probability - vehicle arriving and gap in between them












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$begingroup$


Vehicle arriving at an intersection from one of the approach roads follow the Poisson distribution. The mean rate of arrival is 900 vehicles per hour. If a gap is defined as the time difference between two successive vehicle arrivals (with vehicle assumed to be points), the probability that the gap is greater than 8 seconds is ____.



My approach for solving above question:



If 900 vehicles are passing per hour -> $frac{1}{4}$ vehicles passing per second. So in 4 seconds 1 vehicle passes and in 8 seconds 2 vehicles can pass.



The problem I am facing here is how that gap is related with the question.
It says gap is greater than 8 seconds -> now in 8 seconds 1 vehicle come but earlier in 8 seconds 2 vehicles used to come. so at this point I can say mean of event is 2. i.e. value of lambda is 2. so now are we finding P(X>1) => 1 - P(X=0) i.e. finding probability of more than 1 car will pass in 8 seconds having mean of 2. This last point I am not understanding properly.






probability poisson-distribution







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  • $begingroup$
    maybe the following question and answer can help you: (math.stackexchange.com/questions/1074984/…)
    $endgroup$
    – Falrach
    Dec 20 '18 at 11:53
















0












$begingroup$


Vehicle arriving at an intersection from one of the approach roads follow the Poisson distribution. The mean rate of arrival is 900 vehicles per hour. If a gap is defined as the time difference between two successive vehicle arrivals (with vehicle assumed to be points), the probability that the gap is greater than 8 seconds is ____.



My approach for solving above question:



If 900 vehicles are passing per hour -> $frac{1}{4}$ vehicles passing per second. So in 4 seconds 1 vehicle passes and in 8 seconds 2 vehicles can pass.



The problem I am facing here is how that gap is related with the question.
It says gap is greater than 8 seconds -> now in 8 seconds 1 vehicle come but earlier in 8 seconds 2 vehicles used to come. so at this point I can say mean of event is 2. i.e. value of lambda is 2. so now are we finding P(X>1) => 1 - P(X=0) i.e. finding probability of more than 1 car will pass in 8 seconds having mean of 2. This last point I am not understanding properly.






probability poisson-distribution







share|cite|improve this question









$endgroup$












  • $begingroup$
    maybe the following question and answer can help you: (math.stackexchange.com/questions/1074984/…)
    $endgroup$
    – Falrach
    Dec 20 '18 at 11:53














0












0








0





$begingroup$


Vehicle arriving at an intersection from one of the approach roads follow the Poisson distribution. The mean rate of arrival is 900 vehicles per hour. If a gap is defined as the time difference between two successive vehicle arrivals (with vehicle assumed to be points), the probability that the gap is greater than 8 seconds is ____.



My approach for solving above question:



If 900 vehicles are passing per hour -> $frac{1}{4}$ vehicles passing per second. So in 4 seconds 1 vehicle passes and in 8 seconds 2 vehicles can pass.



The problem I am facing here is how that gap is related with the question.
It says gap is greater than 8 seconds -> now in 8 seconds 1 vehicle come but earlier in 8 seconds 2 vehicles used to come. so at this point I can say mean of event is 2. i.e. value of lambda is 2. so now are we finding P(X>1) => 1 - P(X=0) i.e. finding probability of more than 1 car will pass in 8 seconds having mean of 2. This last point I am not understanding properly.






probability poisson-distribution







share|cite|improve this question









$endgroup$




Vehicle arriving at an intersection from one of the approach roads follow the Poisson distribution. The mean rate of arrival is 900 vehicles per hour. If a gap is defined as the time difference between two successive vehicle arrivals (with vehicle assumed to be points), the probability that the gap is greater than 8 seconds is ____.



My approach for solving above question:



If 900 vehicles are passing per hour -> $frac{1}{4}$ vehicles passing per second. So in 4 seconds 1 vehicle passes and in 8 seconds 2 vehicles can pass.



