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In probability theory, the sample space is a fundamental concept that represents all the possible outcomes of a given experiment. When it comes to rolling two fair dice, the sample space consists of ordered pairs. Each pair shows the result of one die combined with the other. Since each die has 6 faces, you have a total of 6 x 6 = 36 possible outcomes. These can be listed as follows:

• (1,1)
• (1,2)
• (1,3)
• ... through to ...
• (6,6)

Each pair in the list is unique, representing one possible outcome of the two-dice roll. Understanding the sample space helps in calculating probabilities of specific events occurring, as it forms the denominator when determining probabilities such as those for events A and B in our exercise.

Mutually exclusive events are events that cannot occur at the same time. When determining if events are mutually exclusive, we look for overlap between the events. If there is no overlap, the events are mutually exclusive. In the exercise, we evaluated events A and B. Event A was getting a 3 or 4 first, followed by an even number; Event B was having the sum of the rolls being at most 7. Despite each event having specific conditions, there exist outcomes that satisfy both, such as (3,2) and (4,2). This overlap means the events are not mutually exclusive. Because of these overlapping outcomes, it is possible for both events to occur simultaneously, so we consider them not mutually exclusive.

Independent events in probability are those where the occurrence of one event does not affect the probability of the other. To check for independence, we compare the probability of both events occurring together to the product of their individual probabilities. If these two values match, the events are independent.In the exercise, event A was rolling a 3 or 4 first followed by an even number, and event B was the sum of the rolls being 7 or less. We calculated:

• For both A and B together: \[ P(A \cap B) = \frac{2}{36} = \frac{1}{18} \]
• The product of individual probabilities: \[ P(A) \times P(B) = \frac{1}{6} \times \frac{7}{12} = \frac{7}{72} \]

Since \( \frac{1}{18} \) doesn't equal \( \frac{7}{72} \), A and B do not meet the independence criterion.

Conditional probability is the likelihood of an event occurring given that another event has already occurred. To calculate conditional probability, we focus on reducing the sample space to only the outcomes where the given event occurs. We look at the probability of the event in question within this restricted sample space.In our exercise, we were tasked with finding \( P(A|B) \), the probability of event A occurring given event B has occurred. After narrowing the outcomes of event B (sums of 7 or less), we identified how many of those outcomes still meet the criteria for event A.

• The outcomes where both A and B occur were (3,2) and (4,2), resulting in 2 outcomes meeting both criteria.

Thus, the conditional probability was calculated as:\[ P(A|B) = \frac{2}{21} \]This accounting helps tighten predictions by incorporating prior known conditions.