Cars pass an intersection according to a Poisson Process, (N_t)_t greaterthanore

Question

Cars pass an intersection according to a Poisson Process, (N_t)_t greaterthanorequalto 0, at a rate of eight every quarter hour. The probability a car has a single occupant equals .6, the probability it has two occupants equals .3, and the probability it has more than two occupants equals .1. Assume the occupancy status of different vehicles are independent of each other. a . Find the probability that during the first half hour 8 vehicles with one occupant passes the intersection, 6 vehicles with two occupants pass the intersection, and 1 vehicle with more than two occupants pass the intersection. b. Find the probability that the first vehicle to pass the intersection has two occupants. c. Given that 10 vehicles pass the intersection in a half-hour period, find the probability that 4 have no passengers (just the driver), and 5 have two occupants.

Explanation / Answer

Solution

Back-up Theory

If a random variable X ~ Poisson(), i.e., X has Poisson Distribution with mean then

probability mass function (pmf) of X is given by P(X = x) = e – .x/(x!) …………..(1)

where x = 0, 1, 2, ……. ,

Values of p(x) for various values of and x can be obtained by using Excel Function.

If X = number of times an event occurs during period t, Y = number of times the same event occurs during period kt, and X ~ Poisson(), then Y ~ Poisson (k) …………….. (2)

Now, to work out solution,

X = number of vehicles (cars) that pass the intersection per ¼ an hour.

We are given X ~ Poisson(8)……………………………………………………….. (3)

Let Y = number of cars that pass the intersection per ½ an hour. Then, by (2) and (3),

Y ~ Poisson(16)…………………………………………………………………….. (4)

Let Z = number of occupants in a vehicle. Then, we are given:

P(Z = 1) = 0.6, P(Z = 2) = 0.3 and P(Z > 2) = 0.1. …………………………………..(5)     

Also, given that occupancy status of different vehicles is independent, both X and Z and Y and Z are independent. => P(X, Z) = P(X)xP(Z) and P(Y, Z) = P(Y)xP(Z) ………………..(6)

Part (a)

Probability of 8 vehicles with 1 occupant, 6 with 2 and 1 with > 2 in half an hour

= P(Y= 15, Z = 1, Z = 2 and Z > 2)

= P(Y = 15) x P(Z = 1) x P(Z = 2) x P(Z > 2)

= 0.018 x P(Y = 17) = 0.018 x 0.0992 [using Excel Function]

= 0.0018 ANSWER

Part (b)

Given that occupancy status of different vehicles is independent, probability the first vehicle has 2 occupants is the same as any vehicle having 2 occupants = 0.3 (given) ANSWER

Part (c)

Given 10 vehicles pass in half hour, probability 4 with one occupant and 5 with 2 occupants

= P(Z = 1, Z = 2/Y = 10) = P(Z = 1, Z = 2 and Y = 10)/P(Y = 10) = P(Z = 1, Z = 2) [Y and Z are independent] = 0.8 x 0.3 = 0.24 ANSWER

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