The problem I am facing here is how that gap is related with the question.
It says gap is greater than 8 seconds -> now in 8 seconds 1 vehicle come but earlier in 8 seconds 2 vehicles used to come. so at this point I can say mean of event is 2. i.e. value of lambda is 2. so now are we finding P(X>1) => 1 - P(X=0) i.e. finding probability of more than 1 car will pass in 8 seconds having mean of 2. This last point I am not understanding properly.






probability poisson-distribution




probability poisson-distribution






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




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asked Dec 20 '18 at 10:37









swapnilswapnil

335




335












  • $begingroup$
    maybe the following question and answer can help you: (math.stackexchange.com/questions/1074984/…)
    $endgroup$
    – Falrach
    Dec 20 '18 at 11:53


















  • $begingroup$
    maybe the following question and answer can help you: (math.stackexchange.com/questions/1074984/…)
    $endgroup$
    – Falrach
    Dec 20 '18 at 11:53
















$begingroup$
maybe the following question and answer can help you: (math.stackexchange.com/questions/1074984/…)
$endgroup$
– Falrach
Dec 20 '18 at 11:53




$begingroup$
maybe the following question and answer can help you: (math.stackexchange.com/questions/1074984/…)
$endgroup$
– Falrach
Dec 20 '18 at 11:53










1 Answer
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$begingroup$

As you said, the mean rate of arrival is $lambda = frac{1}{4} frac{text{vehicles}}{s}$. By the problem statement, the probability of $n$ cars passing in a time interval $t$ is
$$ P_n(t) = frac{(lambda t)^n}{n!}e^{-lambda t}.$$
If $F_n(t)$ is the probability that $n$ cars have passed at time $t$, the probability that $n$ cars have also passed at time $t + delta t$ is
$$ F_n(t+delta t) = Pr(text{n cars have passed in t})Pr(text{no cars have passed between } t text{ and } t+ delta t)$$
$$ F_n(t+delta t) = F_n(t) (1-lambda delta t).$$
Taking $delta t rightarrow 0$ and using the definition of the derivative gives the differential equation
$$ frac{d}{dt}F_n(t) = -lambda F_n(t)$$
for the probability that only $n$ cars have passed at time $t$. This has the solution (it's separable)
$$ F_n(t) = lambda e^{-lambda t}.$$
Therefore, the distribution of times between cars passing is an exponential distribution with mean $lambda$.
The probability that the gap is greater than $8$ seconds is
$$Pr(t>8) = int_{8}^{infty}frac{1}{4} e^{-frac{1}{4}t},$$
which you can easily evaluate.






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    1 Answer
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    $begingroup$

