I have found in a question that the probability density function of time between vehicles passing a road, where the mean is $1/40 sec^{-1}$ is given by:
$$p(t) = \frac{1}{40}e^{-\frac{1}{40}}$$
I've then found that the probability of waiting for a time $\geq t$ for a vehicle to pass is $$e^{-\frac{t}{40}}$$
The question I'm struggling with is as follows:
I've attached an image of it because I'm confused how to use the Poisson process equation given below the equation. My method of answering this question is as follows, is this correct?:
Probability of a car coming at time $\geq t$ is given by $e^{-\frac{t}{40}}$, therefore by equating this to $\frac{9}{10}$ we can find the time t to cross the road:
$$0.9 = e^{-\frac{t}{40}}$$
$$-40\ln{0.9} = t$$
$$t = 4.21 seconds$$
Is there any way to do this using the equation given and is my method correct? Many thanks
EDIT:
As requested here is the full question (it is just part d I'm struggling with, many thanks for looking):
A certain Poisson process has a mean number, m, of events in a unit time and the
time interval between any two events has a probability density function (p.d.f.) given by
$$p(t) = me^{-mt}\ \mathrm{for}\ t \geq 0$$
Verify that p(t) is correctly normalised.
The time, t seconds, between the arrival of successive vehicles at a particular point on a busy road has a p.d.f given by the $p(t)$ above in which m has the value $1/40 sec^{−1}$.
A pedestrian takes $20$ sec to cross the road and sets off just as one vehicle passes.
(a) Show that the probability of waiting for a time ≥ t for a vehicle to pass is
$$e^{-t/40}$$
(b) Find the probability that the pedestrian will cross the road to the far side before
the next vehicle arrives.
(c) The pedestrian leaves the far side to return to the starting point just as another
vehicle passes. Find the probability that the pedestrian completes each crossing
without a vehicle arriving.
(d) A different pedestrian walks somewhat faster. How long does it take for this latter
pedestrian to cross the road from one side to the other without a vehicle’s arriving,
if the probability of doing so is to be $\geq 9/10$.
(You may assume that for a Poisson process whose mean is µ, the probability $P_n$
of the occurrence of n events is given by $$P_n = \frac{\mu^n}{n!} e^{-\mu}$$
Best Answer
Let $N(t)$ be a Poisson process with rate $m$.
For (a), we have for fixed $T>0$ $$ \mathbb P(N(t)\geqslant T) = e^{-mT}. $$
For (b), let $P$ be the time the pedestrian takes to cross the road, then $$ \mathbb P(T_1 > P) = e^{-mP}, $$ where $T_1=\inf\{t>0:N(t)=1\}$ is the first arrival time of the Poisson process.
For (c), we have $$ \mathbb P(T_1>P, T_2-T_1>P) = \mathbb P(T_1>P)\mathbb P(T_2-T_1>P) = e^{-mP}e^{-mP} = e^{-2mP}. $$
For (d), let $P^*$ be the speed of the pedestrian. Then \begin{align} \mathbb P(N(P^*)=0)\geqslant \frac 9{10} &\iff e^{-\frac{P^*}m}\geqslant \frac9{10}\\ &\iff -\frac{P^*}m \geqslant \log\frac9{10}\\ &\iff P^* \leqslant -m\log\frac9{10} = m\log\frac{10}9. \end{align}