TITLE: Cube and unit cubes
QUESTION [10 upvotes]: Consider a $3\times 3\times 3$ cube consisting of smaller $1\times 1\times 1$ unit cubes. The big cube is painted black on the outside. Suppose we disassemble the cube and pick a random unit cube, look at only one face and see it is black, without looking at the other faces. What is the probability the unit cube we picked is one of the 8 corner cubes?

This one seems simple to me but I am not sure I am right:
The big cube consists of 27 unit cubes of which one only, the middle one, does not have any painted face. All the others (26) have at least one face painted and 8 have 3 faces painted.
Thus the requested probability is $\frac{8}{26}$—is this so?

REPLY [3 votes]: You can also use Bayes' theorem. The prior probability of choosing a corner cube is $8 \over 27$. The probability of a corner cube showing a black face is $1 \over 2$. The overall probability of a black face is ${54 \over 162}=1/3$, so
$P({\rm corner }|{\rm black})={1/2 \over1/3} \times 8/27 =4/9$