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.
