Bayesian update from one slip drawn from a box

Box 1 contains eight slips numbered $1$ through $8$. Box 2 contains eight slips as well: four marked $1$ and four marked $2$. You choose a box by a fair coin flip and draw one slip at random; it reads $2$, and you put it back. You then make $40$ further draws, with replacement, from that same box. What is the expected number of these draws that show a $2$?

$17$. A $2$ is drawn with probability $\tfrac18$ from box 1 and $\tfrac12$ from box 2, so by Bayes the posterior on box 2 is $\tfrac{1/2}{1/2+1/8}=\tfrac45$. The expected count is $\tfrac45\cdot20+\tfrac15\cdot5=17$.