What is the distance to right margin in nested aligned and enumerate environments?

Question

I am using the \aligned environment within an \enumerate setup for formatting my assignment submissions. I have tried using the \align*, \flalign*, and \alignat environments but they didn't behave well in my experiments. I have created a new environ that defaults \item content to math mode (appropriate for my use case). However, when I need to enter longer bits of text, it overflows the right margin.

\text overflowing the right margin in aligned environment:

To address this, I created a \mathbox command that is intended to create a \parbox extending exactly to the right margin from where it is placed. Unfortunately, the calculation seems to be off and the \parbox either overflows or doesn't reach the margin.

\parbox not reaching the right margin:

\parbox not overflowing the right margin:

These are the relevant parts of my code:

\documentclass[10pt]{article}
% ...
\usepackage{geometry}
\geometry{a4paper}

%%% packages - not sure which ones were necessary so I just included them all
\usepackage{graphicx,booktabs,array,paralist,verbatim,subfig,sectsty, makecell,enumitem,titlesec,lmodern,slantsc,listings,setspace,amsmath, environ,showframe,nccmath,xifthen,adjustbox}
\usepackage[fleqn]{mathtools}
% ...
\subsectionfont{\fontsize{11}{13.75}\sffamily\mdseries\scshape}
\renewcommand{\labelenumi}{\textbf{Question \arabic*.}}
\renewcommand{\labelenumii}{(\arabic{enumi}.\arabic{enumii})}
\setlist[enumerate]{align=left, listparindent=2em, parsep=0pt}
% ...
\newenvironment{answergroup}{\begin{enumerate}[parsep=10pt]\begingroup\everymath{\displaystyle}
\setstretch {1.3}}{\endgroup\end{enumerate}}
\newenvironment{answerbody}{\begingroup\everymath{\displaystyle}
\setstretch {1.3}}{\endgroup}
\newcommand{\nextline}{\end{aligned}\\\begin{aligned}[t]&\\&}
\NewEnviron{aligneditem}[1][]{
    \begingroup
    %\allowdisplaybreaks - does nothing in aligned mode
    \setlength{\abovedisplayskip}{0pt}
    \setlength{\belowdisplayskip}{0pt}
    \setlength{\abovedisplayshortskip}{0pt}
    \setlength{\belowdisplayshortskip}{0pt}
    \ifthenelse{\isempty{#1}}
    {\item}
    {\item[#1]}
    $\begin{aligned}[t]
    &\BODY
    \end{aligned}$
    \endgroup
}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% THIS ↓ \dimexpr needs to be fixed somehow %%%%%%%%%%%%%%%
\newcommand{\mathbox}[1]{\boxed{\parbox[t]{\dimexpr\linewidth-\listparindent-\itemindent}{#1}}}%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{document}
\section{Basic Distributions}
\subsection{Binomial Distributions}
\begin{enumerate}
\setcounter{enumi}{3}
\begin{answerbody}
\begin{aligneditem}
\mathbox{
The binomial distribution arises when a random variable is determined to be the number of successes in a series of $n$ independent random trials. In these trials the probability of success is $p$ for each trial, and there are only two possible outcomes: success or failure.
}
\end{aligneditem}
\end{answerbody}
\begin{answerbody}
\begin{aligneditem}
\end{aligneditem}
\end{answerbody}
\begin{answergroup}
\begin{aligneditem}[\unskip]
\mu \geq \frac{5 * 0 + 30 * 1 + 56 * 2 + 15 * 3 + 10 * 4 + 7 * 5}{123} = \frac{262}{123} \\&
\implies \text{Possibly }X \sim \text{Po}(\lambda)\text{ where }\lambda \geq \frac{262}{123}\\&
\mathbox{It appears that sampled stroke patients make at least 2.13 errors on average on this psychometric test.}\\&
\displaystyle\sigma^{2} \geq \sum_{x}{\frac{(x-\mu)^{2}}{N}} \\&
= \frac{1}{123}\left(\begin{multlined}5\left(0 - \frac{262}{123}\right)^{2} + 30\left(1 - \frac{262}{123}\right)^{2} + 56\left(2 - \frac{262}{123}\right)^{2}\\
+ 15\left(3 - \frac{262}{123}\right)^{2} + 10\left(4 - \frac{262}{123}\right)^{2} + 10\left(5 - \frac{262}{123}\right)^{2}\end{multlined}\right)\\&
% ...
\end{aligneditem}
\end{answergroup}
\end{enumerate}
\end{document}

How should I properly calculate the remaining space to the right margin for \parbox?

