Question

If A and B are any two events such that P(B) = P(A and B), then which of the following is correct

• A. P(B|A) = 1
• B. P(A|B) = 1
• C. P(B|A) = 0
• D. P(A|B) = 0

Hint:

The condition $P(B) = P(A \cap B)$ implies that event B is a subset of event A ($B \subseteq A$). If you visualize this with a Venn diagram, the entire circle for B is inside the circle for A. Therefore, if B occurs, A must also occur, making $P(A|B) = 1$.

Answer 1

The Correct Option is B

Step 1: Conceptualization:
Given a relationship between the probabilities of event B and the intersection of events A and B ($A \cap B$), the objective is to compute a conditional probability.
Step 2: Governing Principles:
The definitions of conditional probability are as follows:
\[ P(A|B) = \frac{P(A \cap B)}{P(B)} \quad (\text{where } P(B)>0) \]\[ P(B|A) = \frac{P(A \cap B)}{P(A)} \quad (\text{where } P(A)>0) \]Step 3: Analytical Breakdown:
The provided condition is $P(B) = P(A \cap B)$.
This condition will be applied to determine $P(A|B)$.
Applying the conditional probability formula:
\[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]Substituting the given condition $P(A \cap B) = P(B)$:
\[ P(A|B) = \frac{P(B)}{P(B)} \]Assuming $P(B)>0$ (as $P(A|B)$ is undefined if $P(B) = 0$):
\[ P(A|B) = 1 \]This outcome signifies that if event B occurs, event A is guaranteed to occur. This is logically consistent with the condition $P(B) = P(A \cap B)$, which implies that event B is a subset of event A. Therefore, the occurrence of B necessitates the occurrence of A.
Now, evaluating $P(B|A)$:
\[ P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{P(B)}{P(A)} \]The value of $P(B|A)$ cannot be determined without knowledge of $P(A)$. Consequently, options (1) and (3) are not definitively correct.
Step 4: Conclusion:
The verified statement is $P(A|B) = 1$.