This is the question that I'm stuck with:
An airline knows that overall 3% of passengers do not turn up for flights . The airline decides to adopt a policy of selling more tickets than there are seats on a flight. For an aircraft with 196 seats, the airline sold 200 tickets for a particular flight. By using a suitable approximation, find the probability that there is at least one empty seat on this flight.
Here's what I did:
As p is small, I can use a Poisson approximation
$\therefore X$ ~ $P(6)$ where $X$ is the r.v. "Number of people who don't turn up to flight"
If there is at least one empty seat on the flight, this means there must be at least one person who didn't show up, i.e.
$X \geq 1$
$P(X \geq 1) = 1 - P(X = 0) \\ \space \space\space\space\space\space\space\space\space\space\space\space\space\space \space = 1 - 0.0025 \\ \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space= 0.9975$
However, the mark scheme for this question says:
$P(X > 4) = 1 - P(X \leq 4) \\ \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space = 1 - 0.2851 \\ \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space= 0.7149$
I don't understand where the four in this has come from? Is it something to do with the fact that the difference between actual seats on tickets sold is four?
Thank you :)