I was wondering if someone could help me solve this exponential distribution problem:
A system sends out 60 tickets every 4 seconds on average. Suppose that the time in between two tickets sent out can be modeled as an exponential random variable
Suppose that at time t = 0 a ticket was sent out, what is the probability that at time t = 5 no further tickets were sent out?
My first question was if λ (rate parameter) was equal to 1/15? Since I was thinking the mean was 15 tickets every 1 second and λ = 1 / mean?
Second question, not too sure how to approach the actual probability question...
Thank you in advance!
There is a piece of information missing: Time $t = 0$ means the same thing whether we are talking about seconds, minutes, hours, or years; but $t = 5$ does not specify a unit of measurement. If we are given a rate in the form of $60$ tickets every $4$ seconds on average, we also need to be given a unit of measurement for the time interval of interest. Yes, it is natural to assume that $t = 5$ means $5$ seconds, but it needs to be explicitly stated.
That said, if the rate is $60$ every $4$ seconds, that is equivalent to $15$ tickets per second, or $75$ tickets every $5$ seconds. Any of these will work, but the corresponding unit of time must be adjusted accordingly. For instance, if you use $\lambda = 15$ tickets per second, then the random variable $T$ that denotes the time until the next ticket is sent, will be measured in seconds, and the desired probability is $\Pr[T > 5]$ because the probability that no tickets are sent for the next $5$ seconds is equivalent to the probability that the next ticket sent out takes more than $5$ seconds.
If, however, we choose $\lambda = 75$ tickets every $5$ seconds, then $T$ measures the time in $5$-second intervals, and the desired probability is $\Pr[T > 1]$: the probability that the next ticket sent takes more than $1$ five-second interval. The answer will be the same in each case.
You could even use minutes if you wanted, although I don't know why you would. For example, $60$ tickets every $4$ seconds is equal to $\lambda = 900$ tickets per minute, and you'd calculate $\Pr[T > \frac{1}{12}]$, the probability that the next ticket takes more than $1/12^{\rm th}$ of a minute to be sent out, because $1/12^{\rm th}$ of a minute is $5$ seconds.
I have intentionally omitted the calculation itself because I think that given the above, you should be able to do it yourself.