= Solution
Any two distinct draw positions have the same pair law by <exchangeability>. The <marginal and pair probabilities in sampling without replacement> therefore give
$$
\boxed{\mathbb P(B_2\cap B_3)=\frac{4\cdot3}{7\cdot6}=\frac27.}
$$
Divide this joint <probability> by the relevant marginal <probability> to get the two <conditional probabilities>:
$$
\boxed{\mathbb P(B_2\mid B_3)=\frac{2/7}{4/7}=\frac12,\qquad
\mathbb P(B_3\mid B_2)=\frac{2/7}{4/7}=\frac12.}
$$
Conditioning on the third draw can change the <probability> of the second draw; <conditional probability> does not require the conditioning event to occur earlier in time.
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