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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?
Most Upvoted Answer
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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Community Answer
Air craft of Jet Airways at Ahmedabad airport arrive according to a Po...
Understanding the Problem:
The arrival of aircraft at Jet Airways follows a Poisson process with a rate of 12 per hour. The air traffic controller takes a 2-minute coffee break. We need to find the probability that he will miss one or more arriving aircraft during this break.

Calculating the Arrival Rate:
Given that aircraft arrive at a rate of 12 per hour, we can calculate the average time between arrivals as 1/12 hour per aircraft.

Converting the Break Time:
The controller takes a 2-minute break, which is equivalent to 2/60 = 1/30 hour.

Calculating the Probability:
The probability of missing one or more aircraft during the break can be calculated using the Poisson distribution formula:
P(X >= 1) = 1 - P(X = 0) = 1 - e^(-λ) * λ^0 / 0!
Substitute the values:
λ = 1/30 (break time)
P(X >= 1) = 1 - e^(-1/30) ≈ 0.33
Therefore, the probability that the air traffic controller will miss one or more arriving aircraft during the 2-minute coffee break is approximately 0.33, which corresponds to option A.
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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)0.33b)0.44c)0.55d)0.66Correct answer is option 'A'. Can you explain this answer?
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