Question

At a railway station only one train is handled at atime. The railway yard is sufficient only for 2 trains to wait, while the other is given signal to leave the station. Trains arrives at the station at an average rate of 6 per hours and the railway station can handle them on an average of 6 per hour. Assuming Poisson arrivals and exponential service distribution, find the probability for the number of trains in the system.​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Probability of n trains in the system is Pₙ= (0.50)ⁿ⁺¹ when trains arrive at an average rate of 6 per hour.

Given that,

Only one train is handled at a time at a railroad station. Only two trains can wait in the railway yard while the other is given the all-clear to depart the station. The railway station can accommodate six trains per hour on average, and trains arrive at the station at an average pace of six per hour.

We have to find the probability of how many trains are in the system, assuming Poisson arrivals and an exponential service distribution.​

We know that,

An average of six trains come per hour.

Arrival rate: 6 every hour

An average of 12 can be handled every hour at the railroad station.

Service Rate = μ = 12 per hour

To determine the steady state probabilities for the system's various train counts:

Probability of n trains in the system = Pₙ

Pₙ= $(\frac{\lambda}{\mu})^n (1- \frac{\lambda}{\mu})$

Pₙ= $(\frac{6}{12})^n (1- \frac{6}{12})$

Pₙ= (0.50)ⁿ (0.50)

Pₙ= (0.50)ⁿ⁺¹

Therefore, Probability of n trains in the system is Pₙ= (0.50)ⁿ⁺¹

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