To solve the problem, we need to determine the probability that the function \( f: X \to B \) is bijective, given the sets \( X \) and \( Y \), and the condition \( f(5) = 5 \).
### Step-by-Step Solution:
1. **Define the Sets:**
- Let \( X = \{0, 1, 2, 3, \ldots, 9\} \) which has 10 elements.
- Let \( Y = \{0, 1, 2, \ldots, 100\} \) which has 101 elements.
- Since \( f(5) = 5 \), the element 5 is already mapped to itself.
2. **Define Set B:**
- Let \( B \) be a subset of \( Y \) that includes the element 5. The cardinality of \( B \) is denoted as \( r \) (where \( r \) can range from 1 to 101).
3. **Choosing Elements for Set B:**
- Since 5 is already included in \( B \), we need to select \( r - 1 \) additional elements from the remaining 100 elements in \( Y \) (i.e., from \( Y \setminus \{5\} \)).
- The number of ways to choose \( r - 1 \) elements from 100 elements is given by \( \binom{100}{r - 1} \).
4. **Counting Functions from X to B:**
- Once \( B \) is chosen, we need to determine the number of functions from \( X \) to \( B \).
- Since \( f(5) = 5 \) is fixed, we have 9 remaining elements in \( X \) that can map to any of the \( r \) elements in \( B \).
- Thus, the number of ways to define the function \( f \) is \( r^9 \).
5. **Total Number of Functions:**
- The total number of functions \( f: X \to B \) is then given by:
\[
\sum_{r=1}^{101} \binom{100}{r - 1} r^9
\]
6. **Counting Bijections:**
- For \( f \) to be a bijection, we need to select \( r \) elements from \( Y \) and arrange them.
- Since 5 is already assigned to itself, we need to choose 9 more elements from the remaining 100 elements in \( Y \) (excluding 5).
- The number of ways to choose 9 elements from 100 is \( \binom{100}{9} \).
- The number of ways to arrange these 9 elements is \( 9! \).
- Therefore, the total number of bijections is:
\[
\binom{100}{9} \times 9!
\]
7. **Calculating the Probability:**
- The probability \( P \) that the function \( f \) is bijective is given by the ratio of the number of bijections to the total number of functions:
\[
P = \frac{\binom{100}{9} \times 9!}{\sum_{r=1}^{101} \binom{100}{r - 1} r^9}
\]
### Final Answer:
The final expression for the probability that the function \( f \) is bijective is:
\[
P = \frac{\binom{100}{9} \times 9!}{\sum_{r=1}^{101} \binom{100}{r - 1} r^9}
\]
