Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I've been struggling with a problem that was on one of my Math exams several weeks ago. My professor took off several points for my solution, but I'm still not quite sure why it's wrong.

Here's the problem:

Texas has an average of 110 tornadoes per year. What is the probability that the third tornado of the year will happen on the 4th week of the year?

My approach was to calculate the probability of 0, 1, or 2 tornadoes happening in one week using the Poisson distribution. Then I looked at all of the possible combinations of tornadoes over the first three weeks, (for example there could be two in the first week, one in the first, zero in the third, etc.) and made a table of all the possibilities:

I then used basic multiplication and addition rules to find the final probability.

My final answer was 0.031.

My professor argues, however, that you only need to calculate the probability of one tornado occurring in one week using the Poisson distribution, then you use the negative binomial to calculate the probability of the 3rd tornado happening on the 4th week. (with x=4, k=3, and p=probability of one tornado happening in a week)

Her final answer was 0.037. Our answers are very similar, but not equal.

I went over it with my professor, and I'm still not understanding where my process is wrong. The negative binomial assumes a Bernoulli process, which isn't satisfied here as there could be 0, 1, or 2 tornadoes in a given week. Don't you have into account the differing probabilities given by the Poisson distribution?

Any input would be greatly appreciated. I'm really curious to know where I'm going wrong.

Thanks!

share|improve this question

1 Answer 1

up vote 1 down vote accepted

Let us use a Poisson model, since both of you are doing so, with $\lambda$ somewhere near $\dfrac{110}{52}$.

The negative binomial approach: Call a week bad if the week has at least one tornado. We can ask: what is the probability that the fourth week is the third bad week? The negaive binomial, with $p=1-e^{-\lambda}$, answers this question. But I do not think that is the ordinary english meaning of the question as stated.

I would suggest the following. The number of tornadoes in the first three weeks is Poisson parameter $3\lambda$, where $\lambda$ is as above. Then the event "third tornado in the fourth week" can happen in $3$ ways:

(i) $0$ tornadoes in first three weeks, at least $3$ in fourth week.

(ii) $1$ tornado in first three weeks, at least $2$ in fourth.

(iii) $2$ tornadoes in first three weeks, at least $1$ in the fourth.

Add up.

The calculation is not too bad. The probability of $k$ tornadoes in the first three weeks is $e^{-3\lambda}\dfrac{(3\lambda)^k}{k!}$.

For the probability of the "at least" part, do the usual thing. For example, the probability of at least $1$ in the fourth is $1-e^{-\lambda}$.

Remark: Using a Poisson model is not exactly reasonable! Tornadoes come in bunches. And I do not know about Texas, but in Canada there are very few tornadoes in January.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.