If P(x) = $$\frac{1}{4}$$, P(Y) = $$\frac{1}{3}$$ and $${\text{P}}\left( {{\text{X}} \cap {\text{Y}}} \right) = \frac{1}{{12}},$$ the value of $${\text{P}}\left( {\frac{{\text{Y}}}{{\text{X}}}} \right)$$ is A. $$\frac{1}{4}$$ B. $$\frac{4}{{25}}$$ C. $$\frac{1}{3}$$ D. $$\frac{{29}}{{50}}$$

The correct answer is $\boxed{\frac{1}{3}}$.

The probability of event A happening, given that event B has already happened, is called the conditional probability of A given B, and is denoted by $P(A|B)$. It can be calculated using the following formula:

$$P(A|B) = \frac{P(A \cap B)}{P(B)}$$

In this case, we are given that $P(X) = \frac{1}{4}$, $P(Y) = \frac{1}{3}$, and $P(X \cap Y) = \frac{1}{12}$. We can use these values to calculate $P(Y|X)$ as follows:

$$P(Y|X) = \frac{P(X \cap Y)}{P(X)} = \frac{\frac{1}{12}}{\frac{1}{4}} = \frac{1}{3}$$

Therefore, the probability of event Y happening, given that event X has already happened, is $\frac{1}{3}$.

Option A is incorrect because it is the probability of event X happening, not the probability of event Y happening given that event X has already happened.

Option B is incorrect because it is the probability of event Y happening, but it is not calculated using the correct formula.

Option C is incorrect because it is the probability of event X happening, but it is not the probability of event Y happening given that event X has already happened.

Option D is incorrect because it is the probability of event Y happening, but it is not calculated using the correct formula.