# If the probability of a dog barking one or more times in a given hour is 84%, then what is the probability of a dog barking in 30 minutes?

Poorly worded title but I don't know what the nature of this probability question is called.

I was asked a question: If the probability of a dog barking one or more times in a given hour is 84%, then what is the probability of a dog barking in 30 minutes?

Since I was told that the first solution I wanted to jump to of 42% is incorrect, I was then presented with the following steps:

The chance of the dog not barking in a given hour is 1-84% = 16%

If the chance of a dog not barking over the course of 2 units - 2 half hours for a total of one hour, then x * x = 16%.

Thus, the probability that the does does NOT bark in 30 minutes is $$\sqrt{16%}$$ = 40%. Therefore, the probability of the dog barking in a given 30 minutes is 1-40% = 60%.

Question1: Is this correct?

Question2: Rather than work with the inverse probability 16%, surely I can apply the same logic with just 84% and arrive at the same answer? So, if the probability of the dog barking one or more times in an hour is 84%, then this 84% could also be represented as the probability of two 30 minute instances of a dog barking at least once in each instance. In that case:

p(dog barks in 1st half hour AND dog barks in second half hour) = 84%.

Thus the chance of the dog barking in the first half hour is is $$\sqrt{0.84}$$ = 91.65%.

91.65% does not equal 60% which is what I arrived at by going the negative probability route. I was expecting both numbers to match.

What is the correct way to calculate the probability of a dog barking over 30 minutes if we know that the probability of the dog barking over an hour is 84%?

The answer of 60% relies on the assumption that the event of the dog barking in the first half hour is independent of the dog barking in the second half hour. This is not necessarily true unless it is given (e.g., if the mailman comes exactly once in that hour, and the dog always barks when the mailman comes...). It also assumes the probability of the dog barking is the same in each half hour, which is again not necessarily true unless it is given.

However, given the assumptions above, the 60% answer is correct. You cannot apply the reasoning directly to the 84% probability though. The key is this: the dog does not bark in the whole hour if and only if the dog does not bark in both half-hour intervals. But it is not true that the dog barks during the whole hour if and only if it barks in both half-hour intervals - it need only bark in one of them.

The basic idea here is that the probability of a dog barking in any given window of time should exist and be independent of everything else. Thus, if you want to talk about the probability of a dog not barking in a 1 hour window, this will be exactly equivalent to a dog not barking in two consecutive 30 minute windows, or a dog not barking in 4 consecutive 15 minute windows, and so on.

We were told that the probability of a dog barking at least once during an hour is $$0.84$$. As you have noted, by working with the complementary event -- the dog not barking -- you can arrive at the correct answer. If the dog has a probability of not barking in 1 hour of $$0.16$$, and this is equivalent to a dog not barking barking in two consecutive 30 minute windows (which has a probability of say, $$x$$), then since everything in sight is completely independent we know that $$x^2=0.16$$, or that $$x=.4$$. But that was the probability that a dog doesn't bark, so the complementary probability is $$0.6$$.

If you don't want to use complementary events (which you should really really want to use since they make life easy in a lot of ways), how could you go about getting this answer "directly"? Well, call the probability of the dog barking in a 30 minute window $$y$$. How many ways can we have a dog bark at least once in a full hour? Well, it could not bark in the first half, then bark in the second; or it could bark in the first, and not in the second; or it could bark in both halves. This would be $$(1-y)y+y(1-y)+y^2=.84$$. If you solve that you will find that $$y=0.6$$, as we expected.

The analysis leading to 60% is correct. Your analysis does not take into account possibility of two barks in the same half hour, but none in the other.

Also both attempts at analysis assumes barks are independent. I doubt if most dogs behave that way.