 # 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) Elianna Lawrence 2022-07-14 Answered
Calculating conditional probability given Poisson variable
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 $Pr\left(X=6\mid X=5\right)$.
I know that $Pr\left(X=5\right)={e}^{-6}{5}^{6}/5!.$ Here $\lambda =0.6\cdot 10$ and $X=5$) Similarly for $Pr\left(X=6\right)$. Is Bayes rule useful here: $P\left(Y\mid A\right)=Pr\left(A\mid Y\right)Pr\left(Y\right)/\left(Pr\left(A\mid Y\right)Pr\left(Y\right)+Pr\left(A\mid N\right)Pr\left(N\right)\right)$?
Would appreciate any hints on proceeding with these types of formulations for broadening my understanding.
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Step 1
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.
Step 2
If what you mean is a), 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 $10\cdot 0.6$, 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 $\lambda =25/4$