Question

Choose the correct problem formulation: Hotels, like airlines, often overbook, counting on the fact that some people with reservations will cancel at the last minute. A certain hotel chain finds that 20% of its reservations will not be used. If 15 reservations are randomly selected, what is the probability that more than 8 but less than 12 reservations will be used

Answer 1

Answer:

$P(8 < x \leq 12)=P(X=9)+P(X=10)+P(X=11)+P(X=12)$

$P(X=9)=(15C9)(0.8)^{9} (1-0.8)^{15-9}=0.0429$

$P(X=10)=(15C10)(0.8)^{10} (1-0.8)^{15-10}=0.1032$

$P(X=11)=(15C11)(0.8)^{11} (1-0.8)^{15-11}=0.1878$

$P(X=12)=(15C12)(0.8)^{12} (1-0.8)^{15-12}=0.2501$

$P(8 < x \leq 12)=0.0429+0.1032+0.1878 +0.2501=0.5839$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=15, p=1-0.2=0.8)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

And we want to find this probability:

$P(8 < x \leq 12)=P(X=9)+P(X=10)+P(X=11)+P(X=12)$

$P(X=9)=(15C9)(0.8)^{9} (1-0.8)^{15-9}=0.0429$

$P(X=10)=(15C10)(0.8)^{10} (1-0.8)^{15-10}=0.1032$

$P(X=11)=(15C11)(0.8)^{11} (1-0.8)^{15-11}=0.1878$

$P(X=12)=(15C12)(0.8)^{12} (1-0.8)^{15-12}=0.2501$

$P(8 < x \leq 12)=0.0429+0.1032+0.1878 +0.2501=0.5839$