Do I use chi-square test correctly for such dataset? As a person totally green in statistics I would greatly appreciate your help. I tried to figure this on my own but I'm not sure, so all I ask for is a confirmation or explaining me where my mistake is.
Let's say I have a data on 1096 people. Every person can be characterized by variable X with 5 levels. These people can make 1 of 6 choices: A,B,C,D,E or F.
I would like to test statistically whether these two variables are dependent or independent. Or more specifically, if there is a "preferred choice" (A,B,C,D,E, of F) for each type of person (VS,VA, etc..)
This is my contingency table:

tabXY is my variable in which I store this contingency table in R.
What I did was:
test_tabXY <- chisq.test(tabXY)

My result:
Pearson's Chi-squared test
data:  tabXY
X-squared = 386.92, df = 20, p-value < 2.2e-16

Later I converted the original table into a frequencey table, so that:

and I tested it in the same way. I've got:
Pearson's Chi-squared test
data:  tabXY_freq
X-squared = 220.62, df = 20, p-value < 2.2e-16

First of all, why is the p-value so small? It looks susppicious to me.
And secondly, why do I get different X-squared value when I use the occurance and the frequency table?
I would really appreciate your help
 A: The chi-square test examines whether the columns and rows of your table are independent. With your 1096 total cases and the total numbers of each preferred choice and of each type of person, this is this table you would get (perhaps with some rounding error) if the rows and columns were independent:
  VS VA VT VBT  VB
A 12  9 11  30  27
B 10  7  9  25  22
C 85 63 81 218 192
D  9  7  8  23  20
E 17 13 16  44  38
F 14 10 13  35  31

Now look at your data. Do you think it's very likely that you could, just by chance, have a true distribution like this table based on assuming independence, but still come up with the table you found? Your observed table, in which almost no VS types make A or B choices, while about half of the VA types make such choices? No. The p-value is very low because there is essentially no way that you could have gotten your results if the rows (choices) and columns (person types) were independent.
Your frequency table presents, for each type of person (column),  the fraction making each choice (row). It thus has lost the information about the differences in numbers of each type of person and would be expected to give different results.
