<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\begin{document}
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{\bf Question}
\bigskip

Two fair dice are thrown.  What is the probability of obtaining:
\begin{description}
\item[(a)] A total of exactly 7?
\item[(b)] A total of 7 or more?
\item[(c)] A total of 7 or less?
\end{description}
\vspace{.25in}

{\bf Answer}
\bigskip

\begin{description}
\item[(a)]
 To obtain 7 exactly, one needs the following pairs

\{1,6\}, \{6,1\}, \{2,5\}, \{5,2\}, \{3,4\}, \{4,3\}

each possibility has a probability $\ds \frac{1}{36} =
\left(\frac{1}{6} \right) \times \left(\frac{1}{6} \right)$

There are six possibilities, six elements in the sample space

Hence $\ds p(a_1 + a_2 = 7) = \frac{6}{36} = \frac{1}{6}$

\item[(b)]

\medskip


\begin{eqnarray*} {\rm To\ obtain}\ \  12 &amp;
{\rm need} &amp; \{6,6\}\ \  p = \frac{1}{36}\\  11  &amp;{\rm need} &amp;
\{6,5\}, \{5,6\}\ \   p = \frac{1}{18} \\  10 &amp; {\rm need} &amp;
\{6,4\}, \{4,6\} , \{5,5\}\ \   p = \frac{1}{12} \\  9 &amp; {\rm
need} &amp; \{6,3\}, \{3,6\}, \{4,5\} , \{5,4\}\ \   p = \frac{4}{36}
= \frac{1}{9} \\  8 &amp; {\rm need} &amp; \{6,2\}, \{2,6\}, \{3,5\},
\{5,3\}, \{4,4\}\ \  p = \frac{5}{36}
\end{eqnarray*}

Hence $\ds p(a_1 + a_2 \geq 7) = \frac{1}{36}(1+2+3+4+5+6) =
\frac{21}{36} = \frac{7}{12}$

\item[(c)]

\begin{eqnarray*} p(a_1 + a_2 \leq 7) &amp; = &amp;  1 - p(a_1 + a_2 &gt; 7) \\ &amp;  = &amp; 1 - p(a_1 +
a_2 \geq 7) + p(a_1 + a_2 = 7) \\ &amp; = &amp; 1 - \frac{7}{12} +
\frac{1}{6} \\ &amp; = &amp; \frac{7}{12} \end{eqnarray*}


\end{description}
\end{document}
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