Two distinct numbers are selected at random from the first twelve natural numbers. The probability that the sum will be divisible by 3 is :

Options

A. `1//3`
B. `23//66`
C. `1//2`
D. None of these

Answer

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

### Step 1: Identify the Sample Space
The first twelve natural numbers are:
\[ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 \]

We need to find the total number of ways to select 2 distinct numbers from these 12 numbers. This can be calculated using the combination formula \( \binom{n}{r} \), where \( n \) is the total number of items to choose from, and \( r \) is the number of items to choose.

\[ \text{Total outcomes} = \binom{12}{2} = \frac{12 \times 11}{2 \times 1} = 66 \]

### Step 2: Determine the Conditions for Divisibility by 3
Next, we need to find pairs of numbers whose sum is divisible by 3. To do this, we can categorize the numbers based on their remainders when divided by 3:

- Numbers that give a remainder of 0 when divided by 3: \( 3, 6, 9, 12 \) (4 numbers)
- Numbers that give a remainder of 1 when divided by 3: \( 1, 4, 7, 10 \) (4 numbers)
- Numbers that give a remainder of 2 when divided by 3: \( 2, 5, 8, 11 \) (4 numbers)

### Step 3: Find Valid Pairs
To have a sum that is divisible by 3, we can pair numbers as follows:

1. **(0, 0)**: Both numbers have a remainder of 0.
2. **(1, 2)**: One number has a remainder of 1, and the other has a remainder of 2.

#### Count the pairs for each case:
1. **(0, 0)**:
- We can choose 2 from the 4 numbers that are \( 0 \mod 3 \):
\[ \text{Pairs} = \binom{4}{2} = 6 \]

2. **(1, 2)**:
- We can choose 1 from the 4 numbers that are \( 1 \mod 3 \) and 1 from the 4 numbers that are \( 2 \mod 3 \):
\[ \text{Pairs} = 4 \times 4 = 16 \]

### Step 4: Calculate the Total Valid Pairs
Now, we add the valid pairs from both cases:
\[ \text{Total valid pairs} = 6 + 16 = 22 \]

### Step 5: Calculate the Probability
Finally, we can calculate the probability that the sum of the two selected numbers is divisible by 3:
\[ \text{Probability} = \frac{\text{Total valid pairs}}{\text{Total outcomes}} = \frac{22}{66} = \frac{1}{3} \]

### Conclusion
Thus, the probability that the sum of two distinct numbers selected from the first twelve natural numbers is divisible by 3 is:
\[ \boxed{\frac{1}{3}} \]