\problemdata      
        {probability-intro-5} % latex-friendly label for the prob.
        {mean time to failure, Boole's inequality} % The topic(s) of the problem content.
        {?} % Source (if known)
        {S02 cp9W} % Usage (list ones you are aware of).
        {Albert Meyer, S02} % Last revision info (author, date).


\begin{problem} 
Suppose there is a system with $n$~components, and we know from past
experience that any particular component will fail in a given year with
probability~$p$.  That is, letting $F_i$ be the event that the $i$th
component fails within one year, we have
\[
\pr{F_i} = p
\]
for $1 \leq i \leq n$.  The \emph{system} will fail if \emph{any one} of
its components fails.  What can we say about the probability that the
system will fail within one year?

Let $F$ be the event that the system fails within one year.  Without any
additional assumptions, we can't get an exact answer for $\pr{F}$.
However, we can give useful upper and lower bounds, namely,
\begin{equation}\label{pFnp}
p \le \pr{F} \le np.
\end{equation}
So for example, if $n = 100$ and $p = 10^{-5}$, we conclude that there is
at most one chance in 1000 of system failure within a year and at least
one chance in 100,000.

\bparts

%%FOR FUTURE USE: put part (a) last.

\ppart Prove the inequality~(\ref{pFnp}).
\solution{
$F = \lgunion_{i=1}^n F_i$ so
\begin{align}
p & = \pr{F_1} & \text{(given)}\\
& \le \pr{\lgunion F_i} & \text{(monotonicity)}\\
& \le \sum_{i=1}^n \pr{F_i} & \text{(Boole's inequality)}\\
& = np.
\end{align}
Here, \emph{monotonicity}, refers to the (obvious) fact from 
\hyperref{http://theory.lcs.mit.edu/classes/6.042/spring02/handouts/lectures/ln9.pdf}
{ln9}{monotonicity}{Notes 9}
that
\[
\text{if } A \subseteq B \text{ then } \pr{A} \le \pr{B}.
\]
}

\eparts

Let's model this situation with the sample space $\sspace \eqdef
\power(\set{1, \dots,n})$ of subsets of positive integers $\le n$, where
$s \in \sspace$ corresponds to the numbers of the components which fail
within one year.  For example, $\set{2, 5}$ is the outcome that the second
and fifth components failed within a year and none of the other components
failed.  So the outcome that the system did not fail corresponds to the
emptyset, $\emptyset$.

\bparts

\ppart Show that the probability that the system fails could be as small
as $p$ by describing appropriate probabilities for the sample points.

\solution{There could be a probability $p$ of system failure if the all
individual failures occur together.  That is, let $\pr{\set{1, \dots,n}}
\eqdef p$, $\pr{\emptyset} \eqdef 1-p$, and let the probability of all
other outcomes be zero.  So $F_i = \set{s \in \sspace \suchthat i \in s}$
and $\pr{F_i} = 0 + 0 + \dots + 0 + \pr{\set{1, \dots,n}} = \pr{\set{1,
\dots ,n}} = p$.  Also, the only outcome with positive probability in $F$
is $\set{1, \dots ,n}$, so $\pr{F} = p$, as required.}

\ppart Show that the probability that the system fails could actually
could be as large as $np$ by describing appropriate probabilities for the
sample points.

\solution{Suppose at most one component ever fails at a time.  That is,
$\pr{\set{i}}=p$ for $1 \le i \le n$, $\pr{\emptyset}=1-np$, and
probability of all other points is zero.  The sum of the probabilities of
all the points is one, so this is a well-defined probability space.  Also,
the only sample point in $F_i$ with positive probability is $\set{i}$, so
$\pr{F_i} = \pr{\set{i}} = p$ as required.  Finally, $\pr{F} = np$ because
$F = \set{A \subseteq \set{1,\dots,n} \suchthat A \neq \emptyset}$, so $F$
in particular contains all the $n$ outcomes of the form $\set{i}$.}

\eparts
\end{problem}
