Conditional Probability Equality

We are given two non-mutually exclusive events, A and B, such that the conditional probability of A given B is equal to the conditional probability of B given A.

The condition is stated as: $P(A|B) = P(B|A)$

Conditional Probability Formula

Recall the definition of conditional probability:

• $P(A|B) = \frac{P(A \cap B)}{P(B)}$
• $P(B|A) = \frac{P(B \cap A)}{P(A)}$

Note that the intersection of events is commutative, meaning $P(A \cap B) = P(B \cap A)$.

Applying the Condition

Substitute the formulas into the given condition:

$\frac{P(A \cap B)}{P(B)} = \frac{P(B \cap A)}{P(A)}$

Since $P(A \cap B) = P(B \cap A)$, we can write:

$\frac{P(A \cap B)}{P(B)} = \frac{P(A \cap B)}{P(A)}$

Deriving the Result

For this equality to hold:

• If $P(A \cap B) = 0$, the events would be mutually exclusive, which contradicts the problem statement.
• Therefore, we must have $P(A \cap B) \neq 0$. We can divide both sides of the equation by $P(A \cap B)$: $ \frac{1}{P(B)} = \frac{1}{P(A)} $

This implies that $P(A) = P(B)$.

Conclusion

Given the condition $P(A|B) = P(B|A)$ for non-mutually exclusive events, it must be true that $P(A) = P(B)$.

Option 1 ($A \subset B$ but $A \neq B$) implies $P(A) < P(B)$ (unless $P(A)=0$).
Option 2 ($A = B$) is a specific case where $P(A)=P(B)$ holds, but it's not the general conclusion.
Option 3 ($A \cap B = \phi$) contradicts the non-mutually exclusive condition.
Option 4 ($P(A) = P(B)$) is the direct consequence derived.