PAUBUC1PABC1PA)P(B)P(C1421． （1）P(ABP(AP(B0.80.9（2）P(A∪BP(AP(BP(AB0.80.90.72（3）P(A∪BABP(A∪BP(AB0.980.72故所求概率為P(A|AUB)P(

P(A)P(B)P(

P(A)P(B)P(

400.8511．ABCABC0.3,0.2,0.2，且各元件獨立工ABCABC（1）PAUBUC1PABC1PA)P(B)P(C10.70.80.80.552（2）P(ABCP(AP(BP(C0.30.20.2（3）P(A∪BCP(AP(BCP(ABCP(AP(BP(CP(AP(BP(C=0.3+0.20.20.30.20.2=（1）P(ABCP(AP(BP(C0.10.20.15（2）PAUBUC1PABC1PA)P(B)P(C10.90.80.850.388（3）PABCUABCUABCUABCPABCPABCPABCPP(A)P(B)P(C)P(A)P(B)P(C)P(A)P(B)P(C)P(=0.10.20.15+0.10.20.85+0.10.80.15+0.90.20.15=A1A2A3P(Ai2/3，i1,2,3A1A2A3解：（1）PAUAUA1P(AAA1PA)PA)PA111126 12 2111211122 PA1A2A3UA1A2A3UA1A2A3UA1A2A3P(A1)P(A2)P(A3)P(A1)P(A2)P(A3)P(A1)P(A2)P(A3)P(A1)P(A2)P(A32111211121117 若A與B相互獨立且互不相容，試求min{P(A),P解：因A與B相互獨立且互不相容，有P(AB)=P(A)P(B)且AB=，即P(AB)=0，P(AP(B)=0P(A)=0P(B)=0，min{P(AP(BABAB（3）A（2）ABP(A∪BP(AP(BP(ABP(AP(BP(APP(BPAUBPA)0.90.40.50.83331P( 1 ABP(BP(A∪BABCABCP(AP(BP(CxxP(AP(BP(C1/2P(A∪B∪C9/16P2x2P(AB∪ACP(Axx另一方面，x0.5P(AP(B0.5，P(AB0.25CABUABP(CPABUABPABPABPAPABP(BPAB0.5，P(AB)=0.25=P(A)P(B)，A,B獨立，PACPABPAPAB0.25PA)P(CACP(BCPABP(BPAB0.25P(B)P(CB,CP(AP(BP(C0.5，ABCABCx0.5，故x的最大值等于0.5；（2）P(AP(BP(Cxx12P(A∪B∪CP(AP(BP(CP(ABP(ACP(BCP(ABC=P(A)+P(B)+P(C)P(A)P(B)P(A)P(C)P(B)P(C)=3x3x93x3x2x2x3x1)(x30x1x3x1x14

ABPABPAB1PA)P(BPA)[1P(BPA)P(B1PA)]P(B4P(AP(B)PA)[1PA1，即[PA)]2PA1PA1]20 PA1P(B1 i1,2,3X3P{X11233 解：設(shè)Xn表示n門高射飛機的次數(shù)，且每門高射飛機的概率為p=0.3，則至少命中一次的概率為P{Xn1}=1P{Xn=0}=1(1p)n=10.7n0.99，即0.7n0.01，故nln0.0112.9114，即需要13門高射就能以99%的把握 6P{Xn1}1P{Xn0}15/6)n1/2，即(5/6)n1/2，nln(12)3.8018461/2．80/81，試求該射手進行一次射擊P{X1}1P{X0}11p480，即(1p41 中率為p23解：設(shè)Xn表示n次射擊中次數(shù)，且每次射擊為p=則至少命中一次的概率為P{Xn1}=1P{Xn=0}=1(1p)n=10.8n0.9，即0.8n0.1，故nln0.110.3189，即射擊至少11次才能使至少 一次的概率不小于0.9．P(A10.5PA100PA1PA100PA1PA1UA1A2UA1A2A3PA1PA1A2PA1A2A3P(A)P(A)P(A)P(A)P(A)P(A)1111293 （1）P{X201}=1P{X20=0}=1(1p)20=10.9820=（2）P{Xn1}1P{Xn0}11pn10.98n0.950.98n（2）P(A20.3740.2140.0840.344則三局兩勝制時，甲獲勝的概率為P(A1)=0.36+0.288=0.648；3∶2勝乙的概率為40.630.42P(A20.2160.25920.207360.68256；故P(A1)<P(A2)，五局三勝制時對甲更有利．場為止，此人即為冠軍．而每次比賽雙方取勝的概率都是1/2，現(xiàn)假定甲、乙兩人先比，試求各人得解：設(shè)每局比賽中，甲勝乙、乙勝甲、甲勝丙、乙勝丙、丙勝乙分別記為Ab,Ba,Ac,Ca,Bc,Cb，①AbAc，AbCaBcAbAc，AbCaBcAbCaBcAbAc，……，(AbCaBc)kAbAc②BaCbAcAb，BaCbAcBaCbAcAb，BaCbAcBaCbAcBaCbAcAb，……，(BaCbAc)kAbP(A(0.520.550.58……(0.540.570.510 1

1

2

5P(BPA5①AbCaCb，AbCaBcAbCaCb，AbCaBcAbCaBcAbCaCb，……，(AbCaBc)kAbCaCbBaCbCa，BaCbAcBaCbCa，BaCbAcBaCbAcBaCbCa，……，(BaCbAc)kBaCbCa，……，P(C)=(0.53+0.56+0.59+……)+(0.53+0.56+0.59+……)2

271/2．兩人約定誰先贏得一定的局?jǐn)?shù)就獲得全部賭本．但賭P(A2p221pp231p2p20.5220.5330.540.6875，則乙獲勝的概率P(B2)=1P(A2)=0.3125， n 2 nm因甲獲勝的概率為P(A3)p1(1p)p2(1p)pL m1(1p)m1

n

nm0.510.5n120.5n2Lm10.5nm1 P(B31P

n

nm nm110.510.5n120.5n2L m1

n

nm故甲應(yīng)得賭本的0.510.5n120.5

L m10.5nm1 n

nm nm1乙應(yīng)得賭本的10.510.5n120.5n2L m1 至少贏得n局比賽，故甲獲勝的概率為nm1

nm

nm

p(1p)m1

pn1(1p)m2L

n1

nmnm nm nm

n1L

nm且每個前胎損壞的概率為p1=0.1，每個后胎損壞的概率為p2=0.2，AB相互獨立，P(AP{X1}1P{X0}1(1p1)210.92P(B)=P{Y1}=1P{Y=0}=1(1p2)4=10.84=P(A∪BP(A)+P(B)P(AB0.19+0.59040.190.5904=設(shè)0<P(B)<1，試證A與B獨立的充要條件是P(A|B)P(AB)．證：必要性，若A與B獨立則P(A|B)P(AB)P(A)P(B)P(A)，P(A|B)P(AB)P(A)P(B)P(A) P(B) P(B)PA|BPA|B

P(AB)P(AB)P(A)P(AB)

1P(AB)[1P(BP(B)[P(AP(AB)]P(ABP(AB)P(BP(A)P(BP(B)P故P(AB)=P(A)P(B)，即A與B獨立0P(A1，0P(B1PA|BPA|B1ABPA|BPA|BPAB)PAB)PAB)1PAUB)

1P(AB)1P(A)P(B)P(AB) 1P(B)P(AB)[1P(B)]P(B)[1P(A)P(B)P(P(B)[1P(AB)