table reservation problem I saw a table reservation problem which already being solved in the forum:
Experience shows that 20% of the people reserving tables at a certain restaurant never show up. If the restaurant has 50 tables and takes 52 reservations, what is the probability that it will be able to accommodate everyone?
the answer of this question =1-P(all 52 show up)-P(some set of 51 shows up and the 52-nd person does not)=0.9128 
and I have a further question of it which is:
What is the maximum number of reservations allowed such that the probability of being able to accommodate everyone is at least 70%?
 A: First, some notation and distribution theory. The number $X$ of people
who show is binomial. There are $n$ trials, where $n$ is the
number of reservations taken, and the probability $p$ of Success
is $p = .8.$ 
The probability that everyone who shows is accommodated is
$P(X \le 50),$ where $X \sim Binom(n, p).$ In the first problem
$n = 52.$ So using R software, one obtains $P(X \le 50) = .9999$
(to four places). That small amount of overbooking is unlikely
to inconvenience customers.
 n = 52; p = .8; pbinom(50, n, p)
 ## 0.9998721

You could also use a normal approximation to get essentially the
same result because $E(X) = \mu = np$ and 
$SD(X) = \sigma = \sqrt{np(1-p)}.$
Then you could find the approximation
$P(X < 50.5) \approx P(Z < \frac{50.5 - \mu}{\sigma}).$
where $Z$ is a standard normal random variable. (First, you would
need to compute numerical values of $\mu$ and $\sigma$.)
Now, you want to know the maximum $n$ so that $P(X \le 50) \le .7.$
Using software it is easy to see that $P(X \le 50)$ is a function
of the number of reservations $n$.
 n = 50:100; p = .8; prob = pbinom(52, n, .8)
 plot(n, prob, type="l")
 abline(h = .7, col="red")


We can see from the graph that the probability of accommodating
everyone who shows up drops below 70% when the number of reservations
gets much larger than $n = 60.$ (This can hardly be a shock; just by common sense we ought to
guess that trouble sets in when $.8n > 50.$) It seems the crossover point is just beyond
 $n = 63.$
 max(n[prob > .7])
 ## 63

Even without software, you could find a few points on
this curve using the normal approximation in order to get the
answer.
