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I encountered a set of problems while studying statistics for research which I have combined to get a broader question. I want to know if this is a solvable problem with enough information specifically under what assumptions or approach.

Given that a minesweeper has encountered exactly 5 landmines in a particular 10 mile stretch, what is the probability that he will encounter exactly 6 landmines on the next 10 mile stretch. (Average number of landmines is 0.6 per mile in the 50 mile stretch)

I have figured that the approach involves finding out the Poisson probabilities of the discrete random variable with the combination of Bayes Conditional probability. But am stuck with proceeding on applying the Bayes rule. i.e $\Pr(X=6\mid X=5)$

I know that $\Pr(X=5)=e^{-6}5^6/5!.$ Here $\lambda=0.6\cdot 10$ and $X=5$) Similarly for $\Pr(X=6)$. Is Bayes rule useful here: $P(Y\mid A) = \Pr(A\mid Y)\Pr(Y)/ (\Pr(A\mid Y)\Pr(Y) + \Pr(A\mid N)\Pr(N))$?

Would appreciate any hints on proceeding with these types of formulations for broadening my understanding.

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If one uses Poisson random variables, one assumes that the things happening in disjoint stretches are independent. Hence the probability that there are 6 landmines in stretch 10-20 conditionally on anything about stretch 0-10 is the same as the unconditional probability (in your case, probability that a Poisson(6) is 6). – Did Sep 10 '11 at 9:30
Whatever the formulation, the notation $\Pr(X=6∣X=5)$ does not make much sense since it is always zero. You don't need Bayes law for this. – Srivatsan Sep 10 '11 at 15:07
Shouldn't $\Pr(X=5)=e^{-6}6^5/5!$? – robjohn Sep 10 '11 at 16:35
up vote 5 down vote accepted

The answer to your question depends on what exactly you mean when you say that the "average number of landmines is $0.6$ per mile in the $50$-mile stretch". This could be taken to mean two different things:

a) You have general knowledge about the density of landmines in this general area, which is $0.6$ landmines per mile, but you don't know precisely how many landmines are in this $50$-mile stretch.

b) There are exactly 30 landmines in this $50$-mile stretch, for an average of $0.6$ per mile.

If what you mean is a), then Didier's answer applies: Each $10$-mile stretch can be independently modeled using a Poisson distribution, and since you only know the general density of the landmines but not how many are in this $50$-mile stretch, the fact that $5$ were encountered in the first $10$-mile stretch tells you nothing about how many you'll encountered in the next $10$-mile stretch.

If what you mean is b), then the fact that you encountered only $5$ and not $6$ landmines in the first $10$-mile stretch does make a difference. Whereas without this information, you would use a density of $0.6$ landmines per mile to model the second $10$-mile stretch, and thus the expected value would be $10\cdot0.6$, now you know that there are $30-5=25$ landmines in the remaining $40$ miles, so the expected number for the next $10$-mile stretch is $25/4$ instead of just $24/4$, so the conditional distribution given that information is a Poisson distribution with parameter $\lambda=25/4$.

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There is a third possible meaning, in that 0.6 mines per mile may be some central value (such as the mean) of your prior distribution for the propensity of mines, and seeing 5 in the first 10 miles slightly modifies your opinion, and slightly reduces the expected number for the next ten mile stretch. – Henry Sep 10 '11 at 10:25
@Henry: :-) That was in fact my first reading of the question; but then I thought that a) one wouldn't usually express that by saying "the average number is...", and b) the problem would be underspecified in that case. But you're right, I should have mentioned that interpretation. – joriki Sep 10 '11 at 10:35
Just what I was going to say until I read your answer. (+1) – robjohn Sep 10 '11 at 16:37

Using Poisson random variables is assuming that the things which happen in disjoint stretches are independent. Here, the probability that there are 6 landmines in stretch 10-20 conditionally on anything about stretch 0-10 is the same as the unconditional probability (namely, the probability that a Poisson(6) is 6).

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