\documentclass[12pt]{article}
\usepackage{amsmath, amssymb, graphicx, array, fancyhdr, unicode-math, multirow}

\DeclareMathOperator{\Var}{Var}
\DeclareMathOperator{\sd}{SD}
\DeclareMathOperator{\Pois}{Pois}
\DeclareMathOperator{\Bin}{Bin}
\DeclareMathOperator{\NegBin}{NegBin}
\DeclareMathOperator{\sech}{sech}


% Text: Calibri, all bold
\setmainfont{Calibri}
\renewcommand{\seriesdefault}{b}

% Math: choose a math font and visually embolden it
\setmathfont{TeX Gyre Termes Math}                  % base math font
\setmathfont[FakeBold=1.9]{TeX Gyre Termes Math}    % embolden; adjust 1.6–2.4

% header
\pagestyle{fancy}
\fancyhf{}
\lhead{\resizebox{!}{2cm}{L}}
\chead{\raisebox{0.5cm}{\Huge Independant Events}} % (add title text if you want)
\rhead{\resizebox{!}{2cm}{L}}
\setlength{\headheight}{2.5cm}
\setlength{\headsep}{12pt} % space between header and body

\begin{document}
\thispagestyle{fancy}

% -------- PAGE 1 --------
\begin{table}[h!]
  \renewcommand{\arraystretch}{1.2}

% -------- Single -------------------
\begin{tabular}{m{1.2in} m{4in}}
\multirow{5}{*}{\fbox{\includegraphics[width=1in]{images/single.png}}} 
&$S = {HH, HT, TH, TT} \rightarrow |S| = 4$\\
&$A = {HH, HT}$\\
&$B = {HH, TH}$\\
&$C = {HH, TT}$\\
&\\
\end{tabular}

% -------- Abstract -------------------
\begin{tabular}{m{1.2in} m{4in}}
\multirow{5}{*}{\fbox{\includegraphics[width=1in]{images/abstract.png}}} 
&$A \cap B \cap C = {HH} \rightarrow P(A \cap B \cap C) = \frac14$\\
&$P(A) = \frac{2}{4} = \frac{1}{2}$\\
&$P(B) = \frac{2}{4} = \frac{1}{2}$\\
&$P(C) = \frac{2}{4} = \frac{1}{2}$\\
&$P(A)P(B)P(C) = \frac12 \cdot \frac12 \cdot \frac12 = \frac18$\\
&$P(A \cap B \cap C) \neq P(A)P(B)P(C) \rightarrow  \raisebox{-1.0ex}{\includegraphics[width=0.25in]{images/MutuallyIndependant.png}}(A,B,C) = \mathrel{\raisebox{-0.2\height}{\includegraphics[width=0.25in]{images/thumb_down.png}}}$\\
&\\
\end{tabular}

% -------- Question -------------------
\begin{tabular}{m{1.2in} m{4in}}
\multirow{5}{*}{\fbox{\includegraphics[width=1in]{images/question.png}}} 
&$S = {(i,j) : i,j \subset {1,2,3,4,5,6}} \rightarrow |S| = 36 $\\
&$A = {(1,6), (2,5),(3,4),(4,3),(5,2),6,1)} \rightarrow |A| =6 $\\
&$B = {(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)} \rightarrow |B| =6 $\\
&$C = {(1,4),(2,4),(3,4),(4,4),(5,4),(6,4)} \rightarrow |C| =6 $\\
&$\raisebox{-1.0ex}{\includegraphics[width=0.25in]{images/MutuallyIndependant.png}}(A,B,C) = ?$\\
\end{tabular}
\end{table}




% second page
\newpage
\thispagestyle{fancy}
\begin{table}[h!]
% -------- Answer -------------------
\begin{tabular}{m{1.2in} m{4in}}
\multirow{5}{*}{\fbox{\includegraphics[width=1in]{images/key.png}}} 
&$S = {(i,j) : i,j \subset {1,2,3,4,5,6}} \rightarrow |S| = 36 $\\
&$A = {(1,6), (2,5),(3,4),(4,3),(5,2),6,1)} \rightarrow |A| =6 $\\
&$B = {(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)} \rightarrow |B| =6 $\\
&$C = {(1,4),(2,4),(3,4),(4,4),(5,4),(6,4)} \rightarrow |C| =6 $\\
&$\raisebox{-1.0ex}{\includegraphics[width=0.25in]{images/MutuallyIndependant.png}}(A,B,C) = ?$\\
&\hrule\\
&\\
&$P(A \cap B \cap C) = {(3,4)} \rightarrow P(A \cap B \cap C) = \frac {1}{36}$\\
&\hrule\\
&\\
&$P(A) \times P(B) \times P(C) = \frac {1}{6} \times \frac {1}{6} \times \frac {1}{6} = \frac {1}{216}$\\
&\hrule\\
&\\
&$ \frac {1}{36} \not = \frac {1}{216} \rightarrow \raisebox{-1.0ex}{\includegraphics[width=0.25in]{images/MutuallyIndependant.png}}(A,B,C) = \mathrel{\raisebox{-0.2\height}{\includegraphics[width=0.25in]{images/thumb_down.png}}}$\\
\end{tabular}
\end{table}
 

\end{document}