\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}{D}}
\chead{\raisebox{0.5cm}{\Huge Conditional Probablities}} % (add title text if you want)
\rhead{\resizebox{!}{2cm}{D}}
\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}}} 
&$H = 60$\\
&$U = 100$\\
&$P(M \cap H) = 0.10$\\
&$P(H) = 60/100 = 0.60$\\
&$P(M|H) = ?$\\
&\\
\end{tabular}


% -------- Abstract -------------------
\begin{tabular}{m{1.2in} m{4in}}
\multirow{5}{*}{\fbox{\includegraphics[width=1in]{images/abstract.png}}} 
&$H = 60$\\
&$U = 100$\\
&$P(M \cap H) = 0.10$\\
&$P(H) = 60/100 = 0.60$\\
&$P(M|H) = P(M \cap H) /P(H) = 0.10 / 0.60 = 1/6 \approx 0.17$\\
&\\
\end{tabular}

% -------- Question -------------------
\begin{tabular}{m{1.2in} m{4in}}
\multirow{5}{*}{\fbox{\includegraphics[width=1in]{images/question.png}}} 
&$P(T|F) = 0.40$\\
&$P(F) = 0.01$\\
&$P(T) = 0.20$\\
&$P(T^c | F^c) = ?$\\
\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}}} 
&$P(T|F) = 0.40$\\
&$P(F) = 0.01$\\
&$P(T) = 0.20$\\
&$P(T^c | F^c) = ?$\\
&\hrule\\
&\\
&$P(F^c) = 1 - P(F) = 1 - 0.01 = 0.99$\\
&$P(T^c) = 1 - P(T) = 1 - 0.20 = 0.80$\\
&\hrule\\
&\\
&$P(T \cap F) = P(T|F) \times P(F) = 0.40 \times 0.01 = 0.004$\\
&\hrule\\
&\\
&$P(T \cap F^c) = P(T) - P(T \cap F) = 0.20 - 0.004 = 0.196$\\
&\hrule\\
&\\
&$P(T^c \cap F^c) = P(F^c) - P(T \cap F^c) = 0.99 - 0.196 = 0.794$\\
&\hrule\\
&\\
&$P(T^c | F^c) = \frac{P(T^c \cap F^c)}{P(F^c)} = \frac{0.794}{0.99} \approx 0.80$\\
\end{tabular}

\end{table}


\end{document}