How can I check if two distributions are independent A telephone salesman calls 3 people in a month and sells a product with probability 30%. In each call he sells either 0 or 1 products. He calls three people this month. If he manages to sell the product in his first or second call he gets 100 euro per successful call and 200 if he sells a product in his last call. Let $X$ be number of products he sells and $Y$ the sum of euros he sells. Are $X$ and $Y$ independent?
I managed to compute total distributions and marginal distributions of $X$ and $Y$ but how can I compute $\mathbb{P}(X=x\cap Y=y)?$
 A: Hints (for doing it the hard way):


*

*Let $Z_1$, $Z_2$, $Z_3$ denote the number of sales resulting from the first, second, and
third calls respectively. What kinds of random variables are the $Z_i$'s? (They 
are not Brand X random variables, they have a specific name that is recognized by most
people).

*Express $X$, the total  number of sales, in terms of the $Z_i$'s.

*Express $Y$, the total commission earned, in terms of the $Z_i$'s.

*Now create a table with 5 rows (for the values of $Y$) and $4$ columns
(for the values of $X$) and in each of the $20$ cells fill in the probability
that $X$ and $Y$ have the corresponding values. At most $8$ of the $20$ cells
will have nonzero entries. (Can you tell why?).

*Now you have the joint probability mass function of $X$ and $Y$, and the
marginal mass functions are just the column and row sums of the table entries.
You can
proceed to verify the hard way whether $X$ and $Y$ are independent by checking
whether each and every entry in the table equals the product of the corresponding
row sum and column sum.

Hint for doing it the easy way.
Does $X$ take on value $1$ with nonzero probability? Does $Y$ take on
value $400$ with nonzero probability? Do $X$ and $Y$ take on values
$1$ and $400$ simultaneously with nonzero probability? (It does
not matter in the least what the exact values of the probabilities are).
If the answers to the first
two questions are Yes, and the answer to the third is No, what does that
tell you?
