Differentiating between the union and intersection of probabilities I have the following question:

A box in a supply room contains $15$ compact fluorescent
lightbulbs, of which $5$ are rated $13$-watt, $6$ are rated
$18$-watt, and $4$ are rated $23$-watt. Suppose that three of
these bulbs are randomly selected. What is the probability that one bulb of each type is
selected?

This exercise was solved by finding out the probability that one bulb of each of the three listed types is chosen then adding them together, I had this in mind but I also got confused on why we didn't go for multiplying the probabilities for which lead me to finding out I am not doing well in terms of differentiating between intersection and union of probabilities, both seem reasonable to do here for me which definitely shouldn't be the case.
To make my questions clearer:
-Why not use intersection here? Since intersection means that all events occur during this selection.
-How to be able to differentiate between the two concepts?
 A: This is an intersection since we want one 13-watt bulb and one 18-watt bulb and one 23-watt bulb.
Since there are $5$ 13-watt bulbs, $6$ 18-watt bulbs, and $4$ 23-watt bulbs among the $15$ compact fluorescent bulbs, there are $5 \cdot 6 \cdot 4$ ways to select one of each and $\binom{15}{3}$ ways to select three of the $15$ bulbs.  Hence, the desired probability is
$$\frac{\dbinom{5}{1}\dbinom{6}{1}\dbinom{4}{1}}{\dbinom{15}{3}}$$
Since the choices of the three bulbs are independent, we multiply.  We would add if the events were mutually exclusive.
Let's consider a simpler example.  Suppose we have two blue, one red, and one green ball hidden inside a bag.  What is the probability of selecting one ball of each color when selecting three of the four balls without looking at their colors until after the balls have been removed from the bag?
If we are equally likely to pick any of the four balls, the probability of selecting a blue ball is $1/2$, the probability of picking a red ball is $1/4$, and the probability of picking a green ball is $1/4$.  Adding the probabilities of selecting a blue ball, a red ball, and a green ball would give us $1$, which cannot be the case since the ball we leave in the bag does not have to be blue.
The probability of selecting one ball of each color is
$$\frac{\dbinom{2}{1}\dbinom{1}{1}\dbinom{1}{1}}{\dbinom{4}{3}}$$
since we must select one of the two blue balls, the red ball, and the green ball when selecting three of the four balls in the bag.  Indeed, this is the case since two of the four possibilities are favorable: $\{B_1,G,R\}$ and $\{B_2,G,R\}$, where the possibilities are subsets of three of the four balls and we use the labels $B_1$ and $B_2$ to differentiate between the two blue balls we could have selected.
On the other hand, if we draw one ball from the same bag, the probability of obtaining a blue ball or a red ball or a green ball is
$$\frac{1}{2} + \frac{1}{4} + \frac{1}{4} = 1$$
since there are three mutually exclusive possibilities and one of them must occur.  This is a union of three disjoint events.
