Question
A major hotel chain keeps a record of the number of mishandled bags per 1000 customers. Assume the number of mishandled bags has a poisson distribution.
What is the probability that in the next 1000 customers, the hotel chain will have no more than 3 mishandled bags?
What is the probability that in the next 1000 customers, the hotel chain will have no more than 3 mishandled bags?
Answers
GPT-4o mini
To solve the problem, we need to use the properties of the Poisson distribution.
The Poisson distribution is defined by its parameter \( \lambda \), which represents the average number of events (in this case, mishandled bags) in a fixed interval (in this case, per 1000 customers).
Let’s assume that the hotel chain has an average of \( \lambda \) mishandled bags per 1000 customers. We will use this \( \lambda \) to compute the probabilities.
However, we first need to determine \( \lambda \). If the problem provided a specific average number of mishandled bags per 1000 customers, we would use that. For the sake of this solution, let’s assume \( \lambda = 2 \) (you should replace this with the actual average provided in your data if it is different).
The Poisson probability mass function is given by:
\[
P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}
\]
where \( P(X = k) \) is the probability of observing \( k \) events (mishandled bags), \( e \) is the base of the natural logarithm, \( \lambda \) is the average number of events, and \( k! \) is the factorial of \( k \).
We need to find the probability of having no more than 3 mishandled bags, which is represented as:
\[
P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
\]
Calculating each of these:
1. For \( k = 0 \):
\[
P(X = 0) = \frac{e^{-\lambda} \lambda^0}{0!} = e^{-\lambda}
\]
2. For \( k = 1 \):
\[
P(X = 1) = \frac{e^{-\lambda} \lambda^1}{1!} = e^{-\lambda} \lambda
\]
3. For \( k = 2 \):
\[
P(X = 2) = \frac{e^{-\lambda} \lambda^2}{2!} = e^{-\lambda} \frac{\lambda^2}{2}
\]
4. For \( k = 3 \):
\[
P(X = 3) = \frac{e^{-\lambda} \lambda^3}{3!} = e^{-\lambda} \frac{\lambda^3}{6}
\]
Now, combining these:
\[
P(X \leq 3) = e^{-\lambda} \left( 1 + \lambda + \frac{\lambda^2}{2} + \frac{\lambda^3}{6} \right)
\]
Substituting \( \lambda = 2 \):
\[
P(X \leq 3) = e^{-2} \left( 1 + 2 + \frac{2^2}{2} + \frac{2^3}{6} \right)
= e^{-2} \left( 1 + 2 + 2 + \frac{8}{6} \right)
= e^{-2} \left( 5 + \frac{4}{3} \right)
= e^{-2} \left( \frac{15}{3} + \frac{4}{3} \right)
= e^{-2} \left( \frac{19}{3} \right)
\]
Now, we compute \( e^{-2} \):
\[
e^{-2} \approx 0.1353
\]
Thus, the final probability becomes:
\[
P(X \leq 3) \approx 0.1353 \times \frac{19}{3} \approx 0.1353 \times 6.3333 \approx 0.8575
\]
Therefore, the probability that in the next 1000 customers, the hotel chain will have no more than 3 mishandled bags is approximately \( 0.8575 \) or \( 85.75\% \).
**Note:** Please substitute \( \lambda \) with the correct average if it is different from 2.
The Poisson distribution is defined by its parameter \( \lambda \), which represents the average number of events (in this case, mishandled bags) in a fixed interval (in this case, per 1000 customers).
Let’s assume that the hotel chain has an average of \( \lambda \) mishandled bags per 1000 customers. We will use this \( \lambda \) to compute the probabilities.
However, we first need to determine \( \lambda \). If the problem provided a specific average number of mishandled bags per 1000 customers, we would use that. For the sake of this solution, let’s assume \( \lambda = 2 \) (you should replace this with the actual average provided in your data if it is different).
The Poisson probability mass function is given by:
\[
P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}
\]
where \( P(X = k) \) is the probability of observing \( k \) events (mishandled bags), \( e \) is the base of the natural logarithm, \( \lambda \) is the average number of events, and \( k! \) is the factorial of \( k \).
We need to find the probability of having no more than 3 mishandled bags, which is represented as:
\[
P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
\]
Calculating each of these:
1. For \( k = 0 \):
\[
P(X = 0) = \frac{e^{-\lambda} \lambda^0}{0!} = e^{-\lambda}
\]
2. For \( k = 1 \):
\[
P(X = 1) = \frac{e^{-\lambda} \lambda^1}{1!} = e^{-\lambda} \lambda
\]
3. For \( k = 2 \):
\[
P(X = 2) = \frac{e^{-\lambda} \lambda^2}{2!} = e^{-\lambda} \frac{\lambda^2}{2}
\]
4. For \( k = 3 \):
\[
P(X = 3) = \frac{e^{-\lambda} \lambda^3}{3!} = e^{-\lambda} \frac{\lambda^3}{6}
\]
Now, combining these:
\[
P(X \leq 3) = e^{-\lambda} \left( 1 + \lambda + \frac{\lambda^2}{2} + \frac{\lambda^3}{6} \right)
\]
Substituting \( \lambda = 2 \):
\[
P(X \leq 3) = e^{-2} \left( 1 + 2 + \frac{2^2}{2} + \frac{2^3}{6} \right)
= e^{-2} \left( 1 + 2 + 2 + \frac{8}{6} \right)
= e^{-2} \left( 5 + \frac{4}{3} \right)
= e^{-2} \left( \frac{15}{3} + \frac{4}{3} \right)
= e^{-2} \left( \frac{19}{3} \right)
\]
Now, we compute \( e^{-2} \):
\[
e^{-2} \approx 0.1353
\]
Thus, the final probability becomes:
\[
P(X \leq 3) \approx 0.1353 \times \frac{19}{3} \approx 0.1353 \times 6.3333 \approx 0.8575
\]
Therefore, the probability that in the next 1000 customers, the hotel chain will have no more than 3 mishandled bags is approximately \( 0.8575 \) or \( 85.75\% \).
**Note:** Please substitute \( \lambda \) with the correct average if it is different from 2.