Bayes theorem applied correctly on client churn? Here is a table selected and grouped from table where i store information about client - if he churned(TRUE - he churned, FALSE he stayed) and how many refund he got. CNT counts number of rows per class.

I want calculate the probabilites, so let's prepare the numbers step by step.
(I am inspired by this example)
I will convert the refund into discrete (refund is zero or greater than zero) variable and create a pivot table:

Then I calculate probabilities per class (certain cell divided by sum):

And then relative probabilities per class (I don't know the exact term - for example 0.36% is calculated as 6.03% * 6.02% from previous table):

So now I can calculate the probabilities:
X = Refund is 0;
C = Client churned;
not X = Refund > 0;
not C = Client stayed;
What is the chance that client will leave if he has 0 refund?
Pr(C|X) = Pr(X|C)*Pr(C) / Pr(x) = 0.36%/(0.36%+88.27%)=0.41%
What is the chance that client will stay if he has 0 refund?
Pr(not C|X) = Pr(X|not C)*Pr(not C) / Pr(x) = 88.27%/(0.36%+88.27%)=99.59%
What is the chance that client will leave if he has >0 refund?
Pr(C|not X) = Pr(not X|C)*Pr(C) / Pr(not x) = 0.00%/(0.00%+0.03%)=2.36%
What is the chance that client will stay if he has >0 refund?
Pr(not C|not X) = Pr(not X|not C)*Pr(not C) / Pr(not x)
=0.03%/(0.00%+0.03%)=97.64%
My question is - Can i assume anything if my data have one dominant class like this? Seems like the refund has no influence over churn. But I believe that it does have some effect.
If anyone verify the steps, it would be nice too.
Thank you.
 A: The best way to think about Bayes is conditioning. So when you have a question like this:

What is the chance that client will leave if he has >0 refund?

This question is answered with just data from the refund>0 row. $50/(50+133)\approx 27,3\%$ (Matching the question, I'm using comma as the radix point.) So it is the case that churning and refunds are associated.
Suppose you insist on using Bayes. Here you have a calculation mistake--when you want to determine $P(X|C)$, you need to divide by 6,03 instead of multiplying. (We're conditioning!) 22393 out of 22443 people who churned did not receive any refunds, which is 99,77%, not 0,36%.
(How could it have been obvious that you had a mistake? Your churn chance is lower whether you had a refund or didn't have a refund than it was over the whole population!)
$P(C|not\ X) = P(not\ X|C)*P(C) / P(not\ X) = \frac{50}{22443}\frac{22443}{372246}\frac{372246}{183}=\frac{50}{183}=27,3\%$
(Note why the conditioning view makes sense; all the terms cancel out except for the ones left at the end that correspond to the exact question you asked.)
