Exact and Poisson approximation for a binomial distribution

Question

An airline company sells $200$ tickets for a plane with $198$ seats, knowing that the probability a passenger will not show up for the flight is $0.01$. Assume that the ticket holders act independently of each other.

$a)$ Use the appropriate approximation to compute the probability the airline company will have enough seats for all the passengers who show up.

$b)$ How good is your approximation?

$c)$ Find the exact probability.

What is the question parts $b)$ and $c)$ referring to and how would I solve them? I found $a)$ to be $1-3e^{-2}$ with a Poisson approximation.

Answer 1

It looks like you assumed a Poisson distribution to compute your answer to $a$, which is reasonable, but you should specify that. The Poisson distribution is approximate, being more accurate when you have more events with low probability. For $c$, you are expected to compute the answer using the correct binomial distribution. Then $b$ is the difference between these.

Answer 2

In order to calculate the probability that they will not have enough seats, add up the following:

• The probability that exactly $\color\red{199}$ passengers will arrive, which is:

$$\binom{200}{\color\red{199}}\cdot(1-0.01)^{\color\red{199}}\cdot(0.01)^{200-\color\red{199}}\approx0.27$$

• The probability that exactly $\color\red{200}$ passengers will arrive, which is:

$$\binom{200}{\color\red{200}}\cdot(1-0.01)^{\color\red{200}}\cdot(0.01)^{200-\color\red{200}}\approx0.13$$

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In order to calculate the probability that they will have enough seats, subtract the result from $1$:

$$1-(0.27+0.13)=0.6$$