Conditioning on $B$ makes $B$ the new sample space. For equally likely outcomes:

$$P(A \mid B) = \frac{|A \cap B|}{|B|}$$

Return $P(A \mid B)$ as a float. Assume b is non-empty.

solve({2, 4, 6}, {4, 5, 6})  ->  0.6666666666666666