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 Air craft of Jet Airways at Ahmedabad airport arrive according to a Poisson process at a rate of 12 per hour. All aircraft are handled by one air traffic controller. If the controller takes a 2 – minute coffee break, what is the probability that he will miss one or more arriving aircraft?
  • a)
    0.33
  • b)
    0.44
  • c)
    0.55
  • d)
    0.66
Correct answer is option 'A'. Can you explain this answer?
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Air craft of Jet Airways at Ahmedabad airport arrive according to a Po...
P (miss/or more aircraft) = 1 – P(miss 0) = 1 – P(0 arrive).
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Air craft of Jet Airways at Ahmedabad airport arrive according to a Po...
-minute break every hour, what is the probability that there will be a delay of at least 5 minutes for the arrival of a given aircraft?

We can model the arrival of aircraft as a Poisson process with rate λ = 12 per hour. However, due to the air traffic controller's breaks, the effective arrival rate of aircraft during each hour is reduced to 11 per hour (since the controller is unavailable for 2 minutes out of every 60 minutes).

Let X be the time (in minutes) between successive arrivals of aircraft. Then X follows an exponential distribution with parameter λ/60 = 11/60. The probability of a delay of at least 5 minutes for the arrival of a given aircraft is equal to the probability that the time between successive arrivals exceeds 5 minutes.

Let Y = X - 5 be the time (in minutes) by which the time between successive arrivals exceeds 5 minutes. Then Y follows an exponential distribution with parameter λ/60 = 11/60. The probability of a delay of at least 5 minutes is therefore:

P(Y > 0) = 1 - P(Y ≤ 0) = 1 - F(Y = 0) = 1 - e^(-λt), where t = 5.

Substituting λ = 11/60 and t = 5, we get:

P(Y > 0) = 1 - e^(-11/12) ≈ 0.4677

Therefore, the probability of a delay of at least 5 minutes for the arrival of a given aircraft is approximately 0.4677.
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