Rugby problemDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/