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1A1(1)2__________________(2)532__________________(3)5323____________________(4)____________________2ΩABC(1)P(A)=(2)P(B–A)=P(BA)=(3)P(ABC)=3ABCABC(1)A(2)ABC(3)ABC(4)ABC(5)ABC(6)ABC4ABCP(A)=P(B)=P(C)=1/4P(AC)=1/8P(AB)=P(BC)=0(1)ABC(2)ABC(3)ABC5n123…n(n5)(1)12(2)1(3)1236557A1A2…AnΩA1A2…AnA1A2…An=ΩΩB(1)(2)8ABABC9102(1)(2)(3)(4)100.60.3________11ABP(A)=0.4P(AB)=0.7(1)ABP(B)=(2)ABP(B)=12P(A)=0.5P(B)=0.6P(B|A)=0.8=)(BAPU132480/811ABC3(A)AC=BCA=B(B)A–C=B–CA=B(C)AB=∅BA=∅BA=(D)AC=BCA=B2AB(A)AB=∅BA,(B)AB=∅BA,(C)AB≠∅BA,(D)AB≠∅BA,3ABP(A)0P(B)0(A))()(BPBAP=U(B)AB(C)AB(D))()(BPBAP=4A1A2B0P(B)1P(A1A2|B)=P(A1|B)+P(A2|B)(A)P(A1A2)=P(A1)+P(A2)(B)P(A1A2)=P(A1|B)+P(A2|B)(C)P(A1BA2B)=P(A1B)+P(A2B)(D)P(A1A2|B)=P(A1|B)+P(A2|B)5ABP(AB)=0(A)BA=∅(B)P(A–B)=P(A)(C)P(A)P(B)=0(D)BA≠∅6ABCA⊂CB⊂C(A)P(C)=P(AB)(B)P(C)≤P(A)+P(B)–1(C)P(C)≥P(A)+P(B)–1(D)P(C)=P(AB)70P(A)1P(B)0P(B|A)=P(B|A)(A)P(A|B)=P(A|B)(B)P(A|B)≠P(A|B)(C)P(AB)≠P(A)P(B)(D)P(AB)=P(A)P(B)8A1=A2=A3=A4=(A)(B)321,,AAA432,,AAA(C)(D)321,,AAA432,,AAA9AB(A)AB≠∅AB(B)AB≠∅AB(C)AB=∅AB(D)AB=∅AB10ABP(B)0P(A|B)=1(A)P(AB)P(A)(B)P(AB)P(B)(C)P(AB)=P(A)(D)P(AB)=P(B)1P(AB)=0(1)AB(2)AB(3)AB(4)AB(5)P(A)=0P(B)=0(6)P(A–B)=P(A)2ABP(A)=0.6P(B)=0.7(1)P(AB)(2)P(AB)3AB)()(BAPABP=P(A)=pP(B)4P(A)=0.7P(A–B)=0.3)(ABP55446101103(1)5(2)573412385322295310(01)6/511a220xaxy−x4π1221)(,31)(,41)(===BAPABPAP)(BAPU131041413224411222115ABAB0.02BA0.01AB21AA1641,31,5117ABP(A)=0.4P(AB)=0.7)(ABP.180.60.5190.30.20.10.30.2(1)(2)0.3B1ABCABC=∅,21)()()(==CPBPAP169)(=CBAPUUP(A)2ABC21)()(CBAPABCP=21)()()()(2−++=BCPACPABPABCP30P(A)10P(B)1P(A|B)+1)|(=BAPAB4ABA01)|()|(ABPABP=AB5pp2p(1)(2)610022%10%70.80.90.40.6820.30.35.(1)(2)921