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There are $10^9$ strings of length $9$ that can be formed by using the ten decimal digits with repetition. From these, we must exclude those strings in which at least one odd digit is missing.
Let $A_i$ be the set of outcomes in which the digit $i$ is excluded, where $i \in \{1, 3, 5, 7, 9\}$.
Then, by the Inclusion Principle, the number of strings in which at least one odd digit is missing is
$$\sum_{i} |A_i| - \sum_{i < j} |A_i \cap A_j| + \sum_{i < j < k} |A_i \cap A_j \cap A_k| - \sum_{i < j < k < l} |A_i \cap A_j \cap A_k \cap A_l| + \sum_{i < j < k < l < m} |A_i \cap A_j \cap A_k \cap A_l \cap A_m|$$
$|A_1|$: Since $1$ is excluded each of the nine positions in the string can be filled in nine ways. Hence, $|A_1| = 9^9$. By symmetry,
$$|A_1| = |A_3| = |A_5| = |A_7| = |A_9|$$
$|A_1 \cap A_3|$: Since both $1$ and $3$ are excluded, each of the nine positions in the string can be filled in eight ways. Hence, $|A_1 \cap A_3| = 8^9$. By symmetry,
$$|A_1 \cap A_3| = |A_1 \cap A_5| = |A_1 \cap A_7| = |A_1 \cap A_9| = |A_3 \cap A_5| = |A_3 \cap A_7| = |A_3 \cap A_9| = |A_5 \cap A_7| = |A_5 \cap A_9| = |A_7 \cap A_9|$$
$|A_1 \cap A_3 \cap A_5|$: Since $1$, $3$, and $5$ are all excluded, each of the nine positions in the string can be filled in seven ways. Hence, $|A_1 \cap A_3 \cap A_5| = 7^9$. By symmetry,
$$|A_1 \cap A_3 \cap A_5| = |A_1 \cap A_3 \cap A_7| = |A_1 \cap A_3 \cap A_9| = |A_1 \cap A_5 \cap A_7| = |A_1 \cap A_5 \cap A_9| = |A_1 \cap A_7 \cap A_9| = |A_3 \cap A_5 \cap A_7| = |A_3 \cap A_5 \cap A_9| = |A_3 \cap A_7 \cap A_9| = |A_5 \cap A_7 \cap A_9|$$
$|A_1 \cap A_3 \cap A_5 \cap A_7|$: Since $1$, $3$, $5$, $7$, and $9$ are all excluded, each of the nine positions in the string can be filled in six ways. Hence,
$|A_1 \cap A_3 \cap A_5 \cap A_7| = 6^9$. By symmetry,
$$|A_1 \cap A_3 \cap A_5 \cap A_7| = |A_1 \cap A_3 \cap A_5 \cap A_9| = |A_1 \cap A_3 \cap A_7 \cap A_9| = |A_1 \cap A_5 \cap A_7 \cap A_9| = |A_3 \cap A_5 \cap A_7 \cap A_9|$$
$|A_1 \cap A_3 \cap A_5 \cap A_7 \cap A_9|$: Since $1$, $3$, $5$, $7$, and $9$ are all excluded, each of the nine positions in the string can be filled in five ways. Hence, $|A_1 \cap A_3 \cap A_5 \cap A_7 \cap A_9| = 5^9$.
Thus, by the Inclusion-Exclusion Principle, the number of strings of length $9$ in which at least one odd digit is missing is
$$5 \cdot 9^9 - 10 \cdot 8^9 + 10 \cdot 7^9 - 5 \cdot 6^9 + 5^9$$
Therefore, the number of strings of length $9$ in which no odd digits are missing is
\begin{align*}
10^9 - (5 \cdot 9^9 - 10 \cdot 8^9 + 10 \cdot 7^9 - 5 \cdot 6^9 + 5^9) & = 10^9 - 5 \cdot 9^9 + 10 \cdot 8^9 - 10 \cdot 7^9 + 5 \cdot 6^9 - 5^9\\
& = \binom{5}{0}10^9 - \binom{5}{1}9^9 + \binom{5}{2}8^9 - \binom{5}{3}7^9 + \binom{5}{4}6^9 - \binom{5}{5}5^9\\
& = \sum_{k = 0}^{5} (-1)^{k} \binom{5}{k}(10 - k)^9
\end{align*}
where $\binom{5}{k}$ is the number of ways of excluding $k$ of the $5$ odd digits and $(10 - k)^9$ is the number of ways of filling the nine positions of the string with the remaining $10 - k$ decimal digits.
Best Answer
Use the linearity of expectation. What is the chance that $1$ is present? That is the expected value of an indicator variable.