\problemdata     
{S01-PS9-1}           
{probablility-intro}              
{} 		
{S01-PS9}
{S01 Karger}



\begin{problem}



Suppose that you roll three six-sided dice.

\begin{problemparts}
\problempart What is the probability that the sum of the first two rolls is 7?
\solution{%
  \textbf{Solution:} There are 36 equally-likely outcomes for the
  first two rolls.  In 6 of these outcomes, the sum of the rolls is 7.
  Therefore, the probability is $\frac{6}{36} = \frac{1}{6}$.}
\problempart What is the probability that the sum of the first two
  rolls is 7 and the sum of the last two rolls is 7?
\solution{%
  \textbf{Solution:} There are $6^3 = 216$ equally-likely outcomes for
  all three rolls.  In 6 of these outcomes, the sum of the first two
  rolls is 7 and the third is equal to the first.  Therefore, the
  probability is $\frac{6}{216} = \frac{1}{36}$.}
\problempart What is the probability that every pair of rolls sums up to 7?
\solution{%
  \textbf{Solution:} The probability is zero.  Assume that there were
  a set of rolls such that every pair summed to 7.  Since the first
  rolls sum to 7, one of these rolls must be odd and the other even.
  As a result, the third roll can always be paired with one of the
  same parity to give an even sum.}
\problempart What is the probability that some pair of rolls sums up to 7?
\solution{%
  \textbf{Solution:} Let $A$ be the event that the first two rolls sum
  up to 7, $B$ be the event that the last two rolls sum to 7, and $C$
  be the event that the first and last rolls sum to 7.  The
  probability that some pair of rolls sums up to 7 is $\Pr(A \cup B
  \cup C)$.  We can evaluate this quantity with inclusion exclusion:
  \begin{eqnarray*}
    \Pr(A \cup B \cup C)    & = &   \Pr(A) + \Pr(B) + \Pr(C) \\
                        &   &   \mbox{} - \Pr(A \cap B) - \Pr(A \cap C)
                                - \Pr(B \cap C) \\
                        &   &   \mbox{} + \Pr(A \cap B \cap C) \\
                        & = &   \frac{1}{6} + \frac{1}{6} + \frac{1}{6}
                                - \frac{1}{36} - \frac{1}{36} - \frac{1}{36}
                                + 0 \\
                        & = &   \frac{5}{12}
  \end{eqnarray*}
  All the probabilities were computed in preceding steps.}
\end{problemparts}

\end{problem}
