Prove that if P(A)=a and P(B)=b (with b>0), then P(Amid…

Prove that if $P(A)=a$ and $P(B)=b$ (with $b\gt0$), then $P(A\mid B)\ge\dfrac{a+b-1}{b}$.

Step 1: Bonferroni

\[ P(AB)=P(A)+P(B)-P(A+B)\ge a+b-1. \]

Step 2: Divide by $b$

\[ \boxed{P(A\mid B)=\frac{P(AB)}{b}\ge\frac{a+b-1}{b}.} \quad\blacksquare \]

Step 3: When is the bound useful?

If $a+b\le1$ the bound is $\le0$ and says nothing. It is useful when $A$ is very likely. Example: $a=0.95$, $b=0.4$: $P(A\mid B)\ge\frac{0.35}{0.4}=0.875$ — even after learning that $B$ occurred, $A$ remains likely.

Step 4: Equality

Equality holds iff $P(A+B)=1$. For example, with $X$ uniform in $[0,1]$, $A=\{X\gt0.05\}$, $B=\{X\lt0.4\}$: $P(A\mid B)=\frac{0.35}{0.4}=0.875$ exactly.