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Probability theory is the mathematical framework that helps us quantify uncertainty. It's about measuring the likelihood of different outcomes under various circumstances. One key aspect in this field is the concept of events and their probabilities. When we talk about probability, especially in problems like the one given, we consider how likely a particular event is to occur within a certain context.

The hypergeometric distribution, used in the given problem, is a fine example of applied probability theory. It helps calculate the probability of a certain number of successes in a sample drawn without replacement from a finite population. This is quite different from distributions like the binomial distribution, where the draws are with replacement. In probability theory, the sum of all possible probabilities for a random variable equals one, ensuring each outcome has a defined chance of occurring.

Combinatorics is the branch of mathematics that deals with counting, arranging, and structuring elements within a set. In the context of our problem, it is essential for calculating probabilities via combinations. Combinations, one of the key tools in combinatorics, refer to selecting items from a set without considering the order.

For example, in the hypergeometric distribution formula, we see expressions like \(C(4,2)\), which represent the number of ways to choose 2 items from a set of 4. This can be calculated using the combination formula:

• \(C(n, k) = \frac{n!}{k!(n-k)!}\)

Where \(n!\) ("n factorial") means the product of all positive integers up to \(n\). Combinatorics is pivotal for problems like ours, allowing us to find the different ways to achieve certain outcomes without replacement.

Discrete probability distributions describe the probabilities of outcomes of a discrete random variable, where the outcomes are separate and distinct. In our problem, the hypergeometric distribution is a type of discrete probability distribution.

It fits situations where you have a finite number of objects, and you are interested in the probability distribution of a certain number of "successes" or specific outcomes. Discrete distributions differ from continuous ones, where outcomes can take any value within a range. Here, each outcome has a specific probability, and these probabilities add up to one.

This is crucial in our exercise, as we are determining the likelihoods of achieving specific numbers of successes when drawing without replacement. Once we compute individual probabilities, such as \(P(2)\), \(P(4)\), and \(P(x \leq 1)\), they illustrate how likely each scenario is within the bounds of this discrete context.