Expected value of balls problem The problem goes as follows:

Let there be an urn with 5 white balls and 5 black balls. First, we draw 4 balls (with return). We define the random variable $Y$ to be the number of white balls we drew in the first step.
  Secondly, from the balls we drew in the first step, we draw 3 balls (without returning). We define the random variable $X$ to be the number of white balls in the second step. What is $\mathbb E[X]$?

Now I know there is a really simple solution, that we can define a new random variable named $Z$ to be the number of black balls in the second draw, and then we get $X+Z=3$ and from symmetry $\mathbb E[X]= \mathbb E[Z] = \frac{3}{2}$.
What I want to know is where one of the ways I tried to solve is incorrect. Where was my mistake here?
By using the law of total expectation, Let us calculate $\mathbb E[X] = \mathbb E[\mathbb E[X|Y]]$. We also note that $Y\sim Bin(4,\frac{1}{2})$.
Then we get: $$\mathbb E[X] = \sum_{i=0}^4\mathbb E[X|Y=i]P(Y=i) = (1*\frac{1}{4})(4*(\frac{1}{2})^4) + (1*\frac{1}{2} + 2*\frac{1}{2})(\binom{4}{2}*(\frac{1}{2})^4) + (2*\frac{3}{4}+3*\frac{1}{4})(4*(\frac{1}{2})^4) + (3*1)(\frac{1}{2})^4 = \frac{22}{16}$$
I don't think I've made any calculation mistakes, therefore something is probably wrong in the way itself. Thanks!
 A: You are doing too much busy work and confusing yourself. 
You have identified the distribution of $Y$ as being Binomial, so have its probability mass function.   Then you have attempted to calculate the conditional expectation of $X$ given $Y$ by hand, which isn't wrong per se, when you could simply identify its distribution and use the known formula.
We have $Y\sim\mathcal{Bin}(4,1/2)$, and  $X\mid Y\sim \mathcal{Hypergeometric}(4,Y,3)$.  
So $\mathsf E(Y)= 4\cdot \tfrac 12$ and $\mathsf E(X\mid Y)=3\cdot Y / 4$.
$$\mathsf E(X){=\mathsf E(\mathsf E(X\mid Y)) \\ = \mathsf E(3Y/4) \\ = \tfrac 34\cdot\tfrac 42 \\ = \tfrac 32}$$
Or if you must,
$$\mathsf E(X)~{=~ \sum_{k=0}^4\frac {3\cdot k}4\cdot\binom 4k \frac 1{2^4} \\ = (\tfrac 34\cdot 4+\tfrac 64\cdot \tbinom 42+\tfrac {9}4\cdot\tbinom 43+\tfrac {12}4\cdot\tbinom 44)\times \tfrac 1{16} \\ = \tfrac 32}$$


Thanks, can you please explain the $E[X|Y]=3∗Y/4$ part? – Mickey 

$X$ is the count of favoured items among a sample of size $3$ selected from a population of size $4$ containing $Y$ favoured items.   This is a Hypergeometric distributed random variable, whose (conditional) probability mass function, over the sigma-algebra of $Y$, is as follows.
$$\mathsf P(X=k\mid Y) ~=~\dfrac{\dbinom{Y}{k}\dbinom{4-Y}{3-k}}{\dbinom{4}{3}}$$
The conditional expectation is:$$\begin{align}\mathsf E(X\mid Y)~&=~\sum_{k=0}^{3}\dfrac{k\dbinom{Y}{k}\dbinom{4-Y}{3-k}}{\dbinom{4}{3}}\\[1ex] &=~\dfrac{3Y}{4} \end{align}$$
The derivation of the general closed form of the mean for a hypergeometric distribution is a bit involved, and can be found here.   For this we can do it long hand.
$$\begin{align}\mathsf E(X\mid Y)~&=~\tfrac{\sum_{k=0}^{3}k\tbinom{Y}{k}\tbinom{4-Y}{3-k}}{\tbinom{4}{3}}\\[1ex] &=~\tfrac 1{4}\left({{(0)\mathbf 1_{Y=0}} + {(1\tbinom{1}{1}\tbinom{3}{2})\mathbf 1_{Y=1}} + {(1\tbinom{2}{1}\tbinom{2}{2}+2\tbinom{2}{2}\tbinom{2}{1})\mathbf 1_{Y=2}} + {(2\tbinom{3}{2}\tbinom{1}{1}+3\tbinom{3}{3}\tbinom{1}{0})\mathbf 1_{Y=3}}}+{(3\tbinom{4}{3}\tbinom{0}{0})\mathbf 1_{Y=4}}\right)\\[1ex] &=~\tfrac 1{4}\left({{(0)\mathbf 1_{Y=0}} + {(3)\mathbf 1_{Y=1}} + {(6)\mathbf 1_{Y=2}} + {(9)\mathbf 1_{Y=3}}+{(12)\mathbf 1_{Y=4}}}\right)\\[1ex] &= \tfrac 34Y~~\mathbf 1_{Y\in\{0,1,2,3,4\}} \end{align}$$
Which is a lot of work to explain why you don't have to do a lot of work. $\ddot\smile$
A: I didn't check it all but I think that when you compute $E[X | Y = 1] P[Y=1] = \frac{3}{4}( 3*\frac{1}{2^4})$ you have $\frac{3}{4}$ and not $\frac{1}{4}$. Because, picking 3 balls, is the same as not picking a ball, you have $1$ chance over $4$ to not pick the ball, so $3$ over $4$ to pick it.
