Question

If A and B are two non-mutually exclusive events such that $P(A|B)=P(B|A)$ then

• A. $A\subset B$ but $A\ne B$
• B. $A=B$
• C. $A\cap B=\phi$
• D. $P(A)=P(B)$

Hint:

Conditional probability represents the relative "weight" of the intersection within the sample space of the condition. If the intersection carries the same weight in both $A$ and $B$, then the total weights of $A$ and $B$ must be identical.

Answer 1

The Correct Option is D

We are given two events $A$ and $B$ that are not mutually exclusive, meaning their intersection is not necessarily empty.
The condition provided is the equality of their conditional probabilities.

Step 1: \color{redApply the Definition of Conditional Probability
The formula for conditional probability is $P(X|Y) = \frac{P(X \cap Y)}{P(Y)}$, provided $P(Y) > 0$.
Given $P(A|B) = P(B|A)$, we substitute the definition for both sides:
$\frac{P(A \cap B)}{P(B)} = \frac{P(B \cap A)}{P(A)}$.

Step 2: \color{redSimplify using Set Properties
In set theory, the intersection of sets is commutative, so $A \cap B = B \cap A$.
This implies $P(A \cap B) = P(B \cap A)$.
Our equation becomes: $\frac{P(A \cap B)}{P(B)} = \frac{P(A \cap B)}{P(A)}$.

Step 3: \color{redSolve the Algebraic Equation
Since A and B are non-mutually exclusive, we can assume $P(A \cap B) \neq 0$.
Dividing both sides by $P(A \cap B)$, we get:
$\frac{1}{P(B)} = \frac{1}{P(A)}$.
Taking the reciprocal of both sides:
$P(A) = P(B)$.

Step 4: \color{redEvaluate the Options
The result $P(A) = P(B)$ states that the marginal probabilities of the two events must be equal.
While $A=B$ would satisfy $P(A)=P(B)$, it is not a requirement; two different events can have the same probability.
Therefore, the most general and correct conclusion is $P(A)=P(B)$.
This matches Option (4).