[proofplan] Apply the definition of conditional probability to $\mathbb{P}(B_n \mid A)$, rewrite the numerator $\mathbb{P}(B_n \cap A)$ using the reverse conditional $\mathbb{P}(A \mid B_n)\,\mathbb{P}(B_n)$, and replace the denominator $\mathbb{P}(A)$ using the Law of Total Probability. [/proofplan] [step:Expand $\mathbb{P}(B_n \mid A)$ using the definition and reverse the conditioning] By the definition of conditional probability (valid since $\mathbb{P}(A) > 0$), \begin{align*} \mathbb{P}(B_n \mid A) = \frac{\mathbb{P}(B_n \cap A)}{\mathbb{P}(A)}. \end{align*} Since $\mathbb{P}(B_n) > 0$, we may write $\mathbb{P}(B_n \cap A) = \mathbb{P}(A \mid B_n)\,\mathbb{P}(B_n)$ (again by the definition of conditional probability, now conditioning on $B_n$). Hence \begin{align*} \mathbb{P}(B_n \mid A) = \frac{\mathbb{P}(A \mid B_n)\,\mathbb{P}(B_n)}{\mathbb{P}(A)}. \end{align*} [/step] [step:Replace $\mathbb{P}(A)$ by the Law of Total Probability] Since $(B_n)_{n \ge 1}$ is a partition of $\Omega$ with $\mathbb{P}(B_n) > 0$ for all $n$, the [Law of Total Probability](/theorems/1113) gives \begin{align*} \mathbb{P}(A) = \sum_{k=1}^\infty \mathbb{P}(A \mid B_k)\,\mathbb{P}(B_k). \end{align*} Substituting into the expression from the previous step: \begin{align*} \mathbb{P}(B_n \mid A) = \frac{\mathbb{P}(A \mid B_n)\,\mathbb{P}(B_n)}{\sum_{k=1}^\infty \mathbb{P}(A \mid B_k)\,\mathbb{P}(B_k)}. \end{align*} [/step]