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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 :)

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Yes, as you notice in the last sentence

"Is it something to do with the fact that the difference between actual seats on tickets sold is four?".

The company has sold 200 tickets and the plain has 196 tickets. In order for exactly one seat to be empty on this flight, exactly $5$ persons have to not show up (person 200, 199, 198, 197 and 196). If $6$ persons do not show up, then there will be $2$ empty seats and so on. So, actually you want to calculate the probability $$P(X\ge 5)$$ which however is equivalent to $$P(X>4)$$ as your book indicates, since the number of customers is a discrete number and $>4$ customers is the same as $\ge 5$ customers. Your book chooses the $>4$ because this is then more convenient to be written as $$P(X>4)=1-P(X\le 4)$$ whereas $$P(X \ge 5)=1-P(X<5)=1-P(X\le 4)$$ (you have to be carefull again, that $P(X<5)=P(X\le4)$).

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  • $\begingroup$ Ahh, thank you! It makes sense now :D I completely missed that :) $\endgroup$ – Elise Mar 24 '14 at 13:11
  • $\begingroup$ @Elise you are welcome... :) $\endgroup$ – Jimmy R. Mar 24 '14 at 13:12

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