\documentclass[12pt]{article}
\usepackage{graphicx}
\usepackage{amsmath}

\begin{document}


\begin{figure}
\begin{center}
\includegraphics[width=.4\linewidth]{Poissprobhist.png}
\end{center}
\caption{The probability histogram}
\end{figure}

\noindent
The plot above shows the probability histogram for a \emph{Poisson} random variable $Y$
with $\lambda = 2$.
The heights of the bars are 
\begin{align*}
P(X = 0) = \frac{2^0 e^{-2}}{0!} \doteq .1353 \quad
&P(X = 1) = \frac{2^1 e^{-2}}{1!} \doteq .2707\\
P(X = 2) = \frac{2^2 e^{-2}}{2!} \doteq .2707 \quad
&P(X = 3) = \frac{2^3 e^{-2}}{3!} \doteq .1804\\
P(X = 4) = \frac{2^4 e^{-2}}{4!} \doteq .0902 \quad
&P(X = 5) = \frac{2^5 e^{-2}}{5!} \doteq .0361\\
P(X = 6) = \frac{2^6 e^{-2}}{6!} \doteq .0120 \quad
&P(X = 7) = \frac{2^7 e^{-2}}{7!} \doteq .0034\\
P(X = 8) = \frac{2^8 e^{-2}}{8!}\doteq .0009 \quad
&P(X = 9) = \frac{2^9 e^{-2}}{9!} \doteq .0002\\
P(X = 10) = \frac{2^{10} e^{-2}}{10!} \doteq .00004 \quad & 
\end{align*}
In this case, unlike the binomial probability histogram, the total area
of these boxes is not exactly one.  In fact the sum of the areas of these 11
boxes is $.9999916916$, slightly less than $1$.  The reason is that 
$$P(Y = k) = \frac{2^k e^{-2}}{k!} > 0$$
for all $k \ge 0$.
\end{document}

