Re: More or Less Challenging Problems
 

If you assume that that H=plane lost, T=plane completes mission, then a probability of H=1/18 gives an average number of missions of 18 (17 good missions, and one bad.) To compute the probability of surviving n missions, just do


(17/18)^n.


For 25 missions we get 0.2396 probability of survival.

Re: More or Less Challenging Problems
 

"During one period of the war, the average combat life of the B17 was 18 missions. Assuming a constant probability or success or failure for all missions during the period in question, what is the probability for of success or failure for one mission?"


Bill, to answer your question about a closed solution...


The probability P of being shot down on the nth mission is:


P(n) = p (1-p)^(n-1)


where p is the chance of being shot down on any one mission.


This is the geometric distribution. The geometric distrubution has a mean of:


mu = 1 / p


Since you're given that mu = 18 ... well, it follows that p = 1/18, and so your probability for success on a particular mission is 17/18.


It's not that tough if you remember what the moments of the geometric distribution are.


PP


BTW, if you're curious about how to find the mean, variance, etc. of the geometric distribution from scratch, try this website:


http://mathworld.wolfram.com/GeometricDistribution.html

Re: Your answers their way
 

1) How did you figure out t(n) = [o(n-1)-2*t(n-1)]/2?


We start by determining how many of the outcomes that continue at each stage end in heads, because that is how many will terminate at the next stage. Since the ones that terminate all end in heads, these are removed from the pool of total outcomes. So of the ones that continue, there will be more that end in tails than heads. The difference between the number of continuations that end in tails and the number that end in heads is equal to the number that terminate. The sum of the number that end in tails and the number that end in heads is equal to the number that continue. So if h is the number of continuations that end in heads, and k is the number that end in tails, c is the number that continue, and t is the number that terminate, we have


k+h=c

k-h=t


subtracting gives 2h=c-t so h =(c-t)/2


If o is the number of outcomes, c = o-t, so

h = (o-2t)/2


2) How did you format the table? The last time that I posted actively in this forum, I used to be able to include html key words in these posts. Now those key words don't seem to have an effect.


All you have to do is put (code) before the table and (/code) after the table (except use [] instead of () since I can't do it here) and the table will appear as you type it. To make a number red, use (red) and (/red). These are in posting hints.


I suspected Fibonacci was involved in here somewhere. By the way, did you know that


[1+sqrt(5)]/2 = 1 + 1/(1+1/(1+1/(1+1/(1+....


it is also equal to sqrt(1+sqrt(1+sqrt(1+...


To see this, solve x = 1 + 1/x and

x = sqrt(1+x)


The resulting quadratic gives the z-tranform for the difference equation that yields the Fibonacci series.

Re: More or Less Challenging Problems
 

hello irchans & PseudoSerious,


Thanks for the feed pack for the B17-Germany problem. Say 24% survival -- boy that sure seems like those B17 guys had it rough. I once worked for an ex-B17 pilot back in Ohio in 1955. He mentioned that while flying over Germany many of the B17's beside him would turn into a ball of fire. Later on things got better when the P51 fighters were helping out as escorts.


Thanks


Bill

Off Topic: ex-B17 pilots
 

Bill,


I remember reading a spy novel where the spies had to impersonate veteran fighter pilots that had faced a lot of combat with poor odds of survival. None of the spies succeeded because there was something hard to imitate in the attitude of those veterans that had faced death so many time. Did your ex-B17 pilot boss have a unique attitude?