Maximum and minimum Expected values when taking colored balls We have a sack with $60$ balls.
From them $15$ balls are red, $15$ green, $15$ blue and $15$ yellow.
We take $30$ balls from the sack.
What's the expected number of balls of the color from which the most balls had been taken? And from the color from which the least balls had been taken?
Expressed in the notation I begun to solve this unsuccesfully:
Let $X_i$ be a random event for the number of balls taken of the color $i$.
I look for: $E[\max(X_1,X_2,X_3,X_4)]$ and $E[\min(X_1,X_2,X_3,X_4)]$
I got that $P(X_i=x)=\frac{\binom{15}{x}\binom{45}{30-x}}{\binom{60}{30}}$
 A: The joint distribution of $\boldsymbol X = (X_1, X_2, X_3, X_4)$ that counts the number of balls drawn of each color, is multivariate hypergeometric:  $$\Pr[\boldsymbol X = (x_1, x_2, x_3, x_4)] = \binom{60}{30}^{-1} \prod_{k=1}^4 \binom{15}{x_k}, \quad x_1 + x_2 + x_3 + x_4 = 30.$$  Thus the desired expectation is simply $$\operatorname{E}[\max \boldsymbol X] =  \sum_{\boldsymbol x \in M} \max(\boldsymbol x) \Pr[\boldsymbol X = \boldsymbol x],$$ where $M$ is the set of all possible outcomes of $\boldsymbol X$.  While this is a tedious sum to compute by hand, it is computable using Mathematica:
$$\operatorname{E}[\max \boldsymbol X] = \frac{280571657719508835}{29566145391215356} \approx 9.48963.$$  Similarly,  $$\operatorname{E}[\min \boldsymbol X] = \frac{162920523148721505}{29566145391215356} \approx 5.51037.$$

By request, Mathematica code:
Explicit computation of the sum (I make no claims that it's the most elegant or efficient approach):
Total[ Max[#] (Times @@ Binomial[15, #]) & /@ Select[Append[#, 30 - Total[#]] & /@ Tuples[Range[16] - 1, 3], 0 <= Last[#] <= 15 &]] / Binomial[60, 30]

Using the built-in probability distribution:
Expectation[ Max[x1, x2, x3, x4], {x1, x2, x3, x4} \[Distributed] MultivariateHypergeometricDistribution[30, {15, 15, 15, 15}]]

And of course, altering either code to compute the expectation of the minimum is straightforward.

Minor modification and additions can easily generate a plot of the expectation as a function of the number of balls drawn.  Depicted below is the expectation of the maximum.  It turns out that the explicit calculation (first version) is quite a bit faster than using the built-in distribution for the range of parameters involved.

A: Comment. This plot shows maximum and minimum values from a million runs
of this experiment. Points are randomly 'jittered' $\pm 0.3$
to prevent massive 'overplotting'. A few very rare, but possible
combinations of values at upper-left of the plot did not occur in this particular
simulation. From computations, the respective expected values seem to be about 5.51 and 9.49; the modes are 6 and 9, medians 6 and 9. 
The four $X_i$ of this problem are correlated, so some traditional approaches towards an analytic solution are not available for deriving distributions of the maximum and minimum. Maybe this plot
will suggest possible methods of solution.

