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Rugby problem


Date: 07/25/97 at 04:29:21
From: Peter Holman
Subject: Rugby problem

Here's a question to which I have an answer but do not know how it was
reached:

During the recent tough super 12 rugby series:

  1. 85% of the plyers received a penalty
  2. 80% of the players received bruising
  3. 75% lost some blood
  4. 70% lose all of their hair!

What is the smallest percentage of players who are casualties of all
four of the above ?

I'd appreciate your help.

Peter Holman


Date: 07/25/97 at 11:59:20
From: Doctor Rob
Subject: Re: Rugby problem

The smallest percentage will be obtained when the overlap of these
four categories is as small as possible. If feasible, this happens
when the complements of the four sets of players are disjoint. Since
15% + 20% + 25% + 30% = 90% < 100%, this is feasible.  Those players
not in the union of complements of the four sets would be fourfold
casualties, and their percentage is 100% - 90% = 10%.

-Doctor Rob,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   



Date: 07/25/97 at 13:23:34
From: Doctor Anthony
Subject: Re: Rugby problem

We shall be using the inclusion-exclusion formula:

 n(A or B or C or D) = n(A) + n(B) + n(C) + n(D)
                     - n(A and B) - n(B and C) - n(C and D) - n(D and 
A)
                     - n(A and C) - n(B and D)
                 + n(A and B and C) + n(B and C and D) + n(D and A and 
B)
                 + n(A and C and D)
                     - n(A and B and C and D) 

If you consider just A and B we have

 n(A or B) = n(A) + n(B) - n(A and B)

       100 = 85 + 80 - n(A and B)
       100 = 165 - n(A and B)          so  n(A and B) = 65

so in the worst case with (A or B) covering 100% of players, 
n(A and B) = 65

Similarly we find, comparing other pairs   n(B and C) = 55
                                           n(C and D) = 45
                                           n(D and A) = 55
                                           n(C and A) = 60
                                           n(B and D) = 50

We next consider 3 at a time 

 n(A or B or C) = n(A) + n)B) + n(C) - n(A and B) - n(B and C)
                  - n(C and A) + n(A and B and C)

     100 = 85 + 80 + 75 - 65 - 55 - 60 + n(A and B and C)
     100 = 60 + n(A and B and C)

and so in worst case       n(A and B and C) = 40
              similarly    n(B and C and D) = 25
                           n(C and D and A) = 30
                           n(A and B and D) = 35

We can now complete the formula with all four injuries.  Using the 
values found above we get

  100 = 310 -65-55-45-55-50-60 + 40+25+30+35 - n(A and B and C and D)

This gives  n(A and B and C and D) = 10

So at least 10% of players will have all four injuries. 

-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Permutations and Combinations

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