\problemdata      
        {?} % latex-friendly label for the prob.
        {Probability-Intro} % The topic(s) of the problem content.
        {?} % Source (if known)
        {F01, PS10; F98} % Usage (list ones you are aware of).
        {Christos Kapoutsis, S02} % Last revision info (author, date).

\begin{problem}
Compute the probabilities of the following
three events and determine which one is most likely.  Assume that the
dice are fair, 6-sided, and mutually independent.
\begin{problemparts}
\problempart Rolling at least one 6 when six dice are rolled.

\solution{The probability of rolling a 6 on a particular die is
$1/6$.  The probability of not rolling a 6 on a particular die is
$5/6$, since this is the complementary event.  The probability of
not rolling a 6 on any of six dice is $(5/6)^6$, since the
numbers rolled are mutually independent.  The probability of rolling a 6
on some die is therefore
\[
1 - (\frac{5}{6})^6 = 0.665\ldots.
\]}

\problempart Rolling at least two 6's when twelve dice are rolled.

\solution{The probability of rolling two or more 6's is 1 minus the
probability of rolling zero or one 6's.  The probability of rolling
zero 6's is $(5/6)^{12}$, arguing as before.  The probablity
of rolling exactly one 6 is
\[
12 \cdot \frac{1}{6} \cdot \left(\frac{5}{6}\right)^{11}
\]
since there are 12 ways to choose the die that comes up 6, the probability
of this die coming up 6 is $\cfrac{1}{6}$, and the probability of the
remaining dice all coming up in the range 1 to 5 is $(5/6)^{11}$.
Overall, the probability of rolling two or more sixes is therefore
\begin{eqnarray*}
1 - \left(\frac{5}{6}\right)^{12} -
        12 \cdot \frac{1}{6} \cdot \left(\frac{5}{6}\right)^{11}
                & = &   0.618\ldots.
\end{eqnarray*}
}

\problempart Rolling at least three 6's when eighteen dice are rolled.

\solution{
Arguing as before, the probability of rolling at least three 6's is

\begin{eqnarray*}
1       - \left(\frac{5}{6}\right)^{18}
        - 18 \cdot \frac{1}{6} \cdot \left(\frac{5}{6}\right)^{17}
        - \binom{18}{2} \cdot \left(\frac{1}{6}\right)^2 \cdot
                \left(\frac{5}{6}\right)^{16}
        & = &   0.597\ldots.
\end{eqnarray*}
}

\end{problemparts}

\solution{The first event is most likely.}
\end{problem}
