11. Large Cube. A cube is assembled from 27 playing dice. a) (6 - 11 grades, 1 point) Find the probability that exactly 25 sixes are on the surface of the cube.

b) (7 - 11 grades, 1 point) Find the probability that at least one one is on the surface of the cube.

c) (8 - 11 grades, **1** point) Find the expected number of sixes facing outward.

![](https://cdn.mathpix.com/cropped/2024_05_06_ad4793eab3ce977a151bg-12.jpg?height=377&width=420&top_left_y=200&top_left_x=1503)

d) (8 - 11 grades, 1 point) Find the expected value of the sum of the numbers that are on the surface of the cube.

e) (9 - 11 grades, 3 points) Find the expected value of the random variable: “The number of different digits that ended up on the surface of the cube.”
Solution. 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 of all the cubes showing sixes outward, 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 $\frac{1}{6}$, and with a probability of $\frac{5}{6}$ it shows not a six. If it is located in the middle of an edge, then two of its faces are visible, so a six is visible with a probability of $\frac{2}{6}=\frac{1}{3}$, and a six is not visible with a probability of $\frac{2}{3}$. Finally, if the cube is at a vertex of the large cube, then three of its faces are visible, so with a probability of $\frac{1}{2}$ a six is visible, and with a probability of $\frac{1}{2}$ no six is visible.

The special cube can be in the center of one of the faces with a probability of

$$
\frac{5}{6} \cdot\left(\frac{1}{6}\right)^{5} \cdot\left(\frac{1}{3}\right)^{12} \cdot\left(\frac{1}{2}\right)^{8}
$$

- here the probabilities of independent events are multiplied: "the special cube shows one of the five faces outward, but not a six," "5 other cubes in the centers of the faces show sixes outward," "all 12 cubes in the middle of the edges show sixes," and "all 8 cubes in the vertices show sixes."

The special cube is located on an edge with a probability of

$$
\left(\frac{1}{6}\right)^{6} \cdot \frac{2}{3} \cdot\left(\frac{1}{3}\right)^{11} \cdot\left(\frac{1}{2}\right)^{8}
$$

Finally, this special cube can be at a vertex of the large cube with a probability of

$$
\left(\frac{1}{6}\right)^{6} \cdot\left(\frac{1}{3}\right)^{12} \cdot \frac{1}{2} \cdot\left(\frac{1}{2}\right)^{7}
$$

We need to consider that there are 6 faces, so there are also 6 cubes located in the centers of the faces. The cube has 12 cubes in the middle of the edges and 8 in the vertices. The special cube can be any of them, and all these events are mutually exclusive. Then, by adding their probabilities, we find the probability of event $A$:

$$
\begin{gathered}
P(A)=6 \cdot \frac{5}{6} \cdot\left(\frac{1}{6}\right)^{5} \cdot\left(\frac{1}{3}\right)^{12} \cdot\left(\frac{1}{2}\right)^{8}+12 \cdot\left(\frac{1}{6}\right)^{6} \cdot \frac{2}{3} \cdot\left(\frac{1}{3}\right)^{11} \cdot\left(\frac{1}{2}\right)^{8}+8 \cdot\left(\frac{1}{6}\right)^{6} \cdot\left(\frac{1}{3}\right)^{12} \cdot \frac{1}{2} \cdot\left(\frac{1}{2}\right)^{7}= \\
=\left(\frac{1}{6}\right)^{5} \cdot\left(\frac{1}{3}\right)^{11} \cdot\left(\frac{1}{2}\right)^{7} \cdot\left(6 \cdot \frac{5}{6} \cdot \frac{1}{3} \cdot \frac{1}{2}+12 \cdot \frac{1}{6} \cdot \frac{2}{3} \cdot \frac{1}{2}+8 \cdot \frac{1}{6} \cdot \frac{1}{3} \cdot \frac{1}{2}\right)= \\
=\left(\frac{1}{6}\right)^{5} \cdot\left(\frac{1}{3}\right)^{11} \cdot\left(\frac{1}{2}\right)^{7} \cdot\left(\frac{15+12+4}{6 \cdot 3}\right)= \\
=\frac{1}{6^{5} \cdot 3^{11} \cdot 2^{7}} \cdot \frac{31}{18}=\frac{31}{2^{13} \cdot 3^{18}} \approx 9.76763 \cdot 10^{-12} .
\end{gathered}
$$

b) No ones are on the surface of the cube with a probability of

$$
\left(\frac{5}{6}\right)^{6} \cdot\left(\frac{2}{3}\right)^{12} \cdot\left(\frac{1}{2}\right)^{8}=\frac{5^{6}}{2^{2} \cdot 3^{18}}
$$

