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Arrivals at a telephone booth are considered to be poison, with an average time of 10 minutes between successive arrivals. The length of a phone call is distributes exponentially with mean 3 minutes. The probability that an arrival does not have to wait before service is 
  • a)
    0.3  
  • b)
    0.5  
  • c)
    0.7  
  • d)
    0.9    
Correct answer is 'A'. Can you explain this answer?
Verified Answer
Arrivals at a telephone booth are considered to be poison, with an ave...
Given λ= 1/10 = 0.10 person per minute.
μ = 1/3 = 0.33 person per minute.

(i) Probability that a person arriving at the booth will have to wait,
P (w > 0) = 1 – P0
= 1 – (1 - λ / μ) = λ / μ
= 0.10/0.33 = 0.3
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Most Upvoted Answer
Arrivals at a telephone booth are considered to be poison, with an ave...
Solution:

Given:
The time between successive arrivals follows poison distribution with an average time of 10 minutes.
The length of a phone call follows an exponential distribution with a mean of 3 minutes.

To find:
The probability that an arrival does not have to wait before service.

Method:
We can use the M/M/1 queuing model to solve this problem. M/M/1 stands for Markovian Arrival Process/Markovian Service Process/Single Server Queue.

Assumptions of M/M/1 queuing model:
1. Arrival process follows poison distribution.
2. Service time follows exponential distribution.
3. There is only one server.
4. Service is provided on a first-come-first-serve basis.
5. The queue is of infinite capacity.

Steps to solve the problem using M/M/1 queuing model:
1. Calculate the arrival rate (λ) and service rate (μ).
2. Calculate the utilization factor (ρ) using the formula ρ = λ/μ.
3. Calculate the probability that the server is busy (P0) using the formula P0 = 1 - ρ.
4. Calculate the average number of customers in the system (L) using the formula L = λ/(μ-λ).
5. Calculate the average time spent in the system (W) using the formula W = L/λ.
6. Calculate the probability that an arrival does not have to wait before service (P) using the formula P = e^(-μW).

Calculation:
1. Arrival rate (λ) = 1/10 per minute
Service rate (μ) = 1/3 per minute

2. Utilization factor (ρ) = λ/μ = (1/10)/(1/3) = 0.3

3. Probability that the server is busy (P0) = 1 - ρ = 1 - 0.3 = 0.7

4. Average number of customers in the system (L) = λ/(μ-λ) = (1/10)/[(1/3)-(1/10)] = 0.375

5. Average time spent in the system (W) = L/λ = 0.375/(1/10) = 3.75 minutes

6. Probability that an arrival does not have to wait before service (P) = e^(-μW) = e^(-1/3*3.75) = 0.3

Therefore, the probability that an arrival does not have to wait before service is 0.3, option (a) is correct.
Community Answer
Arrivals at a telephone booth are considered to be poison, with an ave...
I think answer is 0.7, does not have to wait= P0 (idle server). For does have to wait =1-P0 (0.3). 
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Arrivals at a telephone booth are considered to be poison, with an average time of 10 minutes between successive arrivals. The length of a phone call is distributes exponentially with mean 3 minutes. The probability that an arrival does not have to wait before service isa)0.3 b)0.5 c)0.7 d)0.9 Correct answer is 'A'. Can you explain this answer?
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