91Ó°ÊÓ

Understanding pairwise disjoint events is crucial in the exercise. When we say that events \(H_i\) are pairwise disjoint, it means that no two events can happen at the same time. Formally, \(H_i\) and \(H_j\) are pairwise disjoint if for any two events \(H_i\) and \(H_j\) with \(i eq j\), \(H_i \cap H_j = \emptyset\). This implies that the intersection of any two distinct events is empty.

This property is significant because it allows us to confidently decompose and calculate probabilities. In the context of the exercise, because the events are pairwise disjoint, their probabilities can be added together without any overlap.

Here’s a simple example: Imagine you are rolling a fair six-sided die. Define \(H_1\) as rolling a 1, \(H_2\) as rolling a 2, and so on. Each event \(H_i\) is mutually exclusive or pairwise disjoint with each other. If you want the probability of rolling either a 1 or a 2, you simply add \(P(H_1)\) and \(P(H_2)\).

Conditional probability helps us understand the probability of an event given that another event has already occurred. Mathematically, the conditional probability of event \(A\) given event \(B\) is denoted as \(P(A \mid B)\) and is defined as: \[ P(A \mid B) = \frac{P(A \cap B)}{P(B)} \] provided that \(P(B) \eq 0\).

In the context of the exercise, we use conditional probability to express the joint probability \(P(A \cap H_i)\) as the product of \(P(A \mid H_i) P(H_i)\). This transformation is a key step in the proof because it allows us to break down the probability of \(A\) into simpler parts that involve each event \(H_i\) and their corresponding probabilities.

For instance, consider you have a deck of cards, and you want to calculate the probability of drawing an ace given that you drew a red card. Here, event \(A\) is drawing an ace, and event \(B\) is drawing a red card. Since there are 26 red cards and 2 of those are aces, \(P(A \ \backslash \ B)= \frac {2}{26}\).

The sample space, denoted as \(\backslash \Omega\), is the set of all possible outcomes of an experiment. It's crucial to understand the concept of sample space as it provides the foundation for defining events and understanding probabilities. In the context of our exercise, the sample space is composed of the union of the pairwise disjoint events \(H_i\): \[ \bigcup_{i=1}^{\backslash infty} H_{i} = \backslash Omega \]

This means every possible outcome in our sample space can be found in one of our disjoint events \(H_i\). For example, when flipping a coin, the sample space is \(\backslash \ { \backslash ex tt{Heads, Tails}\ \}\). For a die roll, the sample space includes all numbers from 1 to 6.

For our exercise, knowing the sample space is covered by the union of \(H_i\) (each outcome fits into one of these events) is critical because it lets us construct the total probability of an event \(A\) using the given formula. This relationship is a key insight that allows us to apply the law of total probability effectively.