Prove that the condition P(Bmid A)=P(Bmidoverline A) implies…
Prove that the condition $P(B\mid A)=P(B\mid\overline A)$ implies that the events $A$ and $B$ are independent.
Explanation of Method
The total probability formula writes $P(B)$ as a weighted average of $P(B\mid A)$ and $P(B\mid\overline A)$ with weights $P(A)$ and $P(\overline A)$. If the two conditional probabilities are equal, their average equals each of them, so $P(B\mid A)=P(B)$ — independence. (We assume $0\lt P(A)\lt1$, so that both conditional probabilities are defined.)
Step 1: Total probability
\[ P(B)=P(A)P(B\mid A)+P(\overline A)P(B\mid\overline A). \]
Step 2: Use the condition
Let $c=P(B\mid A)=P(B\mid\overline A)$. Then
\[ P(B)=P(A)\,c+P(\overline A)\,c=c\,\big(P(A)+P(\overline A)\big)=c. \]
Step 3: Conclusion
So $P(B\mid A)=P(B)$, which means
\[ \boxed{P(AB)=P(A)P(B\mid A)=P(A)P(B):\ \ A \text{ and } B \text{ are independent.}} \quad\blacksquare \]
Step 4: Meaning and converse
If the chance of $B$ is the same whether or not $A$ occurs, then $A$ carries no information about $B$. The converse also holds: if $A$ and $B$ are independent, then $P(B\mid A)=P(B)=P(B\mid\overline A)$ (see Problem 121).
Result
| Key step | $P(B)=P(A)c+P(\overline A)c=c$ with $c=P(B\mid A)=P(B\mid\overline A)$ |
| Conclusion | $P(AB)=P(A)P(B)$ |