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?
 A: The naive thing is just to accept that $p=0.95$, but you might worry that the group of $10000$ was biased to show up more or less often than the general population.  You can imagine there is a large number of potential passengers, each with the same chance $p$ to show up.  If we are looking for an upper bound to the chance all $100$ will show up, we are looking for an upper bound on $p$.  You took a sample of $10000$ and measured the probability $0.95$.  You could ask the question what is the $p$ that would have more than $9500$ of the $10000$ show up $99\%$ (or your favorite percentage) of the time and claim that is a $99\%$ upper bound on $p$, then use that in the calculation of the chance all $100$ will show up.  You can't get a solid upper bound-there might be a billion potential travelers, only $500$ of whom will not show up, and all $500$ were in you sample of $10000$.
A: A Bayesian approach could be to take a Beta prior distribution for $p$ the probability of somebody turning up, with a density proportional to $p^{\alpha-1}(1-p)^{\beta-1}$.  Common choices are $\alpha=\beta=1$ (a uniform prior), $\alpha=\beta=\frac12$ (a Jeffreys prior) or  $\alpha=\beta=0$ (an improper prior), but with a large number of observations as here, it may make little difference
Assuming individual behaviour is i.i.d., with your observation of $9500$ turning up and $500$ not turning up, you get a posterior distribution for $p$ which is also a Beta density, now proportional to $p^{9500+\alpha-1}(1-p)^{500+\beta-1}$
Given $p$, the probability that the next $100$ all turn up is $p^{100}$.  So combining this with the posterior distribution for $p$ gives a probability of $$\dfrac{\int_0^1 p^{9600+\alpha-1}(1-p)^{500+\beta-1}\,dp}{\int_0^1 p^{9500+\alpha-1}(1-p)^{500+\beta-1}\,dp}=\dfrac{B(9600+\alpha, 500+\beta)}{B(9500+\alpha, 500+\beta)}$$ where the Beta function $B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$ is basically the reciprocal of a generalised binomial coefficient; if you are using a computer to calculate this, you might want to use logarithms to avoid underflow


*

*With $\alpha=\beta=1$ this would give about $0.006019$

*With $\alpha=\beta=\frac12$ this would give about $0.006047$

*With $\alpha=\beta=0$ this would give about $0.006076$

*By comparison, a simple $0.95^{100}$ would give about $0.005921$ 


So they all give about $0.006$
I suspect that the most dubious assumption here is the assumption of i.i.d. individual behaviour: if that is wrong and in fact people tend to turn up or not turn up together then the probability of all $100$ turning up could be much higher 
A: 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 $0.01175$.
