So the question asks: Airlines find that each passenger who reserves a seat fails to turn up with probability 0.01 independently of other passengers. Consequently, Bryanair always sell 100 tickets for their 99 seat aeroplane, and DifficultJet always sell 200 tickets for their 198 seat aeroplane. Which is more over-booked?
At first, I thought I should use the Bernoulli distribution because there are no set trials, with p=0.01, but x has to been either 0 or 1 for this distribution and surely x=100 for Bryanair and y=199,200 for DifficultJet. So, then I tried Binomial, but then surely n=1, so x>n and this is just confusing. Please help!
First of all, you may benefit from being a bit more precise in your thinking. It doesn't mean much to say you're going to "use" a certain distribution. Use it as what, a paperweight? A random variable, that is, a random number, is something that has a distribution. So you should articulate carefully which random number it is which you think has a Bernoulli or a Binomial distribution. For example you could say "the number of passengers who don't cancel on a Bryanair flight follows a Bernoulli distribution", or something like that.
Once you've figured out exactly what it is you're trying to say, you need to start from a conceptual understanding of what the different distributions mean. So:
If you flip a coin (or do any other experiment which can have two different results) the result of the coin flip follows a Bernoulli distribution of parameters $p$ (normally we call one of the results $1$ and the other $0$, and say that $p$ is the probability of getting $1$).
If you flip a coin with probability $p$ of coming up heads (or what-have-you) $n$ times, the number of heads you get follows a Binomial distribution of parameters $n$, $p$.
Can you find a random variable in the problem which follows a Bernoulli distribution? Can you find one which follows a Binomial?