\problemdata      
        {?} % latex-friendly label for the prob.
        {Probability-Conditional} % The topic(s) of the problem content.
        {?} % Source (if known)
        {F01, PS10; F00} % Usage (list ones you are aware of).
        {Christos Kapoutsis, S02} % Last revision info (author, date).

\begin{problem}
\bparts

\ppart Suppose that you are looking in your desk for a letter from some
time ago.  Your desk has eight drawers, and you assess the probability that
it is in any particular drawer as 10\% (so there is a 20\% chance that it
is not in the desk at all). Suppose now that you start searching
systematically through your desk, one drawer at a time.  In addition,
suppose that you have not found the letter in the first $i$ drawers, where
$0 \leq i \leq 7$.  Let $p_i$ denote the probability that the letter will
be found in the next drawer, and let $q_i$ denote the probability that the
letter will be found in some subsequent drawer (both $p_i$ and $q_i$ are
conditional probabilities, since they are based on the assumption that the
letter is not in the first $i$ drawers).  Show that the $p_i$'s increase
and the $q_i$'s decrease.

\hint Observe that if event $A$ implies event $B$ then $\prob{A \mid B} =
\prob{A}/ \prob{B}$.

\solution{

First, we verify the hint.  Saying that $A$ implies $B$ is the same as
saying that that $A \cap B = A$, so
\[
\prob{A \mid B} = \frac{\prob{A \cap B}}{\prob{B}} =
\frac{\prob{A}}{\prob{B}}.
\]

Now, we define the following events:
\begin{eqnarray*}
E_i & \eqdef & \text{letter is in drawer } i,\\
N_i & \eqdef & \text{letter is not in drawers } 1, 2, \ldots i,\\
R_i & \eqdef & \text{letter is in one of drawers } i+1, i+2, \ldots 8.
\end{eqnarray*}
Now we wish to find
\begin{eqnarray*}
p_i & = & \prob{E_{i+1} \mid N_i}\\
&  = & \frac{\prob{E_{i+1} \cap N_i}}{\prob{N_i}}\\
&  = & \frac{\prob{E_{i+1}}}{\prob{N_i}} \qquad
   \text{(from the hint, since $E_{i+1} \Rightarrow N_i$)}\\
& = & \frac{0.1}{0.2 + 0.1(8-i)} = \frac{1}{10-i},
\end{eqnarray*}
which clearly increases as $i$ increases, since the denominator gets
smaller while the numerator remains constant.

Now,
\begin{eqnarray*}
q_i & = & \prob{R_i \mid N_i}\\
& = & \frac{\prob{R_i \cap N_i}}{\prob{N_i}}\\
& = & \frac{\prob{R_i}}{\prob{N_i}} \qquad\text{(from the hint, since $R_i \Rightarrow N_i$)}\\
& = & \frac{0.1(8-i)}{0.2 + 0.1(8-i)}\\
& = & \frac{8-i}{10-i} = \frac{10-i}{10-i} - \frac{2}{10-i} = 1 - \frac{2}{10-i}
\end{eqnarray*}
which decreases as $i$ increases, since the denominator of the fraction
shrinks, so the fraction becomes greater, so the negation of the fraction
becomes lesser.

}
\ppart
The following data appeared in an article in the Wall Street Journal.
For the ages 20, 30, 40, 50, and 60, the probability of a woman in the
U.S. developing cancer in the next ten years is 0.5\%, 1.2\%, 3.2\%,
6.4\%, and 10.8\%, respectively. At the same set of ages, the
probability of a woman in the U.S. eventually developing cancer is
39.6\%, 39.5\%, 39.1\%, 37.5\%, and 34.2\%, respectively.  This seems
strange, but use the previous part of the problem to give an
explanation for these data.

\solution{ We can use the results from a directly by figuring out what
the $p_i$'s and the $q_i$'s are in this situation.  In this case, we care
about if a woman will develop cancer for the ages of 20, 30, 40, 50 and
60.  Therefore, $i$ in this case is such that $i \subseteq \set{20, 30,
40, 50, 60}$.

The $p_i$'s are the probabilities that a woman will develop cancer in the next
10 years. (analogous to the letter being in the next envelope)

The $q_i$'s are the probabilities of a woman in the U.S eventually
developing cancer sometime in the future given that she does not have it
now.  (analogous to the letter being somewhere left in the drawer).

As seen from part a, the $p_i$'s increase while the $q_i$'s decrease.}
\eparts
\end{problem}
