There is a method that can be applied to all of the questions asked, and that is to think in terms of combinations. It may be tempting to view this as a binomial distribution. The key reason why it isn't, is because the marbles are selecCted without replacement. This makes the probabilities change after each selection depending on what had already been chosen. In other words, the selections are not independent.
So let's tackle the first question: the probability that we select 4 marbles, and they are all red. The number of ways that 4 marbles can be selected from 17 total marbles is 17C4, which equals 2380. This will be the denominator representing the total sample space for this problem.
If all 4 marbles are red, then they must have been selected from the 5 red marbles. The number of ways this can happen is 5C4, which equals 5. So the probability of all 4 marbles being red is 5/2380, which is 1/476, or 0.0021.
The probability that exactly 2 are red will have the same denominator, but here we are getting only 2 from the 5 red marbles, and 2 from the 12 non-red marbles. The number of ways to select 2 red and 2 non-red is (5C2 · 12C2), or 10 · 66, which is 660. So this probability is 660/2380, or 33/119, which is about 0.2773.
The probability that none are red is the probability that all 4 come from the 12 non-red marbles. There are 12C4 ways to choose 4 marbles out of 12, or 495. This is also out of 2380, so this probability is 495/2380, or 99/476, which is about 0.208.
