Why is this problem using a Bernoulli random variable instead of a Binomial random variable? My textbook has the following problem:

An airline claims that at most $6\%$ of all lost luggage is never
  found. A random sample of size $200$ is taken. Out of the $200$
  observations (supposed to be independent) $23$ pieces of luggage were
  never found. Test whether the airline's claim is true at a $1\%$
  significance level using the critical value method.

When identifying the probability model for this problem, the textbook uses a random variable with a Bernoulli distribution.
The same textbook defines Bernoulli random variables and Binomial random variables as follows:
Bernoulli Random Variables

A Bernoulli trial is an experiment which has two possible outcomes,
  generally called success and failure.

Binomial Random Variables

Let $X$ be the total number of successes in n repeated, independent
  Bernoulli trials with success probability $p$. $X$ is called the
  binomial random variable with parameters $n$ and $p$, written $X \sim Binom(n, p)$.

It seems to me like the problem should be using a Binomial random variable, since we have many (200) events, and we are seeking to find, out of those 200, how many pieces were lost (failure) and how many were not (success). I certainly do not see how using a Bernoulli random variable is correct in such a situation. 
I would greatly appreciate it if people could please take the time to clarify this for me.
 A: If you have a single experiment with two possible outcomes, the outcome can be modeled as a Bernoulli Random Variable with parameter $p$ as described in your textbook. If you then repeat this experiment, independently, again, and look at the number of "successful" outcomes across each experiment, then that random variable is described by a Binomial random variable.  In other words a binomial "$n$, $p$" random variable is simply the sum of $n$ independent Bernoulli random variables each with success probability $p$.
So in the context of the airline problem, whether or not each "piece of lost luggage is never found" is described by a Bernoulli random variable with success (lost luggage that is never found is the successful event in this case) probability $p=0.06$.  When you look at the total number of lost pieces never found, say, $X$ out of $n=200$ pieces, the random variable $X$ is described by a $Binomial(n=200, p=0.06)$ random variable.
A helpful way to look at this is to consider indicator random variables.  Let  \begin{eqnarray*}
I_{i} & = & \begin{cases}
1 & \mbox{if $i$th independent experiment/trial is a success with probability $p$}\\
0 & \mbox{otherwise with probability $1-p$}
\end{cases}
\end{eqnarray*}
Then $I_i\sim Bernoulli(p)$ for $i=1, 2, . . ., n $.
Now, add up the indicator variables and set the result equal to the new random variable $X$:
\begin{eqnarray*}
\sum_{i=1}^{n}I_{i} & = & X
\end{eqnarray*}
Then $X\sim Binomial(n,p)$ = the number of successes (1's) out of $n$.
