Under the assumption of exponential growth of a population of cells, the population size at time $t$, $N(t)$, is:
$$N(t) = N_0\exp(rt)$$
where $r$ is the rate of division and $t$ is time.
What is the proper way to derive the probability that a cell divides in the time interval $[a, b]$ given a rate $r$?
Intuitively the probability of division in $[a, b]$ should be proportional to the duration of the interval, $\delta t$, and the rate of division $r$, so the probability is:
$$P(\text{division in} [a, b]) = \exp^{-1/r\delta t}$$
Is this right?
There is a rate we can determine from the information given, but it might be valid to assume that the distribution of events is Poisson.
If $\lambda$ is the expected number of events in an interval then using a Poisson distribution, the probability of no events in that interval is $$ \frac{\lambda^0e^{-\lambda}}{0!}=e^{-\lambda} $$ Thus, the probability that at least one event occurs would be $$ 1-e^{-\lambda} $$ If you are actually interested in the probability that exactly one event occurs then that is $$ \frac{\lambda^1e^{-\lambda}}{1!}=\lambda e^{-\lambda} $$
The expected number of events in that time would be $\lambda=N_0\left(e^{br}-e^{ar}\right)$.