## Task A-4.5.

A cube is formed from 27 identical white cubes, and all its outer faces are painted black.

(a) One of these cubes is randomly selected and placed on a table on a randomly chosen face. What is the probability that all five visible faces of the cube will be white?

(b) A cube is on the table with all five visible faces being white. What is the probability that the sixth face of this cube is also white?
## Solution.

After painting the outer sides of the large cube black, 8 small cubes have 3 white faces, 12 small cubes have 4 white faces, 6 small cubes have 5 white faces, and 1 small cube has 6 white faces.

Each of the 27 small cubes can be placed on the table in 6 different ways (depending on which face is on the table), so the total number of events is $27 \cdot 6$.

Only the cubes with 5 or 6 white faces can be placed on the table such that all 5 visible faces are white. The cube with 6 white faces (there is only one such cube) can be placed in 6 ways, while any of the 6 cubes with 5 white faces can be placed on the table in exactly one way.

Therefore, the probability that all five visible faces of the cube are white is

$$
\frac{6 \cdot 1 + 1 \cdot 6}{27 \cdot 6} = \frac{2}{27}
$$

Let $S$ be the event that the selected cube has 6 white faces, and let $H$ be the event that the selected cube is placed on the table such that all 5 visible faces are white. We calculate the conditional probability of event $S$ given that event $H$ has occurred, i.e.,

$$
P(S \mid H) = \frac{P(S \cap H)}{P(H)}
$$

Note that the probability $P(H)$ was calculated in part (a) of the problem.

The probability $P(S \cap H)$ is equal to the probability of event $S$, because if event $S$ occurs, then event $H$ must also occur. This probability is $\frac{1}{27}$ (there is only one completely white cube on the table).

Finally, we have

$$
P(S \mid H) = \frac{P(S \cap H)}{P(H)} = \frac{P(S)}{P(H)} = \frac{\frac{1}{27}}{\frac{2}{27}} = \frac{1}{2}
$$