Analyzing Probability Statements on Events

Let's carefully examine each statement regarding events and their probabilities.

We are given two statements:

1. If A and B are mutually exclusive events, then it is possible that P(A) = P(B) = 0.6.
2. If A and B are any two events such that P(A|B) = 1, then P(B̅|A̅) = 1.

Analyzing Statement 1: Mutually Exclusive Events

Statement 1 says: If A and B are mutually exclusive events, then it is possible that P(A) = P(B) = 0.6.

Definition: Mutually exclusive events A and B are events that cannot occur at the same time. This means their intersection is empty (\(A \cap B = \emptyset\)).

For mutually exclusive events, the probability of their union is the sum of their individual probabilities:

\(P(A \cup B) = P(A) + P(B)\)

The total probability of any event cannot exceed 1. Therefore, \(P(A \cup B) \le 1\).

Let's test the possibility where \(P(A) = 0.6\) and \(P(B) = 0.6\) for mutually exclusive events A and B.

\(P(A \cup B) = P(A) + P(B)\)

\(P(A \cup B) = 0.6 + 0.6\)

\(P(A \cup B) = 1.2\)

Since the probability of the union \(P(A \cup B)\) must be less than or equal to 1, \(1.2 \le 1\) is false.

Thus, it is not possible for two mutually exclusive events to both have a probability of 0.6, because their combined probability would exceed 1.

Conclusion for Statement 1: Statement 1 is incorrect.

Analyzing Statement 2: Conditional Probability Relationship

Statement 2 says: If A and B are any two events such that P(A|B) = 1, then P(B̅|A̅) = 1.

Definition: The conditional probability \(P(A|B)\) is the probability of event A occurring given that event B has occurred. It is defined as \(P(A|B) = \frac{P(A \cap B)}{P(B)}\), provided \(P(B) > 0\).

If \(P(A|B) = 1\) (assuming \(P(B) > 0\)), it means \(\frac{P(A \cap B)}{P(B)} = 1\), which implies \(P(A \cap B) = P(B)\). This condition \(P(A \cap B) = P(B)\) signifies that the occurrence of B always leads to the occurrence of A; essentially, event B is a subset of event A (\(B \subseteq A\)) from a probabilistic perspective (except possibly for sets of probability zero).

We need to evaluate the consequence \(P(B̅|A̅) = 1\). This means \(\frac{P(B̅ \cap A̅)}{P(A̅)} = 1\) (assuming \(P(A̅) > 0\)), which implies \(P(B̅ \cap A̅) = P(A̅)\). This condition \(P(B̅ \cap A̅) = P(A̅)\) signifies that the occurrence of A̅ always leads to the occurrence of B̅; essentially, event A̅ is a subset of event B̅ (\(A̅ \subseteq B̅\)) from a probabilistic perspective.

Let's examine the relationship between \(P(A \cap B) = P(B)\) and \(P(B̅ \cap A̅) = P(A̅)\) for any events A and B.

We use basic probability rules and De Morgan's laws:

• De Morgan's Law: \(B̅ \cap A̅ = (B \cup A)̅\)
• Probability of Complement: \(P(E̅) = 1 - P(E)\)
• Probability of Union: \(P(B \cup A) = P(B) + P(A) - P(A \cap B)\)

Consider the statement \(P(A \cap B) = P(B)\).

From the probability of union rule:

\(P(B \cup A) = P(B) + P(A) - P(A \cap B)\)

Substitute \(P(A \cap B) = P(B)\) into the equation:

\(P(B \cup A) = P(B) + P(A) - P(B)\)

\(P(B \cup A) = P(A)\)

Now consider the statement \(P(B̅ \cap A̅) = P(A̅)\).

Using De Morgan's Law and the probability of complement:

\(P(B̅ \cap A̅) = P((B \cup A)̅) = 1 - P(B \cup A)\)

Using the probability of complement for \(P(A̅)\):

\(P(A̅) = 1 - P(A)\)

So, the statement \(P(B̅ \cap A̅) = P(A̅)\) is equivalent to \(1 - P(B \cup A) = 1 - P(A)\), which simplifies to \(P(B \cup A) = P(A)\).

We have shown that the condition \(P(A \cap B) = P(B)\) is equivalent to the condition \(P(B \cup A) = P(A)\). And the condition \(P(B̅ \cap A̅) = P(A̅)\) is also equivalent to \(P(B \cup A) = P(A)\).

Therefore, the statement \(P(A \cap B) = P(B)\) is mathematically equivalent to the statement \(P(B̅ \cap A̅) = P(A̅)\) for all events A and B.

In standard contexts, the statement "P(X|Y)=1" implies that the event X occurs whenever event Y occurs (except possibly on a set of probability zero). This is equivalent to \(P(X \cap Y) = P(Y)\). Similarly, "P(Z|W)=1" implies \(P(Z \cap W) = P(W)\). The statement in question is thus effectively "If \(P(A \cap B) = P(B)\), then \(P(B̅ \cap A̅) = P(A̅)\)", which we have shown to be true.

Conclusion for Statement 2: Statement 2 is correct.

Summary of Analysis

Based on the analysis, only Statement 2 is correct.

Revision Table: Key Concepts in Probability

Additional Information: Equivalence in Probability

The equivalence \(P(A \cap B) = P(B) \iff P(B̅ \cap A̅) = P(A̅)\) is a powerful result derived purely from the axioms of probability and set theory operations on events. It shows a symmetric relationship between events A and B and their complements A̅ and B̅. If event B is effectively contained within event A (in terms of probability), then the event A̅ is effectively contained within event B̅. This relationship holds regardless of whether the probabilities are zero or non-zero, resolving the potential issues with conditional probability definitions when the denominator is zero.

While conditional probability \(P(A|B)\) is formally defined only when \(P(B) > 0\), statements like "If \(P(A|B)=1\)..." in probability questions often rely on the underlying event relationship \(P(A \cap B) = P(B)\) that \(P(A|B)=1\) implies (when defined). By proving the equivalence of the underlying probability statements \(P(A \cap B) = P(B)\) and \(P(B̅ \cap A̅) = P(A̅)\), the implication in Statement 2 is shown to be universally true for all events A and B.