6.已知P(B)>0,A1A2=∅,则下列式子成立的是( )
①P(A1|B)>0②P(A1∪A2|B)=P(A1|B)+P(A2|B)③P(A1$\overrightarrow{{A}_{2}}$|B)≠0④P($\overline{{A}_{1}{A}_{2}}$|B)=1.
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①P(A1|B)>0②P(A1∪A2|B)=P(A1|B)+P(A2|B)③P(A1$\overrightarrow{{A}_{2}}$|B)≠0④P($\overline{{A}_{1}{A}_{2}}$|B)=1.
A. | ①②③④ | B. | ② | C. | ②③ | D. | ②④ |