Probability Problem in a Lewis Carroll BookDate: 12/13/2008 at 18:09:24 From: James Subject: Probability problem in a Lewis Carroll book In Lewis Carroll's "A Tangled Tale" he asks the following: "Say that 70 per cent of a group of soldiers have lost an eye — 75 per cent an ear — 80 per cent an arm — 85 per cent a leg. What percentage, at least, must have lost all four?" He gives the following answer: "Adding the wounds together, we get 70 + 75 + 80 + 85 = 310, among 100 men; which gives 3 to each, and 4 to 10 men. Therefore the least percentage is 10." This isn't a method I've seen used before and I don't understand the reasoning at all. He discards the solution I would have adopted - using the independent AND rule P(A n B) = P(A).P(B) giving 0.7*0.75*0.8*0.85 = 0.357, so I think that his solution must rely on the "at least" part. His solution seems to be using the independent "OR" rule: P(A u B) = P(A) + P(B), but I can't figure out why. Please help! Thank you. Date: 12/13/2008 at 23:37:50 From: Doctor Peterson Subject: Re: Probability problem in a Lewis Carroll book Hi, James. He isn't really using any such rule. His thinking is something like this. We want to find the SMALLEST number who might have all four wounds. To try to make that happen, we can start distributing wounds to 100 people (let's say we just give them tags identifying a wound, rather than actually doing it!), trying to keep from giving any all four until we have to. There are 310 wounds to distribute. We can use up 300 of them without giving anyone 4 wounds by giving 3 to each man (maybe give out the 70 eyes, then 30 of the 75 ears so each man has one; then start back at the first man giving out the other 45 ears and then the first 55 arms so each has 2; then start again giving out the remaining 25 arms and the first 85 legs). It really doesn't matter how we do it; but seeing that it can be done without giving anyone two of the same wound may make it clearer. We've given 300 wounds to 100 men so that none has more (or less) than 3. Now we have 10 more wounds, which MUST be given to men who already have 3 wounds; so that is the least that can have four wounds. So the percentage (probability) is 10 of 100, or 10%. As for your thoughts, we can't assume anything about independence; and we can't use the formula for mutually exclusive (not independent) "or" because we know they aren't mutually exclusive. You might be able to explain the solution using the general "or" rule, P(AuB) = P(A) + P(B) - P(AnB) which can be generalized into the "inclusion-exclusion rule", though it would probably be more awkward than the simple explanation we just followed. For just two properties (say, eyes and ears), you could do it this way: P(AuB) <= 1 P(A) + P(B) - P(AnB) <= 1 P(A) + P(B) - 1 <= P(AnB) So the LEAST possible value for P(AnB) is the sum of the individual probabilities, minus 1. That looks a lot like what Dodgson did, which could be written as P(AnBnCnD) >= P(A) + P(B) + P(C) + P(D) - 3 If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 12/14/2008 at 05:11:27 From: James Subject: Thank you (Probability problem in a Lewis Carroll book) Thanks very much for your time. That's a lot clearer now. James |
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