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A hotel can accommodate 50 customers, experiences show that $0.1$ of those who make a reservation will not show up. Suppose that the hotel accepts 55 reservations. Calculate the probability that the hotel will be able to accommodate all of the customers that show up.

I only tried to use the Poisson distribution, but I am sure it can not apply here, how this question can be solved easier? thanks

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Use binomial distribution –  Sasha Dec 22 '11 at 6:57
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4 Answers 4

Here is how you can use Poisson distribution: $$ \mathbf P(X=k)=\frac{\lambda^k}{k!}e^{-\lambda}. $$ In you case you should take the "success" as "person does not appear", probability of success is $p=0.1$. The probability that you have more than 5 successes (i.e., you can accommodate all the people) is \begin{align} \mathbf P(X\geq 5)&=1-\mathbf P(X<5)\\ &=1-(\mathbf P(X=0)+\mathbf P(X=1)+\mathbf P(X=2)+\mathbf P(X=3)+\mathbf P(X=4))\\ &=1-e^{-\lambda}\left(1+\lambda+\frac{\lambda^2}{2!}+\frac{\lambda^3}{3!}+\frac{\lambda^4}{4!}\right), \end{align} which can be easily calculated even with a cell phone calculator, noting that $$ \lambda=pn=5.5. $$ The answer is $$ \mathbf P(X\geq 5)=0.6424 $$ (compare with the exact answer by @Marvis)

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In real world the hotel should accept only 50 reservations and the probability will be 1.

Now to increase the revenues the hotel should increase the fixed amount of charges for the customers who does not show up.

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You need to make use of the binomial distribution here. It is not hard to evaluate the answer using binomial distribution itself.

The probability that the hotel will be able to accommodate all of the customers that show up is the probability that at-most $50$ customers show up. Hence, $$\mathbb{P}(\text{at-most }50 \text{ customers show up}) = 1 - \mathbb{P}( \text{number of customers } \geq 51)$$ Hence, \begin{align} \mathbb{P}( \text{number of customers}) & = \binom{55}{55}(0.9)^{55} + \binom{55}{54}(0.9)^{54} (0.1)^1 + \binom{55}{53}(0.9)^{53} (0.1)^2 \\ & + \binom{55}{52}(0.9)^{52} (0.1)^3 + \binom{55}{51}(0.9)^{51} (0.1)^4\\ &\approx 0.345 \end{align} Hence, the required probability is $$\approx 1 - 0.345 = 0.645$$

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For a very similar problem (with $105$ passengers holding reservations on a $100$-seat flight and $0.1$ probability of not showing up), see here and the complete solution is here. It even uses the Poisson approximation that you thought about. Use the same techniques....

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