To solve the problem, we will analyze each statement one by one and determine their truthfulness based on probability concepts.
### Statement 1:
**Statement:** Three natural numbers are taken at random from the set \( A = \{ x : 1 < x \leq 100, x \in \mathbb{N} \} \). The probability that the A.M. of the numbers taken is 25 is equal to \( \frac{{^{74}C_2}}{{^{100}C_3}} \).
**Solution:**
1. **Understanding the Arithmetic Mean (A.M.):**
- The A.M. of three numbers \( x_1, x_2, x_3 \) is given by:
\[
\text{A.M.} = \frac{x_1 + x_2 + x_3}{3}
\]
- For the A.M. to be 25, we have:
\[
x_1 + x_2 + x_3 = 75
\]
2. **Constraints on the Numbers:**
- Since \( x_1, x_2, x_3 \) are natural numbers greater than 1, we can substitute \( x_i = y_i + 1 \) (where \( y_i \) are non-negative integers). Thus:
\[
(y_1 + 1) + (y_2 + 1) + (y_3 + 1) = 75 \implies y_1 + y_2 + y_3 = 72
\]
3. **Finding the Number of Solutions:**
- The number of non-negative integer solutions to the equation \( y_1 + y_2 + y_3 = 72 \) can be found using the stars and bars method:
\[
\text{Number of solutions} = \binom{72 + 3 - 1}{3 - 1} = \binom{74}{2}
\]
4. **Total Outcomes:**
- The total ways to choose any 3 numbers from the set of 100 numbers is:
\[
\binom{100}{3}
\]
5. **Calculating the Probability:**
- Therefore, the probability that the A.M. of the numbers is 25 is:
\[
P = \frac{\binom{74}{2}}{\binom{100}{3}}
\]
**Conclusion for Statement 1:** True.
### Statement 2:
**Statement:** Let \( A = \{2, 3, 4, \ldots, 20\} \). A number is chosen at random from set \( A \) and it is bound to be a prime number. The probability that it is more than 10 is \( \frac{1}{5} \).
**Solution:**
1. **Identifying Prime Numbers in Set A:**
- The prime numbers between 2 and 20 are: \( 2, 3, 5, 7, 11, 13, 17, 19 \).
- Total prime numbers = 8.
2. **Finding Prime Numbers Greater than 10:**
- The prime numbers greater than 10 are: \( 11, 13, 17, 19 \).
- Total prime numbers greater than 10 = 4.
3. **Calculating the Probability:**
- The probability that a randomly chosen prime number from set \( A \) is greater than 10 is:
\[
P = \frac{\text{Number of primes > 10}}{\text{Total number of primes}} = \frac{4}{8} = \frac{1}{2}
\]
**Conclusion for Statement 2:** False.
### Statement 3:
**Statement:** The probability of three persons having the same date and month for their birthday is \( \frac{1}{(365)^2} \).
**Solution:**
1. **Understanding the Problem:**
- The probability that three persons have the same birthday can be calculated by considering one day out of 365 days.
2. **Calculating the Probability:**
- The first person can have their birthday on any of the 365 days. The second and third persons must have their birthday on the same day as the first person.
- Thus, the probability that all three share the same birthday is:
\[
P = \frac{1}{365} \cdot \frac{1}{365} = \frac{1}{365^2}
\]
**Conclusion for Statement 3:** True.
### Final Conclusions:
- Statement 1: True
- Statement 2: False
- Statement 3: True
