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Sorry if my previous question was not so clear. I will delete it later.

Here is a MVE.

\documentclass[12pt,a4paper]{report}

\usepackage{amsmath,amsthm,amssymb}
\usepackage{enumerate}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% Example and Solution

\usepackage{thmtools}

\declaretheoremstyle[
    spaceabove=6pt, spacebelow=6pt,
    headfont=\normalfont\bfseries,
    notefont=\mdseries, notebraces={(}{)},
    bodyfont=\normalfont\itshape,
    postheadspace=1em,
    numberwithin=section
]{exstyle}
%
\declaretheoremstyle[
    spaceabove=6pt, spacebelow=6pt,
    headfont=\normalfont\bfseries,
    notefont=\mdseries, notebraces={(}{)},
    bodyfont=\normalfont,
    postheadspace=1em,
    headpunct={},
    qed=$\blacktriangleleft$,
    numberwithin=section
    %numbered=no
]{solstyle}
\declaretheorem[style=exstyle]{example}
\declaretheorem[style=solstyle]{solution}



\begin{document}
\chapter{ABCDEFGH}
\section{DDDDDDDDDDDDDDDDD}


\begin{example}
    Cars pass a particular point at a rate of 5 cars per minute.
    \begin{enumerate}[(a)]
        \item Find the probability that exactly 4 cars pass the point in a minute.
        \item Find the probability that between at least 3 but fewer than 8 cars pass in a particular minute.
        \item Find the probability that more than 8 cars pass in 2 minutes.
        \item Find the probability that more than 3 cars pass in each of two separate minutes.
    \end{enumerate}


\begin{solution}
    Let $X$ be the number of cars passing in a minute, then $X \sim \text{Po}(5)$.
    \begin{enumerate}[(a)]
        \item
            \begin{equation}
                P(X=4) = \frac{5^4\mathrm{e}^{-5}}{4!} \approx 0.175
            \end{equation}
        \item
            \begin{equation}
                P(3 \leq X \leq 8) 
                = 
                P(X \leq 7) - P(X \leq 2)
                =
                0.8666 - 0.1247
                =
                0.7419
            \end{equation}
        \item
        Poisson distribution assumes a constant rate of occurrence. So in two minutes' time, the rate would be 10. Let $Y$ be the number of cars passing in two minutes. Then $Y \sim \text{Po}(10)$, and
            \begin{equation}
                P(Y>8) = 1 - P(Y \leq 8) = 1 - 0.3328 = 0.6672
            \end{equation}
        \item
            For each one separate minute, we have
            \begin{equation}
                P(X>3) = 1 - P(X \leq 3) = 1 - 0.2650 = 0.7350.
            \end{equation}  \end{enumerate}
            So the probability that more than 3 cars pass in each of two separate minutes is \begin{equation}
                0.7350^2=0.540
            \end{equation}
            to 3.s.f.
\end{solution}
\end{example}




\section{Mean and variance of the Poisson distribution}
We do the same trick as we did in Section~\ref{sec:binom-proof}.
\begin{align}
    \operatorname{E}(X)
    &= \sum_{x=0}^\infty x P(X=x) \\
    &= \sum_{x=1}^\infty x \frac{\mu^x\mathrm{e}^{-\mu}}{x!} \\
    &= \sum_{x=1}^\infty \frac{\mu^x\mathrm{e}^{-\mu}}{(x-1)!} \\
    &= \mu \sum_{x=1}^\infty \frac{\mu^{x-1}\mathrm{e}^{-\mu}}{(x-1)!} \\
    &= \mu \sum_{y=0}^\infty \frac{\mu^y\mathrm{e}^{-\mu}}{y!} \\
    &= \mu.
\end{align}
The expectation of a Poisson distribution is the parameter $\mu$. \\

It can be shown that
\begin{equation}
    \operatorname{E}(X(X-1)) = \mu^2,
\end{equation}
it follows that
\begin{equation}
    \operatorname{Var}(X) 
    = \operatorname{E}[X(X-1)]+\operatorname{E}(X)-\operatorname{E}(X)^2
    = \mu^2 + \mu - \mu^2
    = \mu.
\end{equation}
For a Poisson distribution, the mean equals the variance.


\begin{example}
In producing rolls of cloth there are on average 4 flaws in every 10 metres of cloth.
    \begin{enumerate}[(a)]
        \item Find the mean number of flaws in a 30 metre length.
        \item Find the probability of fewer than 3 flaws in a 6 metre length.
        \item Find the variance of the number of flaws in a 15 metre length.
    \end{enumerate}


    \begin{solution}
    Assuming a Poisson distribution -- flaws in the cloth occur singly, independently, uniformly and randomly.
    \begin{enumerate}[(a)]
        \item If the mean number of flaws in 10 metres is 4, then the mean number of flaws in 30 metre lengths is $3\times4 = 12$.
        \item If there are 4 flaws on average in a 10 metre length there will be $\frac{6}{10}\times4=2.4$ flaws on average in a 6 metre length. If $X$ is the number of flaws in a 6 metre length then $X \sim \text{Po}(2.4)$.
            \begin{align}
                 & P(X<3) \\
                =& P(X=0) + P(X=1) + P(X=2) \\
                =& \mathrm{e}^{-2.4} + 2.4 \times \mathrm{e}^{-2.4} + \frac{2.4^2\times\mathrm{e}^{-2.4}}{2!} \\
                =& (1+2.4+2.4\times1.2)\mathrm{e}^{-2.4} \\
                \approx& 0.570
            \end{align}
        \item If the mean number of flaws in 10 metre lengths is 4, then the mean number of flaws in 15 metre lengths will be
        \begin{equation}
            \mu = \frac{15}{10}\times4=6.
        \end{equation}
        Since, in a Poison distribution, the variance equals the mean the variance is 6.
    \end{enumerate}
    \end{solution}
\end{example}




\end{document}

This compiles fine and I want this to be the teachers' version. What I want for students' version, look like this:

enter image description here enter image description here enter image description here

Overall view

enter image description here

  1. basically, just hide the content that is in solution.
  2. but keep all the equation counters/labels, for example, first equation in Section 1.2 is (1.6), etc...
  3. If possible, I want to keep the current exstyle and solstyle and I can customise them later. These are from here.
  • It's usually preferable to edit the existing question (as long as it doesn't invalidate existing answers). I think it would be best if you put this material on the original question. – Teepeemm May 30 '18 at 13:35
  • You could use the exam class for that. – dexteritas May 30 '18 at 13:51

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