Let $S = \{1, 2, 3, ..., 50\}$. We want to find the probability that $ab$ is divisible by 3, where $a, b \in S$ are chosen at random.

Total number of ways to choose $a$ and $b$ from $S$ is $50 \times 50 = 2500$.

Let $A$ be the event that $ab$ is divisible by 3. Then $A^c$ is the event that $ab$ is not divisible by 3. This happens when neither $a$ nor $b$ is divisible by 3.

The number of elements in $S$ divisible by 3 is $\lfloor \frac{50}{3} \rfloor = 16$.

The number of elements in $S$ not divisible by 3 is $50 - 16 = 34$.

The number of ways to choose $a$ and $b$ such that neither is divisible by 3 is $34 \times 34 = 1156$.

So, the number of ways to choose $a$ and $b$ such that $ab$ is divisible by 3 is $2500 - 1156 = 1344$.

The probability that $ab$ is divisible by 3 is $\frac{1344}{2500} = \frac{336}{625}$.

Now, we need to convert this fraction to have a denominator of 1225. However, $625 \times k = 1225$ has no integer solution for $k$. So, there must be an error in the calculation.

Let's calculate the probability that $ab$ is divisible by 3 directly.

The number of elements in $S$ divisible by 3 is 16.

The number of elements in $S$ not divisible by 3 is 34.

Case 1: $a$ is divisible by 3, $b$ is not divisible by 3. Number of ways = $16 \times 34 = 544$.

Case 2: $a$ is not divisible by 3, $b$ is divisible by 3. Number of ways = $34 \times 16 = 544$.

Case 3: Both $a$ and $b$ are divisible by 3. Number of ways = $16 \times 16 = 256$.

Total number of ways = $544 + 544 + 256 = 1344$.

The probability that $ab$ is divisible by 3 is $\frac{1344}{2500} = \frac{336}{625}$.

Let's try to match the given options. $\frac{1344}{2500} = \frac{x}{1225}$. Then $x = \frac{1344 \times 1225}{2500} = \frac{1344 \times 49}{100} = \frac{65856}{100} = 658.56$. This is not an integer, so the options are incorrect.

However, if the question meant that $a$ and $b$ are chosen independently, then the total number of ways is $50 \times 50 = 2500$.

The probability that $a$ is divisible by 3 is $\frac{16}{50}$. The probability that $a$ is not divisible by 3 is $\frac{34}{50}$.

The probability that $ab$ is divisible by 3 is $1 - P(\text{neither } a \text{ nor } b \text{ is divisible by 3}) = 1 - \frac{34}{50} \times \frac{34}{50} = 1 - \frac{1156}{2500} = \frac{1344}{2500} = \frac{336}{625}$.

Now, let's check the options again. We want to find $x$ such that $\frac{x}{1225} = \frac{336}{625}$. $x = \frac{336 \times 1225}{625} = \frac{336 \times 49}{25} = \frac{16464}{25} = 658.56$.

The options are still incorrect. Let's re-evaluate the number of ways.

Total ways = $50 \times 50 = 2500$.

Number of ways $a$ is divisible by 3 is 16. Number of ways $a$ is not divisible by 3 is 34.

Number of ways $b$ is divisible by 3 is 16. Number of ways $b$ is not divisible by 3 is 34.

Ways $ab$ is divisible by 3: (divisible, not divisible) + (not divisible, divisible) + (divisible, divisible) = $16 \times 34 + 34 \times 16 + 16 \times 16 = 544 + 544 + 256 = 1344$.

Probability = $\frac{1344}{2500} = \frac{336}{625} = \frac{336 \times 1.96}{625 \times 1.96} = \frac{658.56}{1225}$.

Let's try to find the closest option. $\frac{664}{1225} = 0.542, \frac{646}{1225} = 0.527, \frac{527}{1225} = 0.430, \frac{461}{1225} = 0.376$. $\frac{336}{625} = 0.5376$. The closest option is $\frac{646}{1225}$.

Let's assume the question meant to choose $a$ and $b$ without replacement. Then the total number of ways is $50 \times 49 = 2450$.

Number of ways to choose $a$ and $b$ such that neither is divisible by 3 is $34 \times 33 = 1122$.

Number of ways to choose $a$ and $b$ such that $ab$ is divisible by 3 is $2450 - 1122 = 1328$.

Probability = $\frac{1328}{2450} = \frac{664}{1225}$.