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

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

I know that P(X=5) = (e^-6 * 5^6 ) / 5! Here lambda = .6 * 10 and X = 5) Similarly for P(X=6) IS Bayes ru

Question

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

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

I know that P(X=5) = (e^-6 * 5^6 ) / 5! Here lambda = .6 * 10 and X = 5) Similarly for P(X=6) IS Bayes ru

anonymous · Answer

do you have the answer or selected answer from the book?
is this could be
the probability on 5landmines/10miles=.5=P(A), P(B)=.6
the probability of .6landmines/50miles=0.012=P(B/A)
the probability on 6landmines/10miles?=P(A/B)
P(A/B)=P(B/A) P(A)/P(B)=(0.012)(.5)/0.6 =0.01 or 1percent?
                                    =(.012)(.6)/0.5=0.0144 or 1.44percent
how about (1-(.6/50))(.5)/.6=0.823333 or 82.33%