<s>[INST] probability need help on correlation problem A deck of 52 cards is shuﬄed you are dealt 13 cards. Let $X$ and $Y$ denote, respectively, the number of aces and the number of spades in your hand. Show that $X$ and $Y$ are uncorrelated.
I try to write $X=X_1+X_2+\cdots+X_{13}$, such that $X_1\ldots X_{13}$ is the indicator of the $i^{th}$ card. But this doesn't work. please help. 
 [/INST] It will suffice to show that $E(XY)=E(X)E(Y)$. The expected number of Aces is $13\cdot \frac{4}{52}$, and the expected number of spades is $13\cdot\frac{13}{52}$. Thus $E(X)E(Y)=\frac{13}{4}$.
We now compute $E(XY)$.  Note that $X$ and $Y$ are not independent. 
For the calculation, we condition on $X$. The possibilities are $X=0,1,2,3,4$. For each of these values of $X$, we calculate $E(XY|X=x)$. 
It is clear that $E(XY|X=0)=0$.
Next, we find $E(XY|X=1)$. Given that $X=1$, the probability that the lone Ace is the Ace of spades is $\frac{1}{4}$, and the probability it is another Ace is $\frac{3}{4}$. 
If the lone Ace is the Ace of spades, we are choosing $12$ more cards from the $48$ non-Aces, and the expected number of spades is $1+12\cdot \frac{12}{48}$. If the Ace is a non-spade, the same reasoning shows that the expected number of spades is $12\cdot\frac{12}{48}$. Thus 
$$E(XY|X=1)=\frac{1}{4}\cdot 4+\frac{3}{4}\cdot 3=\frac{13}{4}.$$ 
Next we deal with $X=2$. If there are $2$ Aces, the probability that one of them  is the Ace of spades is $\frac{1}{2}$. The same reasoning as in the previous case shows that 
$$E(XY|X=2)=2\cdot\frac{13}{4}=\frac{26}{4}.$$
Next comes $X=3$. If there are $3$ Aces, the probability they include the Ace of spades is $\frac{3}{4}$. If that is the case, we have $1$ spade for sure, and we are choosing $10$ cards from the $48$ non-spades. We obtain
$$E(XY|X=3)=3\cdot\left(\left(1+\frac{10}{4}\right)\frac{3}{4}+\frac{10}{4}\frac{1}{4}                             \right)=\frac{39}{4}.$$  
Finally, 
$$E(XY|X=4)=4\cdot \left(1+\frac{9}{4}\right)=\frac{52}{4}.$$
The probability of $k$ Aces is $\binom{4}{k}\binom{48}{13-k}/\binom{52}{13}$.
Finally, calculate. We get
$$E(XY)=\frac{13}{4}\cdot\frac{1}{\binom{52}{13}}\left(1\binom{4}{1}\binom{48}{12}+2\binom{4}{2}\binom{48}{11}+3\binom{4}{3}\binom{48}{10}+4\binom{48}{9}  \right).$$
This simplifies to $\dfrac{13}{4}$. 
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