Answer 1

I solved this problem by changing the width of the \parbox to \dimexpr\textwidth-\leftmargin-\listparindent+\parskip-\labelsep-\listparindent. I'm not quite sure why this works, but it gets reasonably close to the margin without overflowing, for my use cases. The code after modification is:

\documentclass[10pt]{article}
% ...
\usepackage{geometry}
\geometry{a4paper}

%%% packages - not sure which ones were necessary so I just included them all
\usepackage{graphicx,booktabs,array,paralist,verbatim,subfig,sectsty, makecell,enumitem,titlesec,lmodern,slantsc,listings,setspace,amsmath, environ,showframe,nccmath,xifthen}
\usepackage[fleqn]{mathtools}
% ...
\subsectionfont{\fontsize{11}{13.75}\sffamily\mdseries\scshape}
\renewcommand{\labelenumi}{\textbf{Question \arabic*.}}
\renewcommand{\labelenumii}{(\arabic{enumi}.\arabic{enumii})}
\setlist[enumerate]{align=left, listparindent=2em, parsep=0pt}
% ...
\newenvironment{answergroup}{\begin{enumerate}[parsep=10pt]\begingroup\everymath{\displaystyle}
\setstretch {1.3}}{\endgroup\end{enumerate}}
\newenvironment{answerbody}{\begingroup\everymath{\displaystyle}
\setstretch {1.3}}{\endgroup}
\newcommand{\nextline}{\end{aligned}\\\begin{aligned}[t]&\\&}
\newcommand{\textbox}[1]{\boxed{\parbox[t]{\dimexpr\textwidth-\leftmargin-\listparindent+\parskip-\labelsep-\listparindent}{#1}}}
\NewEnviron{aligneditem}[1][]{
    \begingroup
    %\allowdisplaybreaks - does nothing in aligned mode
    \setlength{\abovedisplayskip}{0pt}
    \setlength{\belowdisplayskip}{0pt}
    \setlength{\abovedisplayshortskip}{0pt}
    \setlength{\belowdisplayshortskip}{0pt}
    \ifthenelse{\isempty{#1}}
    {\item}
    {\item[#1]}
    $\begin{aligned}[t]
    &\BODY
    \end{aligned}$
    \endgroup
}
\begin{document}
\section{Basic Distributions}
\subsection{Binomial Distributions}
\begin{enumerate}
\setcounter{enumi}{3}
\begin{answerbody}
\begin{aligneditem}
\textbox{
The binomial distribution arises when a random variable is determined to be the number of successes in a series of $n$ independent random trials. In these trials the probability of success is $p$ for each trial, and there are only two possible outcomes: success or failure.
}
\end{aligneditem}
\end{answerbody}
\begin{answerbody}
\begin{aligneditem}
\end{aligneditem}
\end{answerbody}
\begin{answergroup}
\begin{aligneditem}[\unskip]
\mu \geq \frac{5 * 0 + 30 * 1 + 56 * 2 + 15 * 3 + 10 * 4 + 7 * 5}{123} = \frac{262}{123} \\&
\implies \text{Possibly }X \sim \text{Po}(\lambda)\text{ where }\lambda \geq \frac{262}{123}\\&
\textbox{It appears that sampled stroke patients make at least 2.13 errors on average on this psychometric test.}\\&
\displaystyle\sigma^{2} \geq \sum_{x}{\frac{(x-\mu)^{2}}{N}} \\&
= \frac{1}{123}\left(\begin{multlined}5\left(0 - \frac{262}{123}\right)^{2} + 30\left(1 - \frac{262}{123}\right)^{2} + 56\left(2 - \frac{262}{123}\right)^{2}\\
+ 15\left(3 - \frac{262}{123}\right)^{2} + 10\left(4 - \frac{262}{123}\right)^{2} + 10\left(5 - \frac{262}{123}\right)^{2}\end{multlined}\right)\\&
% ...
\end{aligneditem}
\end{answergroup}
\end{enumerate}
\end{document}

It produces this output: