已知 $0 < P(B) < 1$ 且 $\mathbf{P}\left[\left(A_1+A_2\right) \mid B\right]=P\left(A_1 \mid B\right)+P\left(A_2 \mid B\right) ，$则下列选项成立的是
A. $\boldsymbol{P}\left[\left(A_1+A_2\right) \mid \bar{B}\right]=P\left(A_1 \mid \bar{B}\right)+P\left(A_2 \mid \bar{B}\right)$
B. $P\left(A_1 B+A_2 B\right)=P\left(A_1 B\right)+P\left(A_2 B\right)$
C. $P\left(A_1+A_2\right)=P\left(A_1 \mid B\right)+P\left(A_2 \mid B\right)$
D. $P(B)=P\left(A_1\right) P\left(B \mid A_1\right)+P\left(A_2\right) P\left(B \mid A_2\right)$