19. (6b, 9-11) A die is rolled six times. Find the expected value of the number of different faces that appear.

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# Solution.

Let $\zeta_{i}$ be a random variable equal to 1 if the face with $i$ points has appeared at least once and 0 if it has never appeared. Then the number of different faces that have appeared is $\zeta=\zeta_{1}+\ldots+\zeta_{6}$. Transitioning to expectations, we get: $\mathrm{M} \zeta=6 \mathrm{M} \zeta_{1}$, since all variables $\zeta_{i}$ are identically distributed. Let's find its expected value

$$
\mathrm{M} \zeta_{1}=0 \cdot\left(\frac{5}{6}\right)^{6}+1 \cdot\left(1-\left(\frac{5}{6}\right)^{6}\right)=1-\left(\frac{5}{6}\right)^{6}
$$

Thus, $М \zeta=6 \cdot\left(1-\left(\frac{5}{6}\right)^{6}\right)=\frac{6^{6}-5^{6}}{6^{5}}$.

Answer: $\frac{6^{6}-5^{6}}{6^{5}}$.