The correct answer is $\boxed{\frac{3}{10}}$.
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 “the selected set contains one red ball and two black balls” and event B is “the first ball selected is red”. We are asked to find the probability of event A happening, given that event B has already happened.
The probability of event A happening is the number of ways to select one red ball and two black balls from a total of 10 balls, divided by the total number of ways to select 3 balls from 10 balls. This is:
$$P(A) = \frac{\binom{4}{1} \binom{6}{2}}{\binom{10}{3}} = \frac{4 \times 15}{120} = \frac{3}{20}$$
The probability of event B happening is the number of ways to select one red ball from a total of 4 red balls, divided by the total number of ways to select 1 ball from 10 balls. This is:
$$P(B) = \frac{\binom{4}{1}}{\binom{10}{1}} = \frac{4}{10} = \frac{2}{5}$$
Therefore, the conditional probability of event A happening, given that event B has already happened, is:
$$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{4 \times 15}{120}}{\frac{2}{5}} = \frac{3}{10}$$