On a Poisson distribution

Question

Here is the Question I'm stuck on:

A petrol station has service areas on both sides of a motorway, one to serve north-bound traffic and the other for south-bound traffic. The number of north— bound vehicles arriving at the station in one minute has a Poisson distribution with mean 1.4, and the number of south-bound vehicles arriving in one minute has a Poisson distribution with mean 2.3 the two distributions being independent.

(i) Find the probability that in a one-minute period (a)exactly three vehicles arrives (b) more than four vehicles arrive at this petrol station giving your answers correct to three places of decimals.

Given that in a particular one-minute period five Vehicles arrive, find:

(ii) the probability that they are all from the same direction

(iii) the most likely combination of north-bound and south-bound

I've completed part i, but don't know how to solve ii and iii. The answers should be: ii. 0.454, and iii. 2 northbound, 3 southbound. Could someone answer this with an explained solution?

Answer 1

I don't get your answers.

Recall the Poisson distribution is:

$$P(k) = e^{-\mu} {\mu^k \over k!}.$$

Here $\mu_1 = 1.4$ and $\mu_2 = 2.3$.

I find:

(i): $\sum\limits_{k=0}^3 P_1(k) P_2(3-k) = 0.20872.$

(ii): $1 - \sum\limits_{k=0}^4 \left( P_1(k) \sum\limits_{j=0}^{4-k} P_2(j) \right) = 0.312781.$

For part 2 i: Calculate the probability of: $(n_1 = 5 \wedge n_2 = 0) \vee (n_1 = 0 \wedge n_2 = 5)$. I find: $0.0143688$.

For part 2 ii: Calculate the $(n_1 = i \wedge n_2 = 5-i)$ for $0 \leq i \leq 5$ and select the maximizing case (indeed $n_1 = 2 \wedge n_2 = 3$, which has overall probability $0.0491325$).