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In probability theory, the sample space refers to the set of all possible outcomes of a particular experiment. In the context of our password problem, the sample space encompasses all combinations of six-character passwords. Each character in these passwords can be either a letter from the 26-letter English alphabet (lowercase) or a digit from 0 to 9, providing 36 options per character. Thus, the sample space size, denoted as \(S\), is calculated by raising the number of options per character to the power of the password's length. Therefore, the total sample space is \(36^6\), which stands for every possible arrangement of characters in the password.

Event probability involves determining the likelihood that a particular scenario (an event) occurs out of all possible scenarios (the sample space). In our example, we are calculating the probabilities for different events such as a password beginning with a vowel (event \(A\)) or ending with an even number (event \(B\)).

• Event A: There are 5 vowels (\(a, e, i, o, u\)) which can start the password. Each following character in the password can be any of the 36 options, repeated 5 more times. Thus, the probability of event A is \( \frac{5 \, \times \, 36^5}{36^6} \), as 5 vowels can start the thousands of combinations of possible five-character continuations.
• Event B: Similarly, event \(B\), where passwords end in one of 5 even digits, also has a probability of \( \frac{36^5 \, \times \, 5}{36^6} \), since each of these is the last character following any set of possible five leading characters.

Combinatorics is the branch of mathematics dealing with counting, arranging, and combining objects. It's essential in calculating probabilities by considering different possibilities without listing each one. The password problem uses combinatorics extensively. For instance, to find how many passwords start with a vowel or end with an even digit, we do not directly count each possibility. Instead, using combinatorial principles, we analyze how we can arrange a desired item (here, a vowel or an even digit) at specified positions and multiply that by arrangements of other positions filled with any other possible character. It simplifies the computational process from an exhaustive enumeration.

In probability, the union of two events refers to the probability of either event happening. The intersection is where both events occur together.

• Intersection (\(A \cap B\)): This is where a password starts with a vowel and ends with an even number. First, set the first character from 5 vowels, the last from 5 even digits, and choose any of the 36 options for the middle four characters, rendering \(5 \times 36^4 \times 5\) combinations. Thus, \(P(A \cap B) = \frac{5 \times 36^4 \times 5}{36^6}\).
• Union (\(A \cup B\)): Calculating this involves combining probabilities from events A and B while subtracting the overlap \(A \cap B\) to avoid double-counting. Formulaically, \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\). This determines the probability of either event happening.