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.