Four red balls, four green balls and four blue balls are put in a box. Three balls are pulled our of the box at random one after another without replacement. The probability that all the three balls are red is A. $$\frac{1}{{72}}$$ B. $$\frac{1}{{55}}$$ C. $$\frac{1}{{36}}$$ D. $$\frac{1}{{27}}$$

The correct answer is $\boxed{\frac{1}{27}}$.

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, event A is “all three balls are red” and event B is “the first ball is red”. We are asked to find the probability of event A given event B.

The probability of event A happening is the number of ways to choose 3 red balls out of 12 balls divided by the total number of ways to choose 3 balls out of 12 balls, which is $\frac{4 \times 3 \times 2}{12 \times 11 \times 10} = \frac{1}{27}$.

The probability of event B happening is the number of ways to choose 1 red ball out of 4 red balls divided by the total number of ways to choose 1 ball out of 12 balls, which is $\frac{4}{12} = \frac{1}{3}$.

Therefore, the probability of event A given event B is $\frac{\frac{1}{27}}{\frac{1}{3}} = \boxed{\frac{1}{27}}$.

Option A is incorrect because it is the probability of event A happening without considering event B.

Option B is incorrect because it is the probability of event B happening without considering event A.

Option C is incorrect because it is the probability of event A and event B happening together.