概率论与数理统计—王松桂(第二版)习题解答

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1、SS11.1(1)ëYÊgÑ·¥,¯u),
Ý;Êg,mΩ1={5,6,7,...};(2)Æfüg,zg*Ñy:êUǑ1,2,3,4,5,6,mΩ2={2,3,...,12};(3)â¢S¹,mΩ3={0,1,2,...};(4)Ω4={(i,j)∣1≤i0

2、,y>0,x+y=l}.1.2(1)ABC;(2)A(B∪C);(3)A∪B∪C;(4)ABC∪ABC∪ABC;(5)AB∪AC∪BC;(6)AB∪AC∪BC½AB∪AC∪BC;(7)ABC;(8)ABC∪ABC∪ABC.1.3(1)AB={x∣0.8

3、(A∪B)=P(A)+P(B)−P(AB),P(A∪B)≤P(A+B);lkP(AB)≤P(A)≤P(A∪B)≤P(A)+P(B).1.7(1)¯W∪E={&ÁÑyí¼½òz5ú«},P(W∪E)=P(W)+P(E)−P(WE)=0.125+0.075−0.025=0.175;(2)¯W−E={&ÁÑyí¼,vkòz5ú«},P(W−E)=P(W)−P(WE)=0.125−0.025=0.1;-1-S(3)¯WE={&ÁÑyí¼,Ǒòz5ú«},P(WE)=P(W∪E)=1−0.175=0.825.1.8

4、(1)ÏǑP(AB)≤P(A)=0.6,P(AB)≤P(B)=0.8,¤±P(AB)=P(A),=A⊂B,P(AB)0.6;(2)ÏǑP(AB)=P(A)+P(B)−P(A∪B)≥P(A)+P(B)−1=0.4,¤±kP(AB)≥0.4,P(A∪B)=1,=A∪B=Ω,P(AB)0.4.1.9¯ABC={¯A,B,CÑu)},¯A∪B∪C={¯A,B,C¥ku)},dP(AB)=0,P(ABC)=0,lkP(A∪B∪C)=P(A)+P(B)+P(C)−P(AB)−P(AC)−P(BC)+P(A

5、BC)=0.2+0.3+0.5−0−0.1−0.2+0=0.7.1.10(1)dA∪B=AB∪BABBp½,P(A∪B)=P(AB)+P(B),P(AB)=P(A∪B)−P(B)=0.6−0.3=0.3;(2)dA=(A−B)∪(AB)A−BABp½,P(A)=P(A−B)+P(AB),P(AB)=P(A)−P(A−B)=0.8−0.4=0.4,P(AB)=1−P(AB)=1−0.4=0.6;(3)dP(AB)=P(AB)9A∪B=AB,P(A∪B)=P(A)+P(B)−P(AB)=P(A)+P(B)−P

6、(AB),P(B)=P(A∪B)−P(A)+P(AB)=P(A∪B)−P(A)+1−P(A∪B)=1−P(A)=0.7.1.11duz¥Ñ±4f¥?Û,¤±z¥ko«{.3ò3¥4f¥
k4«ØÓ{,z«{U5Ó.z«3{Ҵį,įoêǑ4.-Ai={k¥õf¥¥ê´i},Ù¥i=1,2,3,4×3×23P(A1)==;438
3×4×39P(A)=2=;2431641P(A3)==.4316-2-S1.12!¡Æfüg⎧mǑ⎫(1,1)(1,2)...(1,6

7、)⎨(2,1)(2,2)...(2,6)⎬Ω=........................⎩(6,1)(6,2)...(6,6)⎭-A1={ügÑy:êÚǑ3},A2={ügÑy:êÚǑ4},A3={ügÑy:êÚǑ5},21A1={(1,2),(2,1)},P(A1)==,361831A2={(1,3),(2,2),(3,1)},P(A2)==,361241A3={(1,4),(2,3),(3,2),(4,1)},P(A3)

8、==.3691.13
10lê¥?nê,
k3«ØÓ{
.z«{Ñ´U.z10«{´Ä¯.u´Ä¯oêǑ=120.3
(1)44×3¯A={nꥴ5}¹Ä¯êǑ2=2=6,61P(A)==;12020
(2)55×4¯B={nꥴ5}¹Ä¯êǑ2=2=10,101P(B)==.120121.14
12l®¥¥ÅÑü,
k2«ØÓ{
,z«