prob of even

\documentclass[11pt]{article}

\usepackage{fullpage}
\usepackage{microtype}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{enumitem}
\usepackage{relsize}
\usepackage{graphicx}

\begin{document}

    Proof:\\
    \\
    \begin{align*}
    (1-2p)^n &= [(1-p) - p]^n\\
    &= [(-p) + (1-p)]^n\\
    &= \sum_{k = 0}^{n} \binom{n}{k} (-p)^k(1-p)^{n-k}\\
    &= \sum_{k = 0}^{n} \binom{n}{k} (-1)^k \cdot p^k(1-p)^{n-k}\\
    &= \sum_{k \text{ is even}} \binom{n}{k} p^k(1-p)^{n-k} +
    \sum_{k \text{ is odd}} \binom{n}{k} (-1)p^k(1-p)^{n-k}\\
    &= \sum_{k \text{ is even}} \binom{n}{k} p^k(1-p)^{n-k} -
    \sum_{k \text{ is odd}} \binom{n}{k} p^k(1-p)^{n-k}\\
    &= \sum_{k \text{ is even}} \mathbb{P}(X = k) -
    \sum_{k \text{ is odd}} \mathbb{P}(X = k)\\
    &= \mathbb{P}(X = even) - \mathbb{P}(X = odd)
    \end{align*}

    Since $X$ MUST be either odd or even:
    \begin{equation*}
    \mathbb{P}(X = even) + \mathbb{P}(X = odd) = 1
    \end{equation*}
    And so, adding these two equations together:
    \begin{align*}
    2\mathbb{P}(X = even) &= 1 + (1-2p)^n\\
    \therefore \mathbb{P}(X = even) &= \frac{1}{2}\Big[1 + (1-2p)^n\Big]
    \end{align*}

\end{document}