Therefore, at least one one will be with a probability of

$$
1-\frac{5^{6}}{2^{2} \cdot 3^{18}} \approx 0.999989917
$$

c) Consider the cubes located in the centers of the faces. There are six such cubes. Introduce random variables $\xi_{1}, \xi_{2}, \ldots, \xi_{6}$, one for each of these cubes:

$$
\xi_{k}=\left\{\begin{array}{l}
1, \text { if a six is visible on the } k \text {-th cube, } \\
0, \text { if a six is not visible on the } k \text {-th cube. }
\end{array}\right.
$$

Such random variables are called indicators. In this case, they are indicators of the events "a six is visible on the $k$-th central cube."

Similarly, introduce indicators $\eta_{1}, \ldots, \eta_{12}$ for sixes for each of the 12 cubes located in the middle of the edges. Finally, introduce indicators for sixes $\zeta_{1}, \ldots, \zeta_{8}$ for the 8 cubes located at the vertices of the large cube. Then the number of sixes on the surface of the large cube is equal to the sum of all the indicators:

$$
X=\xi_{1}+\xi_{2}+\ldots+\xi_{6}+\eta_{1}+\ldots+\eta_{12}+\zeta_{1}+\ldots+\zeta_{8}
$$

The expectations of the variables $\xi_{k}$ are equal to each other and are equal to $E \xi_{k}=0 \cdot \frac{5}{6}+1 \cdot \frac{1}{6}=\frac{1}{6}$. Similarly, $E \eta_{k}=0 \cdot \frac{4}{6}+1 \cdot \frac{2}{6}=\frac{1}{3}$. Finally, $E \zeta_{k}=0 \cdot \frac{3}{6}+1 \cdot \frac{3}{6}=\frac{1}{2}$.

Therefore,

$$
\mathrm{E} X=6 \mathrm{E} \xi_{1}+12 \mathrm{E} \eta_{1}+8 \mathrm{E} \zeta_{1}=6 \cdot \frac{1}{6}+12 \cdot \frac{1}{3}+8 \cdot \frac{1}{2}=9
$$

d) 54 faces are visible outward, with an average of 3.5 points on each. In total,

$$
54 \cdot 3.5=189
$$

e) Use part b). Introduce for each number of points $k$ from 1 to 6 an indicator:

$$
\xi_{k}=\left\{\begin{array}{l}
1, \text { if the number of points } k \text { is visible on the surface of the cube at least once, } \\
0, \text { if the number } k \text { is absent on the surface of the cube. }
\end{array}\right.
$$

The expectation of each such variable can be found using the result obtained in part b):

$$
\mathrm{E} \xi_{k}=1 \cdot\left(1-\frac{5^{6}}{2^{2} \cdot 3^{18}}\right)+0 \cdot \frac{5^{6}}{2^{2} \cdot 3^{18}}=1-\frac{5^{6}}{2^{2} \cdot 3^{18}}
$$

Then the expectation of the random variable $Y$ "the number of different points visible on the surface of the cube" is

$$
\mathrm{E} Y=\mathrm{E}\left(\xi_{1}+\xi_{2}+\ldots+\xi_{6}\right)=6 \mathrm{E} \xi_{1}=6 \cdot\left(1-\frac{5^{6}}{2^{2} \cdot 3^{18}}\right)=6-\frac{5^{6}}{2 \cdot 3^{17}} \approx 5.99
$$

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) $6-\frac{5^{6}}{2 \cdot 3^{17}} \approx 5.99$.