Starting at time t=0, cars arrive at a car wash according to a Poisson process with rate of λ cars/hour. At any given moment, the car wash is either free or occupied. The car wash is initially free at time t=0. If a car arrives at the car wash when it is free, the car is serviced immediately. Service lasts for 1/4 hours, during which the car wash is occupied. If a car arrives at the car wash when it is occupied, the car is denied service and it leaves the car wash.
Question 1--> Write down the PMF pN(k) of N, the number of cars arriving at the car wash between times 0 and 3, in terms of λ and k.
For k=0,1,…,
Question 2--> Find the probability that a car is accepted for service given that it arrives at time t=1/6. Write down your answer in terms of λ, using the standard notation.
Question 3--> Find the PDF fT2(t) of T2, the time until the second serviced car leaves the car wash.
For t≥0.5,fT2(t) =?
Question 4--> Find the probability that exactly one car is accepted for service between t=0 and t=1.
please can anyone share the answer?
Question 1 --> To find the PMF pN(k) of N, the number of cars arriving at the car wash between times 0 and 3, in terms of λ and k, we can use the fact that N follows a Poisson distribution with rate λ.
The PMF of a Poisson distribution is given by the formula:
pN(k) = (e^-λ * λ^k) / k!
Where e is the base of the natural logarithm (approximately 2.71828).
For k = 0,1,..., the PMF pN(k) is:
pN(k) = (e^-λ * λ^k) / k!
This formula gives the probability of exactly k cars arriving at the car wash between times 0 and 3, given a Poisson process with rate λ.
Question 2 --> To find the probability that a car is accepted for service given that it arrives at time t=1/6, we need to consider two cases.
Case 1: The car arrives when the car wash is free.
In this case, the car will be accepted for service with probability 1, since the car wash is available.
Case 2: The car arrives when the car wash is occupied.
In this case, the car will be denied service and will leave the car wash with probability 0, since the car wash is occupied.
Therefore, the probability that a car is accepted for service given that it arrives at time t=1/6 is 1.
Question 3 --> To find the PDF fT2(t) of T2, the time until the second serviced car leaves the car wash, for t≥0.5, we need to consider the waiting time for the second serviced car.
The waiting time for the second serviced car can be modeled as the sum of two exponential random variables with rate λ/4, as the service time of each car is 1/4 hour.
The PDF of the sum of two exponential random variables with rates μ1 and μ2 is given by the convolution of their individual PDFs, which is the integral of the product of their PDFs.
In this case, the rate of each exponential random variable is λ/4, so the PDF fT2(t) for t≥0.5 can be calculated by integrating the product of two exponential PDFs with rate λ/4 from 0.5 to t:
fT2(t) = ∫[0.5,t] (λ/4 * e^(-λ/4 * x)) * (λ/4 * e^(-λ/4 * (t - x))) dx
Simplifying and solving this integral will give the PDF fT2(t) of T2 for t≥0.5.
Question 4 --> To find the probability that exactly one car is accepted for service between t=0 and t=1, we need to consider the arrival and service of cars during this time interval.
Let's denote the arrival and service times of the cars as A1, S1, A2, S2, ... where Ai represents the arrival time of the ith car and Si represents the service time of the ith car.
To have exactly one car accepted for service between t=0 and t=1, we need the following conditions:
- The arrival time of the accepted car A1 must be within the interval [0, 1].
- The service time of the accepted car S1 must be within the interval [0, 1].
The probability that these conditions are satisfied can be calculated by considering the complement of the event that none or more than one car is accepted for service.
P(exactly one car accepted between t=0 and t=1) = 1 - P(no cars accepted between t=0 and t=1) - P(more than one car accepted between t=0 and t=1)
To calculate these probabilities, we can use the PMF of N from Question 1 and the exponential distribution for the service times.
P(no cars accepted between t=0 and t=1) = pN(0) = e^-λ
P(more than one car accepted between t=0 and t=1) = Sum of pN(k) for k > 1
Substituting these values into the equation, we can find the probability that exactly one car is accepted for service between t=0 and t=1.