To ensure that Callan has at least one marble of each color, let's think about the worst-case scenario because we want to be certain regardless of the order of the marbles Callan draws.
In the worst-case scenario:
- Callan could first draw all the marbles of the most abundant color without obtaining any of the other colors.
- Then, he could draw all the marbles of the second most abundant color.
- Only after these two steps would he be guaranteed to draw a marble of the remaining color.
Callan has 10 blue marbles, 6 green marbles, and 5 red marbles.
Step 1: Draw all blue marbles.
Callan draws all 10 blue marbles and still does not have any green or red marbles.
Step 2: Draw all green marbles.
After drawing the blue marbles, he then draws the 6 green marbles. At this point, he has only blue and green marbles.
Step 3: Get a marble of the last color.
Since Callan has already drawn all the blue and green marbles, the only marbles left are the red ones. With one more draw, he will certainly get a red marble because those are the only ones remaining.
Therefore, the total number of draws Callan needs to ensure he has at least one marble of each color is:
10 (all blue marbles) + 6 (all green marbles) + 1 (for the first red marble) = 17.
Thus, the answer is:
E) 17
