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Suppose p is number of cars per minute passing through a certain road junction between 5
PM and 6PM, and p has a Poisson distribution with mean 3. What is the probability of
observing fewer than 3 cars during any given minute in this interval?
PM and 6PM, and p has a Poisson distribution with mean 3. What is the probability of
observing fewer than 3 cars during any given minute in this interval?
- a)8/(2e3)
- b)9/(2e3)
- c)17/(2e3)
- d)26/(2e3)
Correct answer is option 'C'. Can you explain this answer?
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Poisson Distribution
The Poisson distribution is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space, given the average rate of occurrence of the events. It is often used to model the number of events that occur in a certain period of time, such as the number of cars passing through a road junction in a minute.
The probability mass function (PMF) of a Poisson distribution is given by:
P(X = k) = (e^(-λ) * λ^k) / k!
where:
- X is the random variable representing the number of events
- k is the number of events
- λ is the average rate of events
In this case, the average rate of cars passing through the road junction in a minute is given as λ = 3.
Probability of Observing Fewer than 3 Cars
To find the probability of observing fewer than 3 cars during any given minute in the interval between 5PM and 6PM, we need to calculate the sum of the probabilities of observing 0, 1, and 2 cars.
P(X < 3)="P(X" =="" 0)="" +="" p(x="1)" +="" p(x="" />
Using the Poisson PMF formula, we can calculate each probability:
P(X = 0) = (e^(-3) * 3^0) / 0! = e^(-3) / 1 = 1 / e^3
P(X = 1) = (e^(-3) * 3^1) / 1! = 3 * e^(-3) / 1 = 3 / e^3
P(X = 2) = (e^(-3) * 3^2) / 2! = 9 * e^(-3) / 2 = 9 / (2 * e^3)
Now, we can substitute these values into the equation for P(X < />
P(X < 3)="1" e^3="" +="" 3="" e^3="" +="" 9="" (2="" *="" e^3)="(2" +="" 6="" +="" 9)="" (2="" *="" e^3)="17" (2="" *="" />
Therefore, the probability of observing fewer than 3 cars during any given minute in the interval between 5PM and 6PM is 17 / (2 * e^3).
Therefore, option 'C' (17 / (2 * e^3)) is the correct answer.
The Poisson distribution is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space, given the average rate of occurrence of the events. It is often used to model the number of events that occur in a certain period of time, such as the number of cars passing through a road junction in a minute.
The probability mass function (PMF) of a Poisson distribution is given by:
P(X = k) = (e^(-λ) * λ^k) / k!
where:
- X is the random variable representing the number of events
- k is the number of events
- λ is the average rate of events
In this case, the average rate of cars passing through the road junction in a minute is given as λ = 3.
Probability of Observing Fewer than 3 Cars
To find the probability of observing fewer than 3 cars during any given minute in the interval between 5PM and 6PM, we need to calculate the sum of the probabilities of observing 0, 1, and 2 cars.
P(X < 3)="P(X" =="" 0)="" +="" p(x="1)" +="" p(x="" />
Using the Poisson PMF formula, we can calculate each probability:
P(X = 0) = (e^(-3) * 3^0) / 0! = e^(-3) / 1 = 1 / e^3
P(X = 1) = (e^(-3) * 3^1) / 1! = 3 * e^(-3) / 1 = 3 / e^3
P(X = 2) = (e^(-3) * 3^2) / 2! = 9 * e^(-3) / 2 = 9 / (2 * e^3)
Now, we can substitute these values into the equation for P(X < />
P(X < 3)="1" e^3="" +="" 3="" e^3="" +="" 9="" (2="" *="" e^3)="(2" +="" 6="" +="" 9)="" (2="" *="" e^3)="17" (2="" *="" />
Therefore, the probability of observing fewer than 3 cars during any given minute in the interval between 5PM and 6PM is 17 / (2 * e^3).
Therefore, option 'C' (17 / (2 * e^3)) is the correct answer.
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