What is the distribution of the maximum of independent non identical Binomial variables? If   $Y_{1},Y_{2}, \cdots, Y_{n}$ are independent Binomial random variable with sample size $m$ and different probabilities $p_{i}$ $(i=1,\cdots ,n)$, Could the distribution of $X=\max (Y_{1},Y_{2}, \cdots, Y_{n})$ be an extreme value distribution?
 A: As whuber correctly points out in the comments, the random variable $X$ is discrete with support on the same space as the original random variables.  Hence, the maximum possible value of $X$ is $m$, and it does not make sense to use a normal approximation (or any other approximation) that would allow a larger maximum than this.
The distribution function for $X$ can be obtained by standard methods, but it is an ugly distribution.  For all $x = 0, 1, 2, ..., m$ you have:
$$\begin{equation} \begin{aligned}
F_X(x) \equiv \mathbb{P}(X \leqslant x) 
&= \prod_{i=1}^n \mathbb{P}(Y_i \leqslant x) \\[6pt]
&= \prod_{i=1}^n \sum_{r=0}^x \text{Bin}( r |m, p_i) \\[6pt]
&= \prod_{i=1}^n \sum_{r=0}^x {m \choose r} p_i^r (1-p_i)^{m-r}. \\[6pt]
\end{aligned} \end{equation}$$
Particular values of this distribution function are given by:
$$\begin{equation} \begin{aligned}
F_X(0) &= \prod_{i=1}^n (1-p_i)^m, \\[6pt]
F_X(1) &= \prod_{i=1}^n [ (1-p_i)^m + m p_i (1-p_i)^{m-1}], \\[6pt]
F_X(2) &= \prod_{i=1}^n [ (1-p_i)^m + m p_i (1-p_i)^{m-1} + \tfrac{m(m-1)}{2} p_i^2 (1-p_i)^{m-2}], \\[6pt]
&\text{ } \text{ } \vdots \\[6pt]
F_X(m-1) &= \prod_{i=1}^n [ (1-p_i)^m + m p_i (1-p_i)^{m-1} + \cdots + m p_i^{m-1}(1-p_i) ], \\[6pt]
F_X(m) &= 1. \\[6pt]
\end{aligned} \end{equation}$$
This cannot be simplified any further for the general case.  Even in the case of equal probabilities where $p_1 = \cdots = p_n$ the simplification reduces the product to a simple power, but this still gives a distribution with quite a complicated form.  So long as $m$ is not too large, it should be possible to obtain the probability mass function analytically, which would give all the moments, etc.
