Droping 2 times fair dice, Let $X$ be the number of drops that got even number, Let $Y$ be the number of drops got a number that is divided by $3$ what is true from the following?
A. $\mathbb{E}[X^2Y]= 1$ and $X$ , $Y$ are independant
B.$\mathbb{E}[X^2Y]= \frac{2}{3}$ and $X$ , $Y$ are independant
C.$\mathbb{E}[X^2Y]= \frac{2}{3}$ and $X$ , $Y$ are Not independant
D.$\mathbb{E}[X^2Y]= 1$ and $X$ , $Y$ are Not independant
How should I approach this question? Do I really need to calculate the probability function or there is an efficient way to approach it?
To calculate the expectation is is easy to find out the distribution but it is very immediate...
First observe that
$$\mathbb{P}[\text{Even}]=\frac{1}{2}$$
$$\mathbb{P}[\text{Div 3}]=\frac{12}{36}=\frac{1}{3}$$
and
$$\mathbb{P}[\text{Div 3}|\text{Even}]=\frac{6}{18}=\frac{1}{3}$$
Thus being
$$\mathbb{P}[\text{Div 3}|\text{Even}]=\mathbb{P}[\text{Div 3}]$$
These events are independent.
Given this property also $X\perp \!\!\!\!\!\!\perp Y$ and $X^2\perp \!\!\!\!\!\!\perp Y$
Thus
$$\mathbb{E}[X^2Y]= \mathbb{E}[X^2]\cdot\mathbb{E}[Y]=\Bigg(\frac{2}{4}+1 \Bigg)\Bigg(\frac{4}{9}+\frac{2}{9}\Bigg)=1$$