,1),ω4=(0,1,1,1),ω5=(0,1,0,0),ω6=(0,0,1,0),ω7=(0,1,0,1),ω8=(0,1,1,0),ω9=(1,0,0,0),ω10=(1,0,0,1),ω11=(1,0,1,1),ω12=(1,1,1,1),ω13=(1,1,0,0),ω14=(1,0,1,0),ω15=(1,1,0,1),ω16=(1,1,1,0)}.¯A=”~Ñ”={ω12}.¯B=”2u”={ω4,ω11,ω15,ω16}.¯C=

14、”ü?Ñ”={ω3,ω7,ω8,ω10,ω13,ω14}.¯D=”ØÑ”={ω1,ω2,ω5,ω6,ω9}.)1.2.3.YØ.(1)ABC,(2)A,(3)ABC,(4)ABC,(5)A∪B∪C,(6)ABC∪ABC∪ABC∪ABC,(7)AB∪BC∪CA,(8)ABC,(9)ABC∪ABC∪ABC,(10)ABC∪ABC∪ABC.)1.2.4.(1)A={4,5,6},u´AB={4}.(2)A∪B={2,3,4,5,6}.(3)B−A=BA={4},lB−A={1,2,3,5,6}.(4)BC={4},BC={1,2,3,5,6},ABC={1,2,3},Ï

15、dABC=