
\problemdata     
{S01-PS9-6}           
{probability-conditional}              
{} 		
{S01-PS9}
{S01 Karger}


\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{
\medskip
\textbf{Solution:}
\medskip
The probability of rolling a 6 on a particular die is
$\frac{1}{6}$.  The probability of not rolling a 6 on a particular die
is $\frac{5}{6}$, since this is the complementary event.  The
probability of not rolling a 6 on any of six dice is
$(\frac{5}{6})^6$, 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{
\medskip
\textbf{Solution:}
\medskip
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 $(\frac{5}{6})^{12}$, 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 $\frac{1}{6}$, and the
probability of the remaining dice all coming up in the range 1 to 5 is
$(\frac{5}{6})^{11}$.

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{
\medskip
\textbf{Solution:}
\medskip
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{
\medskip
\textbf{Solution:}
\medskip
The first event is most likely.

}

\end{problem}