In a theater there are 600 seating's and only two cloakroom's. Every person independently from others randomly leaves a cloak in one of them. At least how many ticket's should there be in each of the cloakroom's so the probability of directing guest's to the other one is less or equal to 0.01 ?
So we have 600 Bernoulli trials. $n = 600, p = \frac{1}{2}$ So to obtain a solution I have to estimate this: $P(X > k ) \le 0.01$?
Call the cloakrooms Head and Tail, and let random variable $X$ be the number of customers who choose Head. We assume that the theatre is sold out, everyone uses a cloakroom, and the people choose Head or Tail independently, with probability $\frac{1}{2}$. (These assumptions may not be very realistic.)
Then $\Pr(X\gt k)\approx \Pr(W\gt k)$, where $W$ has normal distribution with the same mean and variance as $X$. This mean and variance are respectively $(600)(1/2)=300$ and $(600)(1/2)(1/2)=150$.
We have $$\Pr(W\gt k)=\Pr\left(Z\gt \frac{k-300}{\sqrt{150}}\right),$$ where $Z$ is standard normal.
From tables, $\Pr(Z\gt 2.33)\approx 0.01$. So we want (approximately) $$\frac{k-300}{\sqrt{150}} \gt 2.33.$$ Calculation gives $k\gt 328.5$, so having $329$ tickets will bring the probability below $0.01$.
Remark: With computers, one can easily do exact calculations for the binomial. It turns out that indeed $329$ is the smallest integer that brings the probability below $0.01$.