Poisson Distribution problem

This can be calculated as P(X >= 3 | X >= 2) = (1 - P(X < 3)) / (1 - P(X < 2)). This simplifies to P(X >= 3 | X >= 2) = (1 - P(X = 2)) / (1 - P(X = 1)). In summary, the probability of having at least one more customer enter the cafe during tea time, given that there are already at least two customers, can be calculated using the formula P(X >= 3 | X >= 2) = (1 - P(X = 2)) / (1 - P(X = 1)).
  • #1
tommyhakinen
36
0

Homework Statement


The number of customers entering a cafe during tea time is known to be poisson distribution with λ = 5.
on a particular day, given that at least 2 customers have entered the cafe during the tea time. what is the probability that at least 1 more customers will enter the cafe during tea time?

The Attempt at a Solution


I was thinking of using this :

[tex]P(X \geq 1 | X \geq 2) = \frac{P(X \geq 1 \cap X \geq 2)}{P(X \geq 2)}[/tex]

However, i soon found that if i do that, the Probability will go to 1. it does not make sense. Please advice. thank you.
 
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  • #2
The conditional probability on the left side represents the probability that X >= 1, given that X >= 2. If X >= 2, then it will certainly be greater than 1. I think you want P(X >= 3 | X >= 2).
 
  • #3
Thanks for the reply. But I still don't quite get it. Any advise? Thanks..
 
  • #4
tommyhakinen said:
Thanks for the reply. But I still don't quite get it. Any advise? Thanks..
If you know that there are two people in the cafe, the probability that there is at least one person is 1. For a different example, if I see that you have a dollar, then I know for certain that you have 50 cents (you might have to change the dollar, though).
what is the probability that at least 1 more customers will enter the cafe during tea time?
You know that there are at least two people in the cafe. "one more customer" means that there are three people. "at least one more customer" means that there are three or more.
What you're trying to find is the probability of 3 or more people in the cafe, given that there already two people there.
 

What is a Poisson Distribution problem?

A Poisson Distribution problem is a statistical problem that involves calculating the probability of a certain number of events occurring within a specific time period or space. It is used to model the number of occurrences of a rare event, such as car accidents or customer arrivals, in a given time or area.

What are the key assumptions of a Poisson Distribution?

The key assumptions of a Poisson Distribution are that the events occur independently of each other, the average rate of occurrence is constant, and the probability of an event occurring is proportional to the length of the time period or the size of the space.

How is a Poisson Distribution different from a Normal Distribution?

A Poisson Distribution is different from a Normal Distribution in several ways. Firstly, a Poisson Distribution is discrete, meaning it only takes on integer values, while a Normal Distribution is continuous. Additionally, a Poisson Distribution is used to model the number of occurrences of a rare event, while a Normal Distribution is used to model the distribution of a continuous variable.

How do you calculate the mean and standard deviation of a Poisson Distribution?

The mean and standard deviation of a Poisson Distribution can be calculated by using the formula: Mean = λ and Standard Deviation = √(λ), where λ is the average rate of occurrence. Alternatively, they can also be calculated using the data set by finding the average of the data and taking the square root of that average.

What are some real-life examples of Poisson Distribution problems?

Poisson Distribution problems can be found in various real-life scenarios, such as tracking the number of earthquakes in a certain region, the number of emails received in a day, or the number of errors in a page of a book. Other examples include the number of births in a hospital in a month or the number of phone calls received by a customer service center in an hour.

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