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        riddles >> medium >> Aircraft Recovery
        
(Message started by: Johno-G on Jan 13^th, 2003, 2:01am)

Title: Aircraft Recovery
Post by Johno-G on Jan 13^th, 2003, 2:01am

A plane flying over a warzone gets downed by the enemy into a large forest. The search and recovery team know that the plane is equaly likeley to have gone down in one of three areas, A, B or C.
If the plane is in reigion A, then the probability of finding it upon a search of that region is 0.9.
If the plane is in reigion B, then the probability of finding it upon a search of that region is 0.95.
If the plane is in reigion C, then the probability of finding it upon a search of that region is 0.85.
What is the probabibilty that the plane is in reigion C, given that searches in regions A and B have been unsuccessful?

Title: Re: Aircraft Recovery
Post by lukes new shoes on Jan 13^th, 2003, 4:12am

0.95

Title: Re: Aircraft Recovery
Post by Johno-G on Jan 13^th, 2003, 4:50am

0.95 ain't the answer. Could you show your working to tell me how you got this?

Title: Re: Aircraft Recovery
Post by towr on Jan 13^th, 2003, 5:26am

actually, .95 was my answer as well..

there is 10% chance they missed the aircraft in A
and 5% chance they missed the aircraft in B.
The apriori chances are equal, 1/3.
so 1-(10%+5%)/3 is the chance the aircraft is in C, which is 95% (0.95)

I don't think this is wrong..

Title: Re: Aircraft Recovery
Post by towr on Jan 13^th, 2003, 6:19am

let's analyze it exhaustively..
(crash an infinite amount of planes, and look which fraction ends up where)

1a) 1/3 of the planes are in A,
1b) 1/3 in B,
1c) 1/3 in C

2a) 90%/3 is found in A,
2b) 95%/3 is found in B,
2c) 85%/3 is found in C,
2d) 10%/3 is in A but not found,
2e) 5%/3 is in B but not found,
2f) 15%/3 is in C but not found.

2a, and 2b aren't a possible end-situation anymore, so their activation get's divided proportionaly to the other end-situations..

2c+2f have (85+15)/(85+15+5+10) = 20/23 (~= 0.87) chance

and now I do think my previous answer was wrong.. (I mean, they can't both be right..)

Title: Re: Aircraft Recovery
Post by redPEPPER on Jan 13^th, 2003, 6:40am

The probability for an event is the number of winning possibilities divided by the number of total possibilities (forgive the poor formulation).

Let's give a possibility of 1 for each sector. Sector C has a possibility of 1, the total possibility is 3, so the probability that the plane is in C is 1/3. Nothing special so far.

Now we search sector A and B. The possibility for the plane to be in sector A goes down to 1 - 0.9 = 0.1. The possibility for B becomes 1 - 0.95 = 0.05. C is still at 1. The total possibility becomes 0.1 + 0.05 + 1 = 1.15

The probability that the plane is in C is therefore 1 / 1.15 = about 0.87

As a bonus, and in order to use the probability provided for sector C, the probability that the plane will be found is 0.85/1.15 = about 0.74

Title: Re: Aircraft Recovery
Post by redPEPPER on Jan 13^th, 2003, 6:41am

Grr, this is what happens when you take too much time to formulate an answer :P

You beat me to it, towr.

Ah well, in order to still have a purpose ;) I'll attempt to explain the error in your previous calculation.

Quote:

so 1-(10%+5%)/3 is the chance the aircraft is in C, which is 95% (0.95)

10%/3 and 5%/3 are the probabilities that the plane is in sector A and B respectively, and will not be found. Those are the probabilities BEFORE any search. So by substracting them from 1, you define 1 as the total probability before any search. The result, 95% is therefore the probability of the remaining cases: the plane is in A or B and will be found, or it is in C.