\problemdata
{s96 ps9}
{expected value of indicator random variable}

\begin{problem}
An airport bus deposits 25 passengers at 7 stops.
Assume that each passenger is as likely to get off at 
any stop as another and that they act independently.
The bus stops only if someone wants to get off.
What is the probability that nobody gets off at the 
third stop?
What is the expected number of stops it will make?

For each individual, the probability he {\em does not} get off at the
third stop is $\frac{6}{7}$.  Therefore, the probability
that no one gets off at the third stop is $(\frac{6}{7})^{25} = 
0.021$.

Let $I_k$ be an indicator variable for whether the bus stops at the
$k$-th stop.  Then, $X=\sum_k I_k$ is the total number of stops.  By
linearity of expectation,
\[
E(X) = E(I_1) + E(I_2) + \dots + E(I_7)
\]

For each $I_k$, the $E(I_k)$ is the probability the bus stop at the
$k$-th stop, or $1-0.021=0.979$.  Therefore, $E(X) = 7(0.979) = 6.85$.

\end{problem}
