Question

Q2. Assume that the number of taxis that arrive at a busy intersection follows a Poisson distribution with a mean of 6 taxis per hour. Let X denote the time between arrivals of taxis at the intersection. (a) What is the mean of X? (b) What is the probability that you wait longer than one hour for a taxi? (c) Suppose that you have already been waiting for one hour for a taxi. What is the probability that one arrives within the next 10 minutes?

Answer 1

Let N(t) be the number of taxis arriving at the intersection and N(t) follows a Poisson process with parameter $\lambda=6$ per hour i.e $\lambda=1/10~ \text{per min}}$ as in a Poisson process rate =mean=variance.

The inter arrival time X hence follows an exponential distribution with parameter $\lambda=1/10$ .

(a) Mean of X=

$\dfrac{1}{\lambda}= 10 ~\text{min}$

(b) Probability of waiting longer than one hour for taxi is

$P(X>1)=P(X>60)=e^{-\lambda t}=e^{-6}=0.00248$

(c) By memoryless property we have

$P(X>60+10|X>60)=P(X>10)\\ =e^{-10.(1/10)}=e^{-1} =0.3679$

(d)

$P(X>x)=0.3\\ \Rightarrow e^{-{(1/10)}x}=0.3\\ \Rightarrow \dfrac{-1}{10}x=\ln 0.3\\ \Rightarrow x=-10 \ln (0.3) = 12.039$

Thus one has to wait for around 12 minutes for waiting probability to be 0.30.