Probability: Poisson distribution involving customer arrivals

In summary, based on the given information, the probability of a non-zero number of customers in store A and zero customers in store B is p(1)*0.5 + p(2)*0.52 + p(3)*0.53 + ..., while the probability of exactly 2 customers in store A and zero customers in store B is p(2)*0.52.
  • #1
sanctifier
58
0

Homework Statement



There are two stores A and B.

Customers can equally enter one of the two stores, i.e., for a specific customer, the probabilities she enters store A or B both are 0.5.

If the total number of customers in two stores has the Poisson distribution of parameter λ, then

Question 1: What is the probability that the number of customers in store A is non-zero and store B has no customers;

Question 2: What is the probability that the number of customers in store A is exactly 2 and store B has no customers.

Homework Equations



Poisson distribution: p(x)=λx/x! * e

The Attempt at a Solution



Answer 1: (1-p(0))*(0.5+0.52+0.53+...)=1-e
Comment: The probability that there are some customer in some store is 1 - p(0), then the probability that x customers entered store A is 0.5x, hence their product should yield the desired answer?

Answer 2: p(2)*0.52
Comment: Similar strategy as answer 1.

Are these correct?

Thank you in advance!
 
Last edited:
Physics news on Phys.org
  • #2
sanctifier said:

Homework Statement



There are two stores A and B.

Customers can equally enter one of the two stores, i.e., for a specific customer, the probabilities she enters store A or B both are 0.5.

If the total number of customers in two stores has the Poisson distribution of parameter λ, then

Question 1: What is the probability that the number of customers in store A is non-zero and store B has no customers;

Question 2: What is the probability that the number of customers in store A is exactly 2 and store B has no customers.


Homework Equations



Poisson distribution: p(x)=λx/x! * e

The Attempt at a Solution



Answer 1: (1-p(0))*(0.5+0.52+0.53+...)=1-e
Comment: The probability that there are some customer in some store is 1 - p(0), then the probability that x customers entered store A is 0.5x, hence their product should yield the desired answer?

Answer 2: p(2)*0.52
Comment: Similar strategy as answer 1.

Are these correct?

Thank you in advance!

You answer to (2) is correct, but your answer to (1) is not.
 
  • #3
Ray Vickson, thank you for your replay.

If answer (2) is correct, then answer (1) is p(1)*0.5 + p(2)*0.52 + p(3)*0.53 + ...

Is this correct?

If it is, then is there a concise form of the summation?
 
  • #4
sanctifier said:
Ray Vickson, thank you for your replay.

If answer (2) is correct, then answer (1) is p(1)*0.5 + p(2)*0.52 + p(3)*0.53 + ...

Is this correct?

If it is, then is there a concise form of the summation?

Yes, and yes. For the latter, see https://www.efunda.com/math/exp_log/series_exp.cfm .
 
  • Like
Likes 1 person

FAQ: Probability: Poisson distribution involving customer arrivals

What is a Poisson distribution?

A Poisson distribution is a probability distribution that represents the number of times an event occurs within a specific time period, given that the event occurs at a constant rate and independently of the time since the last event.

How is a Poisson distribution used in customer arrivals?

A Poisson distribution can be used to model the arrival rate of customers at a business or location. This can help businesses predict and manage staffing needs, wait times, and other aspects related to customer arrivals.

What are the key characteristics of a Poisson distribution?

The key characteristics of a Poisson distribution include a mean or average rate of occurrence, a single parameter (λ) that represents the rate of occurrence, and the assumption that events occur independently and with a constant rate over time.

What is the formula for calculating the probability of a specific number of events occurring in a Poisson distribution?

The formula for calculating the probability of x events occurring in a Poisson distribution is P(x) = (e^-λ * λ^x) / x!, where e is the base of natural logarithms and x! represents x factorial (x * (x-1) * (x-2) * ... * 1).

What are some real-world applications of Poisson distribution in customer arrivals?

Poisson distribution can be used in a variety of real-world applications related to customer arrivals, such as estimating the number of calls or emails a customer service team will receive in a given hour, predicting the number of customers who will visit a store during a specific time period, and forecasting the number of patients who will arrive at a hospital emergency room in a day.

Back
Top