设有如下一组推理规则: r1:IF E1 THEN E2 (0.6) r2:IF E2 AND E3 THEN E4 (0.8) r3:IF E4 THEN H (0.7) r4:IF E5 THEN H (0.9) 且已知CF(E1)=0.5,CF(E3)=0.6,CF(E5)=0.4,结论H的初始可信度一无所知。求CF(H)=?
已知:规则可信度为 r1:IF E1 THEN H1 (0.7) r2:IF E2 THEN H1 (0.6) r3:IF E3 THEN H1 (0.4) r4:IF (H1 AND E4)THEN H2 (0.2) 证据可信度为 CF(E1)=CF(E2)=CF(E3)=CF(E4)=CF(E5)=0.5 H1的初始可信度一无所知,H2的初始可信度CF0(H2)=0.3 计算结论H2的可信度CF(H2)。