Question

If the number of incoming buses per minute at a bus terminus us a random variable having a poisson distribution with $\lambda =0.9$, find the probability that there will be:
(i) exactly 9 incoming buses during a period of 5 minutes.
(ii) fewer than 10 incoming buses during a period fo 8 minutes.
(iii) at least 10 incoming huses during a period of 11 minutes.

Answer 1

Let $E_s=E$ (Journey time for strategy $s$). The journey time is the function of the first arrival time of the rate $\lambda$ Poisson process of bus arrivals. This has Exponential ($\lambda$) distribution (prop 2.2.1). So

$E_s={\int }_0^{\infty }\lambda e^{-\lambda t}\left[\right(t+R\left)1\right(t\le s)+(s+W\left)1\right(t>s\left)\right]dt$

where 1 is the indicator function. Thus

$E_s={\int }_0^s\lambda t{e}^{-\lambda t}dt+R{\int }_0^s\lambda e^{=\lambda t}dt+(s+W){\int }_s^{s+W}\lambda e^{-\lambda t}dt$

$=\frac{1-e^{-\lambda s}}{\lambda }+R(1-e^{-\lambda s})+W{e}^{-\lambda s}$

Suppose we conduct a Poisson experiment, in which the average number of successes within a given region is μ. Then, the Poisson probability is:

$P(x;\mu )=\frac{\left(e^{-\mu }\right)\left({\mu }^x\right)}{x!}$

where x is the actual number of successes that result from the experiment, and e is approximately equal to 2.71828.

i) $\lambda =4.5$

$P(X=9)=\frac{e^{-4.5}\times (4.5{)}^9}{9!}$

ii)
For fewer than 10 incoming buses
$\lambda =7.2$

Required probability $={\sum }_{x=0}^9\frac{e^{-7.2}\times (7.2{)}^x}{x!}$

iii)
For at least 10 incoming buses
$\lambda =9.9$

Required probability $1-{\sum }_{x=0}^{13}\frac{e^{-9.9}\times (9.9{)}^x}{x!}$