Understanding Probability with Mutually Exclusive and Exhaustive Events

The question describes a scenario involving two events, A and B, and their complement events, A' and B'. We are told about two specific events, denoted as AB and ABÌ…, which are stated to be mutually exclusive and exhaustive within the context where event A occurs.

Interpreting the Notation

In probability and set theory notation:

• AB usually represents the intersection of events A and B, written as $A \cap B$.
• AÌ… represents the complement of event A, written as $A^c$ or A'.
• ABÌ… represents the intersection of event A and the complement of event B, written as $A \cap B^c$ or $A \cap B'$.

The statement "AB and ABÌ… are two mutually exclusive and exhaustive events in which the event A can occur" implies that these two events form a partition of A. In other words, event A can be expressed as the union of $A \cap B$ and $A \cap B^c$.

Partitioning Event A

Any event A can be divided into two parts based on whether event B occurs or not:

1. The part of A where B also occurs: This is the intersection $A \cap B$.
2. The part of A where B does not occur: This is the intersection $A \cap B^c$.

These two parts, $A \cap B$ and $A \cap B^c$, are always mutually exclusive. Why? Because if an outcome is in $A \cap B$, it must be in B. If an outcome is in $A \cap B^c$, it must not be in B. An outcome cannot simultaneously be in B and not in B. Thus, there is no overlap between $A \cap B$ and $A \cap B^c$.

Furthermore, the union of these two parts is the event A itself:

$\qquad A = (A \cap B) \cup (A \cap B^c)$

This means that any outcome in A is either in $A \cap B$ or in $A \cap B^c$ (or both, but they are mutually exclusive, so it's one or the other), and any outcome in $(A \cap B) \cup (A \cap B^c)$ must be in A.

Probability of a Partitioned Event

Since event A is the union of two mutually exclusive events, $A \cap B$ and $A \cap B^c$, the probability of A is the sum of the probabilities of these two events.

Using the additive rule for mutually exclusive events:

$\qquad P(A) = P((A \cap B) \cup (A \cap B^c))$

Since $A \cap B$ and $A \cap B^c$ are mutually exclusive:

$\qquad P(A) = P(A \cap B) + P(A \cap B^c)$

Comparing with Options

Let's look at the given options, interpreting the notation as discussed:

1. $P(A) = P(A \cap B) + P(A \cap B^c)$ (using $A^c \cap B$ for AÌ…B and $A \cap B^c$ for ABÌ…) - Incorrect interpretation based on option 2's structure. Let's assume AÌ…B is $A^c \cap B$. Then this is $P(A) = P(A^c \cap B) + P(A \cap B^c)$.
2. $P(A) = P(A \cap B) + P(A \cap B^c)$ (using $A \cap B$ for AB and $A \cap B^c$ for ABÌ…)
3. $P(A) = P(A^c) + P(A \cap B^c)$ (using $A^c$ for AÌ… and $A \cap B^c$ for ABÌ…)
4. $P(A) = 1$

Based on our derivation $P(A) = P(A \cap B) + P(A \cap B^c)$, Option 2 matches this formula exactly when interpreting AB as $A \cap B$ and ABÌ… as $A \cap B^c$. This interpretation is consistent with the partitioning of event A.

Why other options are likely incorrect

• Option 1: $P(A) = P(A^c \cap B) + P(A \cap B^c)$. This relates P(A) to parts involving $A^c$, which doesn't directly follow from partitioning A itself.
• Option 3: $P(A) = P(A^c) + P(A \cap B^c)$. This formula is incorrect as $A$ and $A^c$ are mutually exclusive and exhaustive for the entire sample space, so $P(A) + P(A^c) = 1$. This option suggests $P(A) = 1 - P(A) + P(A \cap B^c)$, which simplifies to $2P(A) = 1 + P(A \cap B^c)$, not a general truth for P(A).
• Option 4: $P(A) = 1$. This would mean A is the sure event (sample space), which is not implied by the problem statement.

Therefore, the formula that correctly represents P(A) based on the stated properties of AB ($A \cap B$) and ABÌ… ($A \cap B^c$) is $P(A) = P(A \cap B) + P(A \cap B^c)$.

Revision Table: Key Probability Concepts

Additional Information: Understanding Event Partitions

The concept of partitioning an event is fundamental in probability. When an event A is partitioned into smaller, mutually exclusive sub-events $E_1, E_2, ..., E_n$, it means two things:

1. The sub-events do not overlap: $E_i \cap E_j = \emptyset$ for any $i \neq j$.
2. The sub-events collectively make up the entire event A: $E_1 \cup E_2 \cup ... \cup E_n = A$.

In our case, event A is partitioned by events related to B: $A \cap B$ and $A \cap B^c$. This is always a valid partition of A because every outcome in A is either in B or not in B, and it cannot be in both B and not B simultaneously. This partition is often used in various probability theorems, such as the Law of Total Probability.

The Law of Total Probability states that if $B_1, B_2, ..., B_n$ are mutually exclusive and exhaustive events that form a partition of the sample space, then for any event A:

$\qquad P(A) = \sum_{i=1}^{n} P(A \cap B_i)$

In our specific problem, the events $B$ and $B^c$ form a partition of the sample space S (assuming $0 < P(B) < 1$). Applying the Law of Total Probability to event A with respect to the partition $\{B, B^c\}$ of the sample space S:

$\qquad P(A) = P(A \cap B) + P(A \cap B^c)$

This confirms our earlier derivation and aligns with the correct option, reinforcing the understanding that $A \cap B$ and $A \cap B^c$ partition the event A.