# 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?

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Sounds like HW... – soandos Jun 3 '11 at 4:03
Does "at any time" mean "in any second"? (If it means "in any millisecond", for instance, the answer will be rather different.) – ShreevatsaR Jun 3 '11 at 4:12
The answer is $1$ if $p = 0$, else it is $0$. – Dan Brumleve Jun 3 '11 at 4:25
Dan, I guess you are right. ShreevatsaR's comment explained it. – dalibocai Jun 3 '11 at 4:30
Dan, what would be the "right" hypothesis to consider for this problem (instead of "at any time, a dog has the probability of p to bark")? – Elliott Jun 3 '11 at 4:32

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 $e^{-pT}$.

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How did you decide the parameter? You missed a p if it is exponential distribution. – dalibocai Jun 3 '11 at 4:23
Nope, this is 1-cdf so I didn't miss a p (you must be thinking of the pdf). The rate is p because that's the probability of barking at any given moment. The time is T because... well that's given. – trutheality Jun 3 '11 at 4:57
Here: answer explained better. – trutheality Jun 3 '11 at 5:18
If an event happens with nonzero probability in infintessimal time, then it is always happening. – Dan Brumleve Jun 3 '11 at 6:03
I guess you're right. I would just write it off as a badly worded problem then. But I bet that what was intended is that $p$ is a barking rate, which I think is a safer assumption than a discrete time assumption, without any further context. – trutheality Jun 3 '11 at 6:20

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).

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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?

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I guess a continuity makes it different. – dalibocai Jun 3 '11 at 4:23
What do you mean? – soandos Jun 3 '11 at 4:29

This should have Poisson distribution. http://en.wikipedia.org/wiki/Poisson_distribution

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