Question

On a highway, vehicles pass according to a Poisson process with rate 1 vehicle per minute. Suppose that 25% of the vehicles are trucks and 75% of the vehicles are cars. Let N_C(t) and N_T(t) denote the number of cars and trucks that pass in t minutes, respectively. Then N(t) = N_C(t) + N_T(t) is the number of vehicles that pass in t minutes.

1. Find E[N(10) | N_T(10)=2]

2. Find P[N_C(10)=14 | N(10)=15]

Answer 1

Here as it is given that vehicles pass according to a Poisson process with rate 1 vehicle per minute. Suppose that 25% of the vehicles are trucks and 75% of the vehicles are cars.

That mean number of trucks passed per miunte will follow poisson process with rate 0.25 trucks per minute. And in t minute, it will follow poisson process with rate 0.25t trucks per minute.

Mean number of trucks passed per minute will follow poisson process with rate 0.75 cars per minute . And in t minute, it will follow poisson process with rate 0.75t car per minute.

(a) Here the notation are N(t) Total vehicles passed, wheras, NC(t) are number of cars and NT(t) are number of trucks passed in t minutes

we have to find

E[N(10) l NT(10) = 2] that means we have to find expected number of vehicles passed in 10 minutes where there are two trucks passed in 10 minutes. So, here it is clearly given that Number of trucks are 2 in 10 minutes, so that means we have to only find the expected number of cars in 10 minutes.

As we know

N(10) = NC(10) + NT(10)

E[N(10) l NT(10)= 2] = E [NC(10) + NT(10) l NT(10) = 2] = E[NC(10) l NT(10) = 2] + E[NT(10) l NT (10) = 2]

E[NC(10) l NT(10) = 2] = E[NC(10)] as number of cars are unaffected by number of trucks.

E[NC(10) l NT(10) = 2] = E[NC(10)] = 7.5

E[NT(10) l NT (10) = 2] = 2  

so,

E[N(10) l NT(10)= 2] = 7.5 + 2 = 9.5

(b) P[NC(10) = 14 l N(10) = 15]

that means the queestion is asking here that what is the probability that there are 14 cars in 10 minutes when it is given that there are 15 vehicles comes in 10 minutes. That means there are 15 - 14 = 1 truck arrive in 10 mins.

As,

N(10) = NC(10) + NT(10)

so here if N(10) = 15 , NC(10) = 14 then

NT(10) = 15 - 14 = 1

P[NC(10) = 14 l N(10) = 15] = P[NT(10) = 1]

so here expected number of cars in 10 mins = 10 * 0.25 = 2.5

P[NT(10) = 1] = POISSON (x = 1 ; t = 2.5)

= e^-2.5 * 2.5¹/1!= 0.2052