Problem 7 - A Computational Introduction to Number Theory and Algebra

Question

Let \(\mathcal{B}\) be an event with \(\mathrm{P}[\mathcal{B}] \neq 0,\) and let \(\left\\{\mathcal{C}_{i}\right\\}_{i \in I}\) be a finite, pairwise disjoint family of events whose union contains \(\mathcal{B}\). Again, generalizing the law of total probability, show that for every event \(\mathcal{A},\) if \(I^{*}:=\left\\{i \in I: \mathrm{P}\left[B \cap C_{i}\right] \neq 0\right\\},\) then we have $$ \mathrm{P}[\mathcal{A} \mid \mathcal{B}]=\sum_{i \in I^{*}} \mathrm{P}\left[\mathcal{A} \mid \mathcal{B} \cap C_{i}\right] \mathrm{P}\left[\mathcal{C}_{i} \mid \mathcal{B}\right] $$

Step-by-Step Solution

Answer

Answer: The derived equation for conditional probability is: $$\mathrm{P}[\mathcal{A} \mid \mathcal{B}] = \sum_{i \in I^{*}} \mathrm{P}[\mathcal{A} \mid \mathcal{B} \cap \mathcal{C}_{i}] \mathrm{P}[\mathcal{C}_{i} \mid \mathcal{B}]$$

Step 1: Write the definition of conditional probability for \(\mathrm{P}[\mathcal{A} \mid \mathcal{B}]\)

We will first write the definition of the conditional probability: $$\mathrm{P}[\mathcal{A} \mid \mathcal{B}] = \frac{\mathrm{P}[\mathcal{A} \cap \mathcal{B}]}{\mathrm{P}[\mathcal{B}]}$$

Step 2: Apply the law of total conditional probability for \(\mathrm{P}[\mathcal{A} \cap \mathcal{B}]\)

Since the events \(\mathcal{C}_i\) are pairwise disjoint and their union contains \(\mathcal{B}\), we can rewrite the probability of the intersection \(\mathcal{A} \cap \mathcal{B}\) as follows: $$\mathrm{P}[\mathcal{A} \cap \mathcal{B}] = \sum_{i \in I} \mathrm{P}[\mathcal{A} \cap (\mathcal{B} \cap \mathcal{C}_{i})]$$

Step 3: Apply the definition of conditional probability for \(\mathrm{P}[\mathcal{A} \mid \mathcal{B} \cap \mathcal{C}_{i}]\) and \(\mathrm{P}[\mathcal{C}_{i} \mid \mathcal{B}]\)

Using the definition of conditional probability, we rewrite each term in the summation: $$\mathrm{P}[\mathcal{A} \mid \mathcal{B} \cap \mathcal{C}_{i}] = \frac{\mathrm{P}[\mathcal{A} \cap (\mathcal{B} \cap \mathcal{C}_{i})]}{\mathrm{P}[\mathcal{B} \cap \mathcal{C}_{i}]}$$ $$\mathrm{P}[\mathcal{C}_{i} \mid \mathcal{B}] = \frac{\mathrm{P}[\mathcal{B} \cap \mathcal{C}_{i}]}{\mathrm{P}[\mathcal{B}]}$$

Step 4: Combine the definitions from Step 3 with the equation from Step 2 and rearrange terms

Now we substitute the definitions of conditional probabilities from Step 3 into the equation from Step 2, and rearrange the terms: $$\mathrm{P}[\mathcal{A} \cap \mathcal{B}] = \sum_{i \in I} \mathrm{P}[\mathcal{A} \mid \mathcal{B} \cap \mathcal{C}_{i}] \mathrm{P}[\mathcal{B} \cap \mathcal{C}_{i}]$$ $$\mathrm{P}[\mathcal{A} \cap \mathcal{B}] = \sum_{i \in I} \mathrm{P}[\mathcal{A} \mid \mathcal{B} \cap \mathcal{C}_{i}] \mathrm{P}[\mathcal{C}_{i} \mid \mathcal{B}] \mathrm{P}[\mathcal{B}]$$

Step 5: Replace summation index set I with \(I^{*}\) and multiply by \(\mathrm{P}[\mathcal{B}]\)

Now, we replace the summation index set \(I\) with \(I^{*}\) (since the terms with zero probability do not contribute to the sum) and multiply both sides of the equation by \(\mathrm{P}[\mathcal{B}]\): $$\mathrm{P}[\mathcal{A} \cap \mathcal{B}] = \sum_{i \in I^{*}} \mathrm{P}[\mathcal{A} \mid \mathcal{B} \cap \mathcal{C}_{i}] \mathrm{P}[\mathcal{C}_{i} \mid \mathcal{B}] \mathrm{P}[\mathcal{B}]$$

Step 6: Replace \(\mathrm{P}[\mathcal{A} \cap \mathcal{B}]\) with the definition of conditional probability

We now replace \(\mathrm{P}[\mathcal{A} \cap \mathcal{B}]\) with the definition of conditional probability from Step 1: $$\mathrm{P}[\mathcal{A} \mid \mathcal{B}] = \sum_{i \in I^{*}} \mathrm{P}[\mathcal{A} \mid \mathcal{B} \cap \mathcal{C}_{i}] \mathrm{P}[\mathcal{C}_{i} \mid \mathcal{B}]$$ This is the equation we wanted to show.

Key Concepts

Understanding Probability Theory

Probability theory is the mathematical study of uncertainty. It provides a framework for quantifying how likely it is that certain events will occur.

At the heart of probability theory is the concept of an event, which is any collection of outcomes from some activity or experiment that is well-defined. For example, when flipping a coin, the possible events include getting a 'heads' or a 'tails'.

Understanding the outcomes of different events and how they relate to each other is crucial for calculating probabilities. These calculations often require different tools and principles, such as the law of total probability and concepts surrounding disjoint events, which are fundamental to tackling a wide range of problems in the field.

Law of Total Probability

The law of total probability is an important rule in probability that relates marginal probabilities to conditional probabilities. It is particularly useful when dealing with a set of disjoint events that covers the entire sample space.

The law states that if we have a collection of events that are mutually exclusive and exhaustive, the probability of any event can be found by summing the probabilities of the intersections of that event with the collection's events, each weighted by the probability of the corresponding collection's event.

\[\mathrm{P}(A) = \sum_{i} \mathrm{P}(A \cap B_{i})\]Where \(B_{i}\) are the disjoint events that cover the entire sample space. The exercise provided utilizes this principle to derive the conditional probability \(\mathrm{P}(\mathcal{A} \mid \mathcal{B})\) by taking into account the disjoint and collective nature of \(\mathcal{C}_{i}\) events.

Disjoint Events in Probability

Disjoint events, also known as mutually exclusive events, are events that cannot occur simultaneously.

For instance, when rolling a six-sided die, the events of rolling a 'three' and rolling a 'four' are disjoint because you cannot roll both at the same time. This characteristic significantly simplifies the calculation of probabilities involving such events, especially when combined with the law of total probability.

When considering the conditional probability expression in the exercise, the disjoint nature of the events \(\mathcal{C}_i\) is vital. It ensures that the probabilities of the intersections \(\mathrm{P}(\mathcal{B} \cap \mathcal{C}_{i})\) do not overlap, allowing for a clear and precise calculation of the sum detailed in the exercise solution.