Show that if the conditional probability P(Amid B) is…

Show that if the conditional probability $P(A\mid B)$ is greater than the unconditional probability $P(A)$, then the conditional probability $P(B\mid A)$ is also greater than the unconditional probability $P(B)$.

Step 1: Assumptions

The conditional probabilities $P(A\mid B)=\frac{P(AB)}{P(B)}$ and $P(B\mid A)=\frac{P(AB)}{P(A)}$ require $P(A)\gt0$ and $P(B)\gt0$.

Step 2: Translate the hypothesis

\[ P(A\mid B)\gt P(A) \iff \frac{P(AB)}{P(B)}\gt P(A) \iff P(AB)\gt P(A)P(B). \]

(Multiplying by $P(B)\gt0$ keeps the direction of the inequality.)

Step 3: Read it back

Dividing $P(AB)\gt P(A)P(B)$ by $P(A)\gt0$:

\[ \frac{P(AB)}{P(A)}\gt P(B), \quad\text{i.e.}\quad \boxed{P(B\mid A)\gt P(B).} \quad\blacksquare \]

Step 4: Meaning

The condition $P(AB)\gt P(A)P(B)$ says that $A$ and $B$ are positively correlated. It is symmetric: if $B$ makes $A$ more likely, then $A$ makes $B$ more likely. The same argument works with "$\lt$" (negative correlation) and "$=$" (independence).

Step 5: Example

Roll a die, $A$ = "even" $=\{2,4,6\}$, $B$ = "at least $4$" $=\{4,5,6\}$. Then $P(A)=P(B)=\frac12$, $P(AB)=P(\{4,6\})=\frac13\gt\frac14$.

\[ P(A\mid B)=\tfrac23\gt\tfrac12=P(A), \qquad P(B\mid A)=\tfrac23\gt\tfrac12=P(B). \ \checkmark \]