= Solution
For $\mathbb P(B)>0$, the <conditional probability> is
$$
\mathbb P(A\mid B)
=\frac{\mathbb P(A\cap B)}{\mathbb P(B)}.
$$
Hence
$$
\mathbb P(B_j\mid A)
=\frac{\mathbb P(A\mid B_j)\mathbb P(B_j)}
{\mathbb P(A)}.
$$
Because the $B_k$ form a partition, the law of total probability gives
$$
\mathbb P(A)
=\sum_{k=1}^n\mathbb P(A\mid B_k)\mathbb P(B_k).
$$
Substitution proves <Bayes' theorem> in the required form:
$$
\boxed{
\mathbb P(B_j\mid A)
=\frac{\mathbb P(A\mid B_j)\mathbb P(B_j)}
{\sum_{k=1}^n\mathbb P(A\mid B_k)\mathbb P(B_k)}}.
$$
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