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Vehicular arrival at an isolated intersection follows the Poisson distribution. The mean vehicular arrival rate is 2 vehicle per minute. The probability (round off to two decimal places) that at least 2 vehicle will arrive in any given 1- minute interval is ______.
Correct answer is '0.27'. Can you explain this answer?
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Vehicular arrival at an isolated intersection follows the Poisson dist...
Given :
λ = 2 Vehicle/min = 2 Vehical/60 sec = 1/30 Veh/sec
n = 2 and t = 1min
Hence, the probability that atleast 2 vehicles will arrive in any given 1 min interval is 0.2706.
λ = 2 Vehicle/min = 2 Vehical/60 sec = 1/30 Veh/sec
n = 2 and t = 1min
Hence, the probability that atleast 2 vehicles will arrive in any given 1 min interval is 0.2706.
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Vehicular arrival at an isolated intersection follows the Poisson dist...
**Poisson Distribution and Vehicular Arrival**
The Poisson distribution is commonly used to model the arrival of events in a fixed interval of time. In this case, we are considering the arrival of vehicles at an isolated intersection. The mean vehicular arrival rate is given as 2 vehicles per minute. We need to find the probability that at least 2 vehicles will arrive in any given 1-minute interval.
**Poisson Distribution Formula**
The probability mass function (PMF) of the Poisson distribution is given by the formula:
P(x; λ) = (e^-λ * λ^x) / x!
Where:
- P(x; λ) is the probability of observing x events in a given interval,
- λ is the average rate of events (mean),
- e is the base of the natural logarithm (approximately 2.71828),
- x is the number of events.
**Calculating the Probability**
To find the probability that at least 2 vehicles will arrive in a 1-minute interval, we need to calculate the complement of the probability that 0 or 1 vehicle will arrive.
**Probability of 0 Vehicles**
The probability of 0 vehicles arriving in a 1-minute interval can be calculated using the Poisson distribution formula:
P(0; 2) = (e^-2 * 2^0) / 0! = e^-2 ≈ 0.1353
**Probability of 1 Vehicle**
The probability of 1 vehicle arriving in a 1-minute interval can also be calculated using the Poisson distribution formula:
P(1; 2) = (e^-2 * 2^1) / 1! = 2e^-2 ≈ 0.2707
**Probability of at Least 2 Vehicles**
To find the probability of at least 2 vehicles arriving, we subtract the sum of the probabilities of 0 and 1 vehicle from 1:
P(at least 2) = 1 - P(0) - P(1) ≈ 1 - 0.1353 - 0.2707 = 0.594
Therefore, the probability that at least 2 vehicles will arrive in any given 1-minute interval is approximately 0.59 or 59%.
However, the correct answer provided is 0.27. It is possible that there was an error in the calculation or the given answer is incorrect.
The Poisson distribution is commonly used to model the arrival of events in a fixed interval of time. In this case, we are considering the arrival of vehicles at an isolated intersection. The mean vehicular arrival rate is given as 2 vehicles per minute. We need to find the probability that at least 2 vehicles will arrive in any given 1-minute interval.
**Poisson Distribution Formula**
The probability mass function (PMF) of the Poisson distribution is given by the formula:
P(x; λ) = (e^-λ * λ^x) / x!
Where:
- P(x; λ) is the probability of observing x events in a given interval,
- λ is the average rate of events (mean),
- e is the base of the natural logarithm (approximately 2.71828),
- x is the number of events.
**Calculating the Probability**
To find the probability that at least 2 vehicles will arrive in a 1-minute interval, we need to calculate the complement of the probability that 0 or 1 vehicle will arrive.
**Probability of 0 Vehicles**
The probability of 0 vehicles arriving in a 1-minute interval can be calculated using the Poisson distribution formula:
P(0; 2) = (e^-2 * 2^0) / 0! = e^-2 ≈ 0.1353
**Probability of 1 Vehicle**
The probability of 1 vehicle arriving in a 1-minute interval can also be calculated using the Poisson distribution formula:
P(1; 2) = (e^-2 * 2^1) / 1! = 2e^-2 ≈ 0.2707
**Probability of at Least 2 Vehicles**
To find the probability of at least 2 vehicles arriving, we subtract the sum of the probabilities of 0 and 1 vehicle from 1:
P(at least 2) = 1 - P(0) - P(1) ≈ 1 - 0.1353 - 0.2707 = 0.594
Therefore, the probability that at least 2 vehicles will arrive in any given 1-minute interval is approximately 0.59 or 59%.
However, the correct answer provided is 0.27. It is possible that there was an error in the calculation or the given answer is incorrect.
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Vehicular arrival at an isolated intersection follows the Poisson distribution. The mean vehicular arrival rate is 2 vehicle per minute. The probability (round off to two decimal places) that at least 2 vehicle will arrive in any given 1- minute interval is ______.Correct answer is '0.27'. Can you explain this answer?
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