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# sol11 - AMS 301(Spring 2011 Homework Set 11 Solution Notes...

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AMS 301 (Spring 2011) Homework Set # 11 – Solution Notes Problem A: Let M be the math club and C the chess club, then N ( M ) = 15, N ( C ) = 12. Since 13 members belong to exactly one of the clubs, we know that N ( M C ) N ( M C ) = 13. Also, N ( M C ) = N ( M )+ N ( C ) N ( M C ). Subtracting N ( M C ) from both sides of the equation we get N ( M C ) N ( M C ) = N ( M ) + N ( C ) 2 N ( M C ). Plugging in the values we know, we get: 13 = 15 + 12 2 N ( M C ), so N ( M C ) = (15 + 12 13) / 2 = 7. # 16, 8.1: (a). Yes. If 20 like all 3, then 25 = 45 20 like any pair, (but not all three) therefore, 60 20 25 25 = 10 like only tennis, and this is impossible! (b). Let A be the percent that like all. We must have that A 45, and then the number of people that like only chess is 50 45 45 + A = A 40 0, so A 40. # 20, 8.1: Let A i be the event that digit i is missing, for i = 1 , 2 , 3. Then N ( A i ) = 9 n since there are 9 digits other than i . N ( A i N j ) = 8 n , and N ( A 1 A 2 A 3 ) = 7 3 .
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