91Ó°ÊÓ

In probability and statistics, we often talk about 'sample space'. The sample space, usually denoted by the symbol \(S\), is the set of all possible outcomes in a probability experiment. Imagine you're rolling a die. The sample space for this experiment would be \(\{1, 2, 3, 4, 5, 6\}\) because these are all the possible results you can get from a single roll. Each outcome in the sample space is equally likely if the die is fair.
For the given exercise, our sample space is \(S=\{1,2,3,4,5,6,7,8,9,10,11,12\}\). This tells us there are 12 possible outcomes when the probability experiment is conducted. Knowing the sample space helps us understand the full range of possible outcomes and serves as the foundation for calculating probabilities.

The next step in solving our probability problem is calculating probabilities themselves. Probability is a measure of how likely an event is to occur, ranging from 0 (impossible) to 1 (certain). For a simple event from our sample space, the probability is calculated by dividing the number of favorable outcomes by the total number of possible outcomes in the sample space.
For example, if event \(E=\{2,3,4,5,6,7\}\), event \(F=\{5,6,7,8,9\}\), event \(G=\{9,10,11,12\}\), and event \(H=\{2,3,4\}\), these sets represent subsets of the sample space. To find the probability of one of these events occurring, we use the formula:
\ P(Event) = \frac{\text{Number of outcomes in the event}}{\text{Total number of outcomes in the sample space}} \
For event \(E\), the probability is:
\ P(E) = \frac{|E|}{|S|} = \frac{6}{12} = \frac{1}{2} \
This means there is a 50% chance that event \(E\) will occur.

The General Addition Rule in probability helps us to find the probability of either of two events happening. If you want to calculate the probability of either event \(E\) or event \(H\) happening, we need to be careful not to count the outcomes that could be common to both events twice. The rule is given by:
\ P(E \text{ or } H) = P(E) + P(H) - P(E \cap H) \
This rule subtracts the intersection of \(E\) and \(H\) since it is counted twice otherwise. For our problem, event \(E\) has 6 outcomes, and event \(H\) has 3 outcomes. The intersection set \(E \cap H = \{2,3,4\}\) with 3 outcomes. Let's calculate it step-by-step:
\ P(E) = \frac{|E|}{|S|} = \frac{6}{12} = \frac{1}{2} \
P(H) = \frac{|H|}{|S|} = \frac{3}{12} = \frac{1}{4} \
P(E \cap H) = \frac{|E \cap H|}{|S|} = \frac{3}{12} = \frac{1}{4} \
Substitute these values into the General Addition Rule:
\ P(E \text{ or } H) = \frac{1}{2} + \frac{1}{4} - \frac{1}{4} = \frac{1}{2} \
So, the probability of either \(E\) or \(H\) occurring is \(\frac{1}{2}\) or 50%.