Suppose you draw 4 gumballs from the jar without replacement. Is it more likely that you will draw exactly 1 marble of each colour or is it more likely that you will not draw exactly 1 of each colour?

Lacking data. How many colors/marbles?

To determine whether it is more likely to draw exactly 1 marble of each color or not, we can calculate the probabilities of each scenario.

Let's assume that the jar contains two different colors of marbles: red and blue. Since we are drawing 4 marbles without replacement, we have the following possibilities:
1) Drawing exactly 1 marble of each color: RBBR, RBRB, BRRB, BRBR, BBRR, RRBB (where R represents a red marble and B represents a blue marble).
2) Not drawing exactly 1 marble of each color: RRRR, BBBB.

Now, let's calculate the probabilities for each scenario:
1) Drawing exactly 1 marble of each color: There are 6 possible arrangements, and each has the same probability of occurring. The probability of drawing 1 marble of each color is therefore 6/20 (since there are 20 possible outcomes when drawing 4 marbles without replacement).

2) Not drawing exactly 1 marble of each color: There are 2 possible arrangements (RRRR and BBBB), and each has the same probability of occurring. The probability of not drawing exactly 1 marble of each color is therefore 2/20.

Comparing the probabilities, we can see that the probability of drawing exactly 1 marble of each color is larger (6/20) than the probability of not drawing exactly 1 marble of each color (2/20). Therefore, it is more likely to draw exactly 1 marble of each color when drawing 4 marbles without replacement from the jar.