Probability that everyone shows up for flight? 
The probability of a flight reservation being a no-show is unknown but
  after observing $10000$ flight reservations we found that $95\%$ of
  those people showed up.
If we consider a new sample of $100$ flight reservations, what is the
  chance that each of the people shows up? Can we find a useful upper bound to it?

I assume that all show-ups are i.i.d.
If we write $p$ for the probability of a passenger showing up for her flight, then the probability that everyone shows up is simply $p^{100}$, but of course $p$ is not known.
My idea was to use the information that we had $95\%$ show-ups in the sample with $10000$ reservations to bound likely values of $p$.
How can we proceed?

I will propose here a classical solution, assuming the probability of no-show is i.i.d.
From a frequentist approach, we may consider $p = \bar{p} = 0.95$ the best estimate for the probability for showing. Since the variable is i.i.d., the discrete probability distribution function of $n$ show events in $N$ bookings is binomial, $e.g.$,
$$
P(n) = p^n(1-p)^{N-n}\binom{N}{n}
$$
The variance of the binomial is $\sigma = \sqrt{Np(1-p)}$, so the estimate of the variance is $\bar{\sigma} = \sqrt{N\bar{p}(1-\bar{p})}$. This gives $\bar{\sigma} =21.79$, or $\bar{\sigma}_p =0.002179$.
The probability of 100 show events in a 100 sample would simply be given by $0.95^{100}=0.00592$. The upper limit can be computed by using a 3-sigma upper limit for the probability $p$, $p_{ul}=p+3\sigma_p=0.95654$. Computing now $p_{ul}^{100}$ this gives the upper limit of $0.01175$.
