已知 $0 < P(B) < 1$, 且 $P\left[\left(A_1+A_2\right) \mid B\right]=P\left(A_1 \mid B\right)+P\left(A_2 \mid B\right)$, 则下列选项成立的是
$\text{A.}$ $P\left[\left(A_1+A_2\right) \mid \bar{B}\right]=P\left(A_1 \mid \bar{B}\right)+P\left(A_2 \mid \bar{B}\right)$.
$\text{B.}$ $P\left(\Lambda_1 B+A_2 B\right)=P\left(A_1 B\right)+P\left(A_2 B\right)$.
$\text{C.}$ $P\left(A_1+A_2\right)=P\left(A_1 \mid B\right)+P\left(A_2 \mid B\right)$.
$\text{D.}$ $P(B)=P\left(A_1\right) P\left(B \mid A_1\right)+P\left(A_2\right) P\left(B \mid A_2\right)$.
$\text{E.}$
$\text{F.}$