Three natural numbers are taken at random from the set of first 100 natural numbers. The probability that their A.M. is 25, is

Options

A. `(.^(77)C_(2))/(.^(100)C_(3))`
B. `(.^(25)C_(2))/(.^(100)C_(3))`
C. `(.^(74)C_(72))/(.^(100)C_(97))`
D. none of these

Explanation

To find the probability that the arithmetic mean (A.M.) of three randomly selected natural numbers from the first 100 natural numbers is 25, we can follow these steps:

### Step 1: Understanding the Problem
We need to find three natural numbers \( A, B, C \) such that their arithmetic mean is 25. This implies:
\[ \frac{A + B + C}{3} = 25 \]
Multiplying both sides by 3 gives:
\[ A + B + C = 75 \]

### Step 2: Setting Up the Sample Space
The total number of ways to choose 3 natural numbers from the first 100 natural numbers can be calculated using combinations:
\[ \text{Total ways} = \binom{100}{3} \]

### Step 3: Finding Favorable Outcomes
We need to find the number of ways to select \( A, B, C \) such that \( A + B + C = 75 \). Since \( A, B, C \) must be natural numbers, they must each be at least 1.

To simplify, we can define new variables:
\[ A = A' + 1, \quad B = B' + 1, \quad C = C' + 1 \]
where \( A', B', C' \) are non-negative integers. Then:
\[ (A' + 1) + (B' + 1) + (C' + 1) = 75 \]
This simplifies to:
\[ A' + B' + C' = 72 \]

### Step 4: Using Stars and Bars
The problem now is to find the number of non-negative integer solutions to the equation \( A' + B' + C' = 72 \). This can be solved using the "stars and bars" theorem:
\[ \text{Number of solutions} = \binom{n + r - 1}{r - 1} \]
where \( n \) is the total we want (72) and \( r \) is the number of variables (3). Thus:
\[ \text{Number of ways} = \binom{72 + 3 - 1}{3 - 1} = \binom{74}{2} \]

### Step 5: Calculating the Probability
Now, we can calculate the probability:
\[ P(\text{A.M.} = 25) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{\binom{74}{2}}{\binom{100}{3}} \]

### Step 6: Final Calculation
Now we need to compute \( \binom{74}{2} \) and \( \binom{100}{3} \):
\[ \binom{74}{2} = \frac{74 \times 73}{2} = 2701 \]
\[ \binom{100}{3} = \frac{100 \times 99 \times 98}{3 \times 2 \times 1} = 161700 \]
Thus, the probability is:
\[ P(\text{A.M.} = 25) = \frac{2701}{161700} \]

### Conclusion
The probability that the arithmetic mean of three randomly selected natural numbers from the first 100 natural numbers is 25 is:
\[ \frac{2701}{161700} \]