= Solution
By the definition of <conditional probability>,
$$
\mathbb P(B_i\mid A)=\frac{\mathbb P(A\cap B_i)}{\mathbb P(A)}
=\frac{\mathbb P(A\mid B_i)\mathbb P(B_i)}{\mathbb P(A)}.
$$
Since the $B_j$ form a disjoint partition of the <sample space>, the <law of total probability> gives $\mathbb P(A)=\sum_j\mathbb P(A\mid B_j)\mathbb P(B_j)$. Substitution proves <Bayes' theorem> in the required form:
$$
\boxed{\mathbb P(B_i\mid A)=
\frac{\mathbb P(A\mid B_i)\mathbb P(B_i)}
{\sum_j\mathbb P(A\mid B_j)\mathbb P(B_j)}.}
$$
The hypotheses make the denominator positive and each conditioning probability well defined.
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