The time between successive cars on a certain road is exponentially distributed and the probability is $1/2$ that the next car will arrive within two minutes. Assume the time between and particular pair of cars is independent of the times between all other pairs of cars.

  1. What is the probability the next car will arrive within one minute?

  2. What is the expected time until the next car will arrive?

So I let T=the time (in minutes) between car arrivals

and this is exponentially distributed, so $f(t)=\lambda e^{-\lambda t}$, $t>0$.

Also, I let N(t)= the number of car arrivals in a time interval t and this has a Poisson distribution.
So to find $\lambda$, I set N(2) (I plugged in $t=2$) equal to $\frac{1}{2}$ to solve for $\lambda$, then use that to find the probability for part $1$...is that correct?


It is not clear what you mean by $N(2)$. The number of arrivals in $2$ minutes indeed has Poisson distribution with parameter $2\lambda$, where $\lambda$ is the parameter of the exponential. What $\lambda$ we would get from your calculation can only be clear if the calculation is shown. If $X$ is the Poisson with parameter $2\lambda$, and you put $\Pr(X\ge 1)=\frac{1}{2}$, or equivalently $\Pr(X=0)=\frac{1}{2}$, you will get the right $\lambda$.

In the answer below, we calculate $\lambda$, so that you can verify whether your procedure gave the right number.

The interarrival time $T$ has exponential distribution, with parameter say $\lambda$. Thus $$\Pr(T\le 2)=\int_0^2 \lambda e^{-\lambda t}\,dt=1-e^{-2\lambda}.$$ We are told that this probability is $\frac{1}{2}$, and therefore $$1-e^{-2\lambda}=\frac{1}{2}.$$ Solve. We get $\lambda=\frac{\ln 2}{2}$.

Now that we know $\lambda$, the computations for questions 1 and 2 are routine.

  • 1
    $\begingroup$ You are welcome. The relationship between the exponential and the Poisson is very important, but this problem is best viewed purely in terms of the exponential. $\endgroup$ – André Nicolas Dec 17 '13 at 7:19

I feel as though you can solve this just by sticking to the exponential distribution.

You could solve for $\lambda$ with the following equation:

$$\frac{1}{2} = \int_{0}^2 \lambda e^{-\lambda t}dt$$ and then once you have that, you can solve for the expected time with $$E = \lambda\int_0^{\infty} t e^{-\lambda t} dt$$ using the definition of expected value.


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