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Combinatorial analysis is a powerful tool in probability theory that helps us determine the number of ways to choose items from a larger set. In the context of the egg problem, we are determining how many ways four eggs can be selected from eighteen. This is a classic "combinations" problem, where the order of selection doesn't matter.
The formula for combinations is given by:

• \[ C(n, k) = \frac{n!}{k!(n-k)!} \]

Here, \( n \) is the total number of items, and \( k \) is the number of items to choose. Factorials (\( n! \)) are simply the product of all positive integers up to \( n \). For example, \( 4! = 4 \times 3 \times 2 \times 1 = 24 \). This formula tells us how many different groups of \( k \) items we can select from \( n \) items.
Applying this to our problem, we find there are 3060 ways to choose 4 eggs from the carton of 18 eggs. This overall number serves as the basis for calculating specific probabilities.

In this problem, the distinction between spoiled and unspoiled eggs is essential for calculating different probabilities. Given that there are 7 spoiled eggs and 11 unspoiled eggs in the carton, we can predict the likelihood of selecting certain numbers of unspoiled eggs.
To find the probability that a selection of eggs includes all unspoiled ones, we must calculate how many ways we can select 4 unspoiled eggs out of the 11 available. This is another combinations problem:

• \[ C(11, 4) = \frac{11!}{4!(11-4)!} = 330 \]

This tells us there are 330 combinations where we select only unspoiled eggs.
50Since these eggs are a subset of all possible selections (3060 total), we can establish the probability for exact numbers of unspoiled eggs in the chosen group.

Calculating probabilities involves determining the proportion of favorable outcomes to the total number of outcomes, both of which can be identified using combinatorial analysis. To illustrate:

• The probability of selecting exactly four unspoiled eggs (from 11 available) is \( \frac{330}{3060} = 0.1078 \).
• For outcomes involving two or fewer unspoiled eggs, we calculate separately for 0, 1, and 2 unspoiled eggs, add their probabilities, allowing us to account for all such outcomes.
• No unspoiled eggs could be picked by selecting spoiled ones, calculated via \( C(7, 4) \).
• One unspoiled egg would mean \( C(11, 1) \times C(7, 3) \).
• Two unspoiled is \( C(11, 2) \times C(7, 2) \).
Summing these probabilities gives the total for two or fewer unspoiled eggs.

The calculation for more than one unspoiled egg utilizes the complement rule: subtract the probability of picking none or just one from 1 (total certainty). Each probability calculation cements how these events can be anticipated based on the configuration of eggs.