When an event can be reached through several separate routes, multiply along each route and add the routes:

$$P(A) = \sum_{i=1}^{n} P(A \mid B_i)\,P(B_i)$$

Return $P(A)$. The two lists are aligned and the same length.

solve([0.6, 0.4], [0.02, 0.05])  ->  0.032
# Factory 1 makes 60% with a 2% defect rate; Factory 2 makes 40% at 5%.

Note this is a weighted average, not the plain average of 2% and 5%.