    As you said, the mean rate of arrival is $lambda = frac{1}{4} frac{text{vehicles}}{s}$. By the problem statement, the probability of $n$ cars passing in a time interval $t$ is
    $$ P_n(t) = frac{(lambda t)^n}{n!}e^{-lambda t}.$$
    If $F_n(t)$ is the probability that $n$ cars have passed at time $t$, the probability that $n$ cars have also passed at time $t + delta t$ is
    $$ F_n(t+delta t) = Pr(text{n cars have passed in t})Pr(text{no cars have passed between } t text{ and } t+ delta t)$$
    $$ F_n(t+delta t) = F_n(t) (1-lambda delta t).$$
    Taking $delta t rightarrow 0$ and using the definition of the derivative gives the differential equation
    $$ frac{d}{dt}F_n(t) = -lambda F_n(t)$$
    for the probability that only $n$ cars have passed at time $t$. This has the solution (it's separable)
    $$ F_n(t) = lambda e^{-lambda t}.$$
    Therefore, the distribution of times between cars passing is an exponential distribution with mean $lambda$.
    The probability that the gap is greater than $8$ seconds is
    $$Pr(t>8) = int_{8}^{infty}frac{1}{4} e^{-frac{1}{4}t},$$
    which you can easily evaluate.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      As you said, the mean rate of arrival is $lambda = frac{1}{4} frac{text{vehicles}}{s}$. By the problem statement, the probability of $n$ cars passing in a time interval $t$ is
      $$ P_n(t) = frac{(lambda t)^n}{n!}e^{-lambda t}.$$
      If $F_n(t)$ is the probability that $n$ cars have passed at time $t$, the probability that $n$ cars have also passed at time $t + delta t$ is
      $$ F_n(t+delta t) = Pr(text{n cars have passed in t})Pr(text{no cars have passed between } t text{ and } t+ delta t)$$
      $$ F_n(t+delta t) = F_n(t) (1-lambda delta t).$$
      Taking $delta t rightarrow 0$ and using the definition of the derivative gives the differential equation
      $$ frac{d}{dt}F_n(t) = -lambda F_n(t)$$
      for the probability that only $n$ cars have passed at time $t$. This has the solution (it's separable)
      $$ F_n(t) = lambda e^{-lambda t}.$$
      Therefore, the distribution of times between cars passing is an exponential distribution with mean $lambda$.
      The probability that the gap is greater than $8$ seconds is
      $$Pr(t>8) = int_{8}^{infty}frac{1}{4} e^{-frac{1}{4}t},$$
      which you can easily evaluate.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        As you said, the mean rate of arrival is $lambda = frac{1}{4} frac{text{vehicles}}{s}$. By the problem statement, the probability of $n$ cars passing in a time interval $t$ is
        $$ P_n(t) = frac{(lambda t)^n}{n!}e^{-lambda t}.$$
        If $F_n(t)$ is the probability that $n$ cars have passed at time $t$, the probability that $n$ cars have also passed at time $t + delta t$ is
        $$ F_n(t+delta t) = Pr(text{n cars have passed in t})Pr(text{no cars have passed between } t text{ and } t+ delta t)$$
        $$ F_n(t+delta t) = F_n(t) (1-lambda delta t).$$
        Taking $delta t rightarrow 0$ and using the definition of the derivative gives the differential equation
        $$ frac{d}{dt}F_n(t) = -lambda F_n(t)$$
        for the probability that only $n$ cars have passed at time $t$. This has the solution (it's separable)
        $$ F_n(t) = lambda e^{-lambda t}.$$
        Therefore, the distribution of times between cars passing is an exponential distribution with mean $lambda$.
        The probability that the gap is greater than $8$ seconds is
        $$Pr(t>8) = int_{8}^{infty}frac{1}{4} e^{-frac{1}{4}t},$$
        which you can easily evaluate.






        share|cite|improve this answer











        $endgroup$



        As you said, the mean rate of arrival is $lambda = frac{1}{4} frac{text{vehicles}}{s}$. By the problem statement, the probability of $n$ cars passing in a time interval $t$ is
        $$ P_n(t) = frac{(lambda t)^n}{n!}e^{-lambda t}.$$
        If $F_n(t)$ is the probability that $n$ cars have passed at time $t$, the probability that $n$ cars have also passed at time $t + delta t$ is
        $$ F_n(t+delta t) = Pr(text{n cars have passed in t})Pr(text{no cars have passed between } t text{ and } t+ delta t)$$
        $$ F_n(t+delta t) = F_n(t) (1-lambda delta t).$$
        Taking $delta t rightarrow 0$ and using the definition of the derivative gives the differential equation
        $$ frac{d}{dt}F_n(t) = -lambda F_n(t)$$
        for the probability that only $n$ cars have passed at time $t$. This has the solution (it's separable)
        $$ F_n(t) = lambda e^{-lambda t}.$$
        Therefore, the distribution of times between cars passing is an exponential distribution with mean $lambda$.
        The probability that the gap is greater than $8$ seconds is
        $$Pr(t>8) = int_{8}^{infty}frac{1}{4} e^{-frac{1}{4}t},$$
        which you can easily evaluate.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Feb 21 at 21:17

























        answered Dec 22 '18 at 3:15









        kevinkayakskevinkayaks

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