A probability question about a dog At any time, a dog has the probability of p to bark. What's the probability that this dog did not bark in the past T seconds?
 A: This is phrased as a continuous time question: the probability $p$ isn't referring to barking within a minute, or second, or microsecond, but at any time. That indicates continuous time.
The distribution that describes the probability of an event that occurs at a constant rate is the exponential distribution.
Having a probability $p$ of barking at a "moment" -- an infinitesimal unit of time, means that:
$p = \lim_{t \rightarrow 0} P($Bark at time $<t)/t = \lim_{t \rightarrow 0} F(t)/t = f(0) = \lambda e^{-\lambda 0} = \lambda$
For the exponential. So the rate parameter is $p$.
The question asks for the probability of not barking in an interval of time $T$.
That is given by
$1 - F(T) = 1-(1-e^{-pT}) = e^{-pT}$.
That makes the answer $e^{-pT}$.
A: It should be (1-P)^T.Because 1-P is the probability that the dog has not barked in the last 1 sec (assuming that p is the probability that it does not bark in a given second). 
A: Hint: What are the odds that an event happens twice in a row? three times? in relation to it happening once?
Also, what are the odds of an event not happening, as opposed to it happening?
A: This should have Poisson distribution.
http://en.wikipedia.org/wiki/Poisson_distribution
