Meaning of $\frac{P(X\cap Y)}{P(X)P(Y)}$ Imagine that we have a set $\Omega$ and $X$ and $Y$ are events that can happen, I mean, $P(X),P(Y)>0$. Then, what does it mean the ratio $\frac{P(X\cap Y)}{P(X)P(Y)}$?
I know that $\frac{P(X\cap Y)}{P(X)P(Y)}=\frac{P(X|Y)}{P(X)}=\frac{P(Y|X)}{P(Y)}$ and if that ratio is equal to 1 then $X,Y$ are independent events, but I can't figure out what exactly it means... please give simple examples.
I found this when reading about lift-data mining.
 A: The ratio $\frac{P(X\cap Y)}{P(X)P(Y)}=\frac{P(X|Y)}{P(X)}$ is called lift, and it is a measure of how good the occurrence of event $Y$ is at predicting the occurrence of event $X$. We interpret the lift as the ratio in which the probability of $X$ increases after the occurrence of $Y$. Notice that $\frac{P(X^c|Y)}{P(X^c)} = \frac{1 - P(X|Y)}{1 - P(X)} = \frac{1/P(X) - \text{lift}}{1/P(X) - 1}$, which is 1 precisely when lift $= 1$, and increases as lift increases. This is consistent with predicting the non occurrence of $X$ after $Y$ when lift $< 1$.
Example: $X$ is the event that this answer receives an up-vote. $Y$ is the event that I provide an example. In this case we expect lift $> 1$, as $Y$ is presumably associated with an increased probability of $X$.
Source: Wikipedia
A: As @JMoravitz noted, it's $\frac{P(X|Y)}{P(X)} = \frac{P(Y|X)}{P(Y)}$.
Let's call this ratio $r$. So if  $r>1$, then $P(X|Y) = r P(X) > P(X)$ and $P(Y|X) = r P(Y) > P(Y)$, and reverse inequalities hold if $r < 1$. Intuitively, if $r > 1$ and $Y$ has been observed, the probability of $X$ increases by this factor.
Example: Let $X$ be the event that you get COVID while spending an hour on the exercise bicycle in the gym and let $Y$ be the event that a person with asymptomatic COVID uses the bicycle next to you during this hour. Then $r$ measures the increase in risk due to the exposure to somebody with COVID who exercises next to you, and presumably $r > 1$.
A: As @Chrisjanjigian has noted
Cov(1,1)=(,)−()()
So when covariance is zero P(A,B)=P(A)P(B)
And correlation is cov(AB)/[stdev(A)*stdev(B)]
The long answer:

We have balls in 2 pools and the square outside the pools. 40 balls are not in the pools, 25 in pool A, 35 in pool B. A total of 85 unique balls.
Here is the calculation to see if the events are independent or not. This example uses no measurements simply probabilities of 1 and 0 for each ball to calculate the covariance. Cov(AB)=P(AB)-P(A)P(B) = E[(A-P(A)(B-P(B)] =
Average((A-P(A)(B-P(B)) where A and B are either 1 or 0 for each of the 85 balls in the average based on whether it is or isnt in A or B

