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+%This command provides the text of the last column (Probability)
+%on page 3
+%
+%The macro has one parameter:
+% 1) The width of the text
+%
+\newcommand\TThreeProb[1]{%
+ \parbox[t]{#1}{%
+ \DisplaySpace{\TThreeDisplaySpace}{\TThreeDisplayShortSpace}
+
+
+ %Formula 9
+ \TThreeTitle{Normal (Gaussian) distribution:}
+ \begin{DisplayFormulae}{1}{0pt}{2ex plus 1ex minus 1ex}{\BigChar}{\StyleWithoutNumber}
+ \Fm{p(x) = \frac{1}{\sqrt{2 \pi} \sigma} e^{-(x-\mu)^2/2\sigma^2}},
+ \Fm{\E[X] = \mu}
+ \end{DisplayFormulae}
+
+ %Formula 10
+ \TThreeTitle{Continuous distributions:}%
+ If $\Pr[a<X<b] = \int_{a}^b p(x)\dx$,
+ then $p$ is the probability density function of $X$.
+
+ If $\Pr[X<a] = P(a)$,
+ then $P$ is the distribution function of $X$.
+
+ If $P$ and $p$ both exist then
+ $P(a) = \int_{-\infty}^a p(x)\dx$.
+
+ %Formula 11
+ \TThreeTitle{Expectation:}
+ If $X$ is discrete
+ $\E[g(X)] = \sum_x g(x) \Pr[X=x]$.
+
+ \begin{DisplayFormulae}{1}{0pt}{3ex plus 1ex minus 1ex}{\SmallChar}{\StyleWithoutNumber}
+ \unskip
+ If $X$ continuous then
+ \def\FirstPart{\E[g(X)]\mbox{}}
+ \FmPartA{\FirstPart = \int_{-\infty}^{\infty} g(x) p(x)\dx}
+ \FmPartB{\FirstPart}{= \int_{-\infty}^{\infty} g(x) \, d P(x)}.
+ \end{DisplayFormulae}
+
+ %Formula 12
+ \TThreeTitle{Variance, standard deviation:}
+ \begin{DisplayFormulae}{1}{0pt}{3ex plus 1ex minus 1ex}{\BigChar}{\StyleWithoutNumber}
+ \Fm{\Var[X] = \E[X^2] - \E[X]^2},
+ \Fm{\sigma = \sqrt{\Var[X]}}
+ \end{DisplayFormulae}
+
+ %Formula 13
+ \TThreeTitle{For events $A$ and $B$:}%
+ \begin{DisplayFormulae}{1}{0pt}{3ex plus 1ex minus 1ex}{\BigChar}{\StyleWithoutNumber}
+ \Fm{\Pr[A \Or B] = \Pr[A] + \Pr[B] - \Pr[A \And B]}
+ \FmPartA{\MathRemark[\relax]{\text{iff $A$ and $B$ are independent:}}}
+ %Small initial space to show that the remark is only for
+ %this equation
+ \FmPartB{xxxx}{\Pr[A \And B] =\Pr[A] \cdot \Pr[B]}
+ \Fm{\Pr[A \vert B] = \frac{\Pr[A \And B]}{\Pr[B]}}
+ \end{DisplayFormulae}%
+
+ %Formula 14
+ \TThreeTitle{For random variables $X$ and $Y$:}%
+ \begin{DisplayFormulae}{1}{0pt}{3ex plus 1ex minus 1ex}{\BigChar}{\StyleWithoutNumber}
+ \FmPartA{\MathRemark[\relax]{\text{if $X$ and $Y$ are independent:}}}
+ %Small initial space to show that the remark is only for
+ %this equation
+ \FmPartB{xxxx}{\E[X \cdot Y] = \E[X] \cdot \E[Y]}
+ \Fm{\E[X + Y] = \E[X] + \E[Y]}
+ \Fm{\E[c X] = c \E[X]}
+ \end{DisplayFormulae}
+
+ %Formula 15
+ \TThreeTitle{Bayes' theorem:}%
+ \begin{DisplayFormulae}{1}{0pt}{3ex plus 1ex minus 1ex}{\BigChar}{\StyleWithoutNumber}
+ \Fm{\Pr[A_i\vert B] =
+ \frac{\Pr[B\vert A_i] \Pr[A_i]}{\sum_{j=1}^n \Pr[A_j] \Pr[B\vert A_j]}}
+ \end{DisplayFormulae}
+
+ %Formula 16
+ \TThreeTitle{Inclusion-exclusion:}
+ \begin{DisplayFormulae}{1}{0pt}{3ex plus 1ex minus 1ex}{\BigChar}{\StyleWithoutNumber}
+ \raggedright
+ \def\FirstPart{\Pr\Big[\bigvee^n_{i=1} X_i \Big] = \mbox{}}
+ \FmPartA{\FirstPart \sum^n_{i=1} \Pr[X_i] +}
+ \FmPartB{\FirstPart}{\sum_{k=2}^n (-1)^{k+1} \sum_{\smash{i_i<\cdots <i_k}}
+ \Pr\Big[\bigwedge^k_{j=1} X_{i_j}\Big]}
+ \end{DisplayFormulae}
+
+ %Formula 17
+ \TThreeTitle{Moment inequalities:}
+ \begin{DisplayFormulae}{1}{0pt}{3ex plus 1ex minus 1ex}{\BigChar}{\StyleWithoutNumber}
+ \Fm{\Pr\big[\vert X\vert \geq \lambda \E[X]\big] \leq \frac{1}{\lambda}},
+ \Fm{\Pr\Big[\big\vert X - \E[X]\big\vert \geq \lambda \cdot \sigma \Big]
+ \leq \frac{1}{\lambda^2}}
+ \end{DisplayFormulae}
+
+ %Formula 18
+ \TThreeTitle{Geometric distribution:}%
+ \begin{DisplayFormulae}{1}{0pt}{3ex plus 1ex minus 1ex}{\BigChar}{\StyleWithoutNumber}
+ \Fm{\Pr[X = k] = pq^{k-1}\MathRemark{q = 1-p}},
+ \Fm{\E[X] = \sum^\infty_{k=1} kpq^{k-1} = \frac{1}{p}}
+ \end{DisplayFormulae}
+
+ \AdjustSpace{2ex plus 1ex minus .5ex}
+ \noindent
+ The ``coupon collector'':
+ We are given a random coupon each day,
+ and there are $n$ different types of coupons.
+ The distribution of coupons is uniform.
+
+ The expected number of days to pass before we to collect all $n$ types is
+ $n=H_n$.
+ }%
+}