Three distinct numbers are chosen at random from the first 15 natural numbers. The probability that the sum will be divisible by 3 is

Options

A. `(30)/(91)`
B. `(31)/(91)`
C. `(60)/(91)`
D. none of these

Explanation

To solve the problem of finding the probability that the sum of three distinct numbers chosen from the first 15 natural numbers is divisible by 3, we can follow these steps:

### Step 1: Determine the Total Number of Ways to Choose 3 Numbers
The total number of ways to choose 3 distinct numbers from the first 15 natural numbers can be calculated using the combination formula:

\[ \text{Total ways} = \binom{15}{3} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455 \]

### Step 2: Classify the Numbers by Remainders
Next, we classify the first 15 natural numbers based on their remainders when divided by 3:

- **Remainder 0**: 3, 6, 9, 12, 15 (5 numbers)
- **Remainder 1**: 1, 4, 7, 10, 13 (5 numbers)
- **Remainder 2**: 2, 5, 8, 11, 14 (5 numbers)

### Step 3: Identify Favorable Outcomes
We need to find the combinations of numbers that yield a sum divisible by 3. This can happen in two scenarios:

1. **All three numbers from the same group**:
- Choosing 3 numbers from the group with remainder 0: \(\binom{5}{3}\)
- Choosing 3 numbers from the group with remainder 1: \(\binom{5}{3}\)
- Choosing 3 numbers from the group with remainder 2: \(\binom{5}{3}\)

Each of these combinations can be calculated as:
\[ \binom{5}{3} = \frac{5 \times 4}{2 \times 1} = 10 \]
So, the total for this scenario is:
\[ 10 + 10 + 10 = 30 \]

2. **One number from each group**:
- Choosing 1 number from each of the three groups: \(\binom{5}{1} \times \binom{5}{1} \times \binom{5}{1}\)
\[ \binom{5}{1} = 5 \quad \text{(for each group)} \]
So, the total for this scenario is:
\[ 5 \times 5 \times 5 = 125 \]

### Step 4: Calculate Total Favorable Outcomes
Adding both scenarios gives us the total favorable outcomes:
\[ \text{Total favorable outcomes} = 30 + 125 = 155 \]

### Step 5: Calculate the Probability
Now we can calculate the probability that the sum of the chosen numbers is divisible by 3:
\[ \text{Probability} = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{155}{455} \]

### Step 6: Simplify the Probability
To simplify \(\frac{155}{455}\), we can divide both the numerator and the denominator by their greatest common divisor (GCD), which is 5:
\[ \frac{155 \div 5}{455 \div 5} = \frac{31}{91} \]

### Final Answer
Thus, the probability that the sum of the three distinct numbers chosen from the first 15 natural numbers is divisible by 3 is:
\[ \frac{31}{91} \]