From 27 dice, a cube is formed.

a) Find the probability that exactly 25 sixes are on the surface of the cube.

b) Find the probability that at least one one is on the surface of the cube.

c) Find the expected value of the number of sixes facing outward.

d) Find the expected value of the sum of the numbers that ended up on the surface of the cube.

e) Find the expected value of the random variable: "The number of different digits that ended up on the surface of the cube."
a) One of the 27 cubes is in the center and therefore not visible at all. The other 26 cubes are visible. Thus, the required event \( A=\{25 \text{ sixes} \} \) consists in all the cubes showing sixes outwardly, except for one - let's call it the special cube.

Consider all the cubes. If a cube is in the center of one of the faces, it "shows" a six with a probability of 1/6. If it is in the middle of an edge, two of its faces are visible, so a six is visible with a probability of \( 2 / 6 = 1 / 3 \). Finally, if a cube is at a vertex of the large cube, three of its faces are visible, so a six is visible with a probability of 1/2.

The special cube can be in the center of a specific face with a probability of \( 5 / 6^5 / 6^5 \cdot (1 / 6)^5 \cdot (1 / 3)^{12} \cdot (1 / 2)^8 \) - here the probabilities of independent events "the special cube shows a face other than a six outwardly," "5 other cubes in the centers of faces show sixes outwardly," "sixes are visible on all 12 cubes in the middle of edges," and "sixes are visible on all 8 cubes at the vertices" are multiplied.

The special cube is located in the middle of a specific edge with a probability of \( (1 / 6)^6 \cdot 2 / 3 \cdot (1 / 3)^{11} \cdot (1 / 2)^8 \).

Finally, the special cube can be at a specific vertex of the large cube with a probability of \( (1 / 6)^6 \cdot (1 / 3)^{12} \cdot (1 / 2)^8 \). The cube has 6 faces, 12 edges, and 8 vertices. Therefore,

\[
P(A) = (6 \cdot 5 + 12 \cdot 2 + 8 \cdot 1) \left( \frac{1}{6} \right)^6 \left( \frac{1}{3} \right)^{12} \left( \frac{1}{2} \right)^8 = \frac{62}{6^6 \cdot 3^{12} \cdot 2^8} = \frac{31}{2^{13} \cdot 3^{18}} \approx 9.76763 \cdot 10^{-12}
\]

b) No ones are visible on the surface of the cube with a probability of \( \left( \frac{5}{6} \right)^6 \left( \frac{2}{3} \right)^{12} \left( \frac{1}{2} \right)^8 = \frac{5^6}{2^2 \cdot 3^{18}} \).

Therefore, at least one one will be visible with a probability of \( 1 - \frac{5^6}{2^2 \cdot 3^{18}} \approx 0.999989917 \).

c) Consider the cubes located at the centers of the faces. There are six such cubes. Introduce indicators \( \xi_1, \xi_2, \ldots, \xi_6 \): \( \xi_k = 1 \) if a six is visible on the \( k \)-th cube, \( \xi_k = 0 \) if a six is not visible on the \( k \)-th cube.

Similarly, introduce indicators \( \eta_1, \eta_2, \ldots, \eta_{12} \) for sixes on each of the 12 cubes located in the middle of the edges. Finally, introduce indicators \( \zeta_1, \zeta_2, \ldots, \zeta_8 \) for sixes on the 8 cubes located at the vertices of the large cube. Then the number of sixes on the surface of the large cube is the sum of all indicators: \( X = \xi_1 + \ldots + \xi_6 + \eta_1 + \ldots + \eta_{12} + \zeta_1 + \ldots + \zeta_8 \).

The expectations \( E \xi_k \) of the quantities \( \xi_k \) are equal to each other and equal to \( 0 \cdot 5 / 6 + 1 \cdot 1 / 6 = 1 / 6 \). Similarly, \( E \eta_k = 1 / 3 \), \( E \zeta_k = 1 / 2 \). Therefore, \( E X = 6 E \xi_1 + 12 E \eta_1 + 8 E \zeta_1 = 9 \).

d) 54 faces are visible outwardly, with an average of 3.5 points on each face. Therefore, \( 54 \cdot 3.5 = 189 \).

e) Introduce for each number of points \( k \) from 1 to 6 an indicator \( \theta_k: \theta_k = 1 \) if the number of points \( k \) is visible on the surface of the cube at least once, and 0 otherwise.

The expectation of each such quantity is found in part b): \( E \theta_k = 1 - \frac{5^6}{2^2 \cdot 3^{18}} \).

Therefore, the expectation \( E Y \) of the random variable \( Y \) "the number of different points visible on the surface of the cube" is

\[
E (\theta_1 + \theta_2 + \ldots + \theta_6) = 6 E \theta_1 = 6 \left( 1 - \frac{5^6}{2^2 \cdot 3^{18}} \right) = 6 - \frac{5^6}{2 \cdot 3^{17}} \approx 5.99
\]

## Answer

a) \( \frac{31}{2^{13} \cdot 3^{18}} \approx 9.77 \cdot 10^{-12} \)
b) \( 1 - \frac{5^6}{2^2 \cdot 3^{18}} \approx 0.99998992 \);
c) \( 9 \); d) 189; e
\( 6 - \frac{5^6}{2 \cdot 3^{17}} \approx 5.99 \)

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