I have two variables, X and Y. I also have a dice. I roll the dice 10 times. X is defined as the number of times I get a result greater than 3, meaning 4,5,6 and Y is defined as the number of times I get a roll less than 3, meaning 1,2. I noticed that X and Y are Geometricly distributed, with n = 10 and p = 0.5 for X and 0.33 for Y. As a result I know E(X) and E(Y). I have sat and tried but I have no idea how to calculate E(XY). I would very much appreciate it if someone could give a hint or direction as to how I would calculate E(XY). I have thought of trying to sum over all x and y of xyP(x,y) but that seems way too long for this question.
Thank you in advance.
Edit: Thanks to everyone who pointed out they are not geometrically distributed but rather binomial. I meant to write binomial distribution but made a mistake as English is not my first language and the names are different in my language.
Well, unfortunately they are not geometrically distributed.. in fact, since the 10 rolls are indipendents between each other we have: $X\sim B(10,\frac{1}{2})$ and $Y\sim B(10,\frac{1}{3})$ , where $B(n,p)$ is the binomial distribution It would have been nicer if $X,Y$ were indipendents (to istantly conclude that $\mathbb{E}(XY) = \mathbb{E}(X)\mathbb{E}(Y)$), but they are not.
We can write instead $X = \sum_{i=1}^{10}X_i$ and $Y = \sum_{j=1}^{10}Y_j$ where $X_i\sim B(1,\frac{1}{2})$ and $Y_j\sim(1,\frac{1}{3})$ so that: $$ \mathbb{E}(XY) = \mathbb{E}(\Bigr(\sum_{i=1}^{10}X_i\Bigr)\cdot\Bigr(\sum_{j=1}^{10}Y_j\Bigr)) = \mathbb{E}\Bigl(\sum_{i,j=1}^{10}X_iY_j\Bigr) = \sum_{k=1}^{10}\mathbb{E}(X_kY_k) +\sum_{i\not= j}^{10}\mathbb{E}(X_iY_j) $$ $$ \mathbb{E}(XY) = 10\mathbb{E}(X_1Y_1) + 90\mathbb{E}(X_1Y_2) = 10a+90b $$ Since $\mathbb{E}(X_kY_k) = a\in\mathbb{R}$ for all $k\in\{1,\dots,10\}$ and similarly
$\mathbb{E}(X_iY_j) = b\in\mathbb{R}$ for all $(i,j)\in\{1,\dots,10\}^2$ with $i \not= j$.
Now: $$ a = \mathbb{E}(X_1Y_1) = \sum_{x,y=0}^{1}xyf_{(X_1,Y_1)}(x,y) = f_{(X_1,Y_1)}(1,1) = P(X_1=1,Y_1=1) = 0 $$ because it's impossible that the first roll is a number less than 3 and at the same time bigger than 3.
$$ b = \mathbb{E}(X_1Y_2) = \sum_{x,y=0}^{1}xyf_{(X_1,Y_2)}(x,y) = f_{(X_1,Y_1)}(1,1) = P(X_1=1,Y_2=1) = P(X_1 = 1)P(Y_2 = 1) = \frac{1}{2}\cdot\frac{1}{3} = \frac{1}{6} $$
because $X_1$ and $Y_2$ are indipendents! (different roll). Finally we can conclude that:
$$ \mathbb{E}(XY) = 10a+90b = 10\cdot 0 + 90\cdot \frac{1}{6} = 15 $$ Just for fun we can compute the covariance between $X$ and $Y$ as following: $$ Cov(X,Y) = \mathbb{E}(XY)-\mathbb{E}(X)\mathbb{E}(Y) = 15 - (10\cdot\frac{1}{2})(10\cdot\frac{1}{3}) = -\frac{5}{3}. $$