When you enter your bank, you find that there are only two tellers, both busy serving other customers, and that there are no other customers in line. Assume that the service times for you and for each of the customers being served are independent identically distributed exponential random variables. Also assume that after a service completion, the next customer in line immediately begins to be served. What is the probability that you will be the last to leave? Hint: Think of the situation at the time that you start getting serve.
Answer is 1/2
To calculate the probability that you will be the last to leave, we need to analyze the situation when you start getting served.
Let's break down the problem step by step:
Step 1: The time it takes for the customer in front of you to finish being served. This time follows an exponential distribution. Let's call it T1.
Step 2: The time it takes for the second customer to finish being served. This time also follows an exponential distribution. Let's call it T2.
Step 3: The time it takes for you to finish being served. This time also follows an exponential distribution. Let's call it T3.
To be the last to leave, both T1 and T2 need to be larger than T3. This means that T3 needs to be the smallest of the three exponential random variables.
The probability that T3 is the minimum of three exponential random variables can be calculated using the order statistics of exponential distributions.
For exponential random variables, the minimum of n exponential random variables with a rate parameter λ is also an exponential random variable with a rate parameter nλ. In this case, we have 3 exponential random variables with rate parameters λ, λ, and λ.
The rate parameter λ for each random variable depends on the average service rate. If the average service rate is μ, then λ = 1/μ.
Let's denote the probability that T3 is the minimum of the three exponential random variables as P(T3 is minimum).
Using the order statistics of exponential random variables, we can calculate P(T3 is minimum) as:
P(T3 is minimum) = 1/(1 + 2)
Since there are 2 other customers in front of you, the probability that you will be the last to leave is:
P(you are last to leave) = P(T3 is minimum)
= 1/(1 + 2)
= 1/3
Therefore, the probability that you will be the last to leave is 1/3.
To find the probability that you will be the last to leave the bank, we need to consider the situation at the time you start getting served.
Since the service times for you and each of the customers are exponentially distributed, we can use the memoryless property of the exponential distribution. This property states that the time it takes for a service to be completed does not depend on how long it has already taken.
In this case, when you start getting served, both tellers are already busy serving other customers. Therefore, the amount of time it will take for you to complete your service will depend on the remaining time of the customers currently being served.
To calculate the probability that you will be the last to leave, we need to consider two scenarios:
1. The service times of the two current customers being served are shorter than the time it takes for you to be served.
2. At least one of the current customers' service time is longer than the time it takes for you to be served.
Let's denote the service time of the first teller as T1 and the service time of the second teller as T2. Also, denote the time it takes for you to be served as T.
For the first scenario, the probability can be calculated as follows:
P(Scenario 1) = P(T1 < T) * P(T2 < T)
Since the service times are independent and identically distributed, the probability of T1 < T is the same as P(T2 < T), which we can denote as p.
Therefore, P(Scenario 1) = p * p = p^2
For the second scenario, the probability can be calculated as follows:
P(Scenario 2) = 1 - P(Scenario 1)
Now, the probability that you will be the last to leave can be calculated as:
P(Last to leave) = P(Scenario 2)
With this information, you can calculate the probability that you will be the last to leave the bank. Note that the exact values of p and the specific exponential distribution parameters would be needed to obtain a numerical answer.