# Thread: binomial distribution using an poisson approximation

1. ## binomial distribution using an poisson approximation

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.
(a) Write down a suitable model for the number of passengers who do not turn up
for this flight after buying a ticket.

By using a suitable approximation, find the probability that
(b) more than 196 passengers turn up for this flight,

(c) there is at least one empty seat on this flight.
a) = Let X = “number of passengers who do not show up” then X ~ Bin(200, 0.03)

As p is very small, the Poisson approximation may be used. X ~ Po(6)

b) = P(X < 4 ) = 0.1512 (from Cumulative binomial Tables)

c) I stuck on part 'c', because the correct answer is P(X>4) , which does not make sense to me, because if your looking for the probability that there is at least one empty seat, than that must mean that 199 people turned up for their flight and your probability would be the = P (X>1) = 1-P(x<1).

Am I correct or the book is right? If so can you please explain why it is p(x>4) ?

thanks!

2. You have 200 passengers and 196 seats.

If you want to have 1 or more empty seats, then at least 5 passengers must not turn up.

Since the number of people turning up must be an integer (whole number)
$\displaystyle P(x ~at~ least~ 5) = p(X > 4)$

The difference betwee the question's answer and your approach is that the question looks at the 200 passengers and you are trying to look at the 196 seats