Question

Show that if $P\left(A\right)>0$, then

$P(AB\mid A)\ge P(AB\mid A\cup B)$

Short answer

We proved that $P(AB\mid A)\ge P(AB\mid A\cup B)$by applying conditional probability as$P\left(A\right)>0$.

Step-by-step solution

Given Information

If$P\left(A\right)>0$, We have to prove that$P(AB\mid A)\ge P(AB\mid A\cup B)$.

Explanation

$AB\subseteq (A\cup B)\Rightarrow AB\cap (A\cup B)=AB$

The conditional probability $P[AB\mid (A\cup B\left)\right]$can be reduced to

$P[AB\mid (A\cup B\left)\right]=\frac{P[AB\cap (A\cup B\left)\right]}{P(A\cup B)}=\frac{P\left(AB\right)}{P(A\cup B)}$

Again, from set arithmetic

$A\subseteq A\cup B\Rightarrow P\left(A\right)\le P(A\cup B)$

Finally

$P(AB\mid A)=\frac{P\left(AB\right)}{P\left(A\right)}\ge \frac{P\left(AB\right)}{P(A\cup B)}=P[AB\mid (A\cup B\left)\right]$

Final Answer

$P(AB\mid A)$- is the percentage of $A$that is in role="math" localid="1647853157806" $AB$

$P[AB\mid (A\cup B\left)\right]$- is the percentage of $A\cup B$that is in $AB$.

And since $A\cup B$ is larger,$P[AB\mid (A\cup B\left)\right]\le P(AB\mid A)$.