Suppose cars pass a toll station following a Poisson process with a rate of = 0.
ID: 2936883 • Letter: S
Question
Suppose cars pass a toll station following a Poisson process with a rate of = 0.5 per minute. The time starts exactly when a car passes the station. This car at time zero is not counted in all the following questions. (a) Find the expected time until the second car passes the station. (b) Find the probability that the next car is the first one whose waiting time (starting from the passing time of the previous car) exceeds 1 minute. (c) Find the probability that the 4th car is the second one whose waiting time exceeds I minute. (d) Find the probability that exactly two cars pass the station in the first 2 minutes, given that exactly one car passes the station in the first minute.Explanation / Answer
Solution- Let X be the random variable denoting the number of cars passing oer minute which follows poisson distribution with u = 0.5
(a) Expected time until next train passes must be equal to 1 / u
as expected 0.5 train passes in one minute then it must be the case one car passes in 2 minutes.
(b) Let T be the waiting time which follows exponential distribution with parameter u = 0.5
P( T > 1 )
= e^ (-1 * 0.5) [ Using formula of cumulative distribution function of exponential distribution ]
= 0.607
(c) We have to obtain the probability such that 4th car is the 2nd car to come in more than 1 minute.
SO out of first three cars exactly one car must come after more than 1 minute. This happens with probability =
3c1 * ( 0.607)1 * ( 1-0.0.607 )^2
=0.282
Fourh car comes after one minute with probability = 0.607
Hnce required probability =0.282 * 0.607
= 0.1709
(d) Let Y random variablewhich denotes the number of cars passing in 2 minutes .
Require probability = P( Y =2 | X = 1 )
= P (X=1) * P(X=1)
= 0.092
Answers
TY!