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Probability Problem in a Lewis Carroll Book

```Date: 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?"

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.

```

```
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%.

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
```
Associated Topics:
High School Probability

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