Updating the probability of the next observation

Coming to the probability of White in the second extraction, it is now clear why $15/25=3/5=60\%$ is wrong: it assumed the remaining five boxes equally likely,while they are not. Also in this case maieutics helps: it becomes suddenly clear that we have to assign a higher `weight' to the compositions we consider more likely. That is, in general and remembering that the weights $P(B_i\,\vert\,I)$ sum up to unity,

$$P(\mbox{W}\,\vert\,I) = \sum_i P(\mbox{W}\,\vert\,B_i,I)\cdot P(B_i\,\vert\,I)\,.$$ (10)

After the observation of White in the first extraction we then get

$$P(\mbox{W}^{(2)}\,\vert\,\mbox{W}^{(1)},I) = \sum_i P(\mbox{W}^{(2)}\,\vert\,B_i,\mbox{W}^{(1)},I)\cdot P(B_i\,\vert\,\mbox{W}^{(1)},I)$$

$$= \sum_i P(\mbox{W}\,\vert\,B_i,I)\cdot P(B_i\,\vert\,\mbox{W}^{(1)},I)\,,$$ (11)

where $P(\mbox{W}^{(2)}\,\vert\,B_i,\mbox{W}^{(1)},I)$ has been rewritten as $P(\mbox{W}\,\vert\,B_i,I)$ since, assuming a particular composition, the probability of White is the same in every extraction. Moreover, since $\pi_i = P(\mbox{W}\,\vert\,B_i)$, we can rewrite Equation (11), in analogy with Equation (7), i.e. replacing $B_i$ by $\pi_i$, as

$$P(\mbox{W}^{(2)}\,\vert\,\mbox{W}^{(1)},I) = \sum_i \pi_i\cdot P(\pi_i\,\vert\,\mbox{W}^{(1)},I)\,,$$ (12)

which will deserve comments later.