mm1_solutions_part2

mm1_solutions_part2 - n n-r Proof ˆ n r = n r n-r = n n...

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Problem 3.3 S = { 1 , 2 , 3 , 4 , 5 , 6 , 7 } E = { 1 , 3 , 5 , 7 } F = { 7 , 4 , 6 } G = { 1 , 4 } (a) EF = E T F = { 7 } (b) E S FG = E S ( F T G ) = { 1 , 3 , 5 , 7 } S { 4 } = { 1 , 3 , 4 , 5 , 7 } (c) EG c = { 1 , 3 , 5 , 7 } T { 2 , 3 , 5 , 6 , 7 } = { 3 , 5 , 7 } (d) EF c S G = ( { 1 , 3 , 5 , 7 } T { 1 , 2 , 3 , 5 } ) S { 1 , 4 } = { 1 , 3 , 5 } S { 1 , 4 } = { 1 , 3 , 4 , 5 } (e) E c ( F S G ) = { 2 , 4 , 6 } T { 1 , 4 , 6 , 7 } = { 4 , 6 } (f) EG S FG = ( E T G ) S ( F T G ) = { 1 } S { 4 } = { 1 , 4 } 1
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Problem 3.7 (a) E S E c = S (b) E T E c = (c) ( E S F ) T ( E S F c ) = E T E S E T F c S F T E S F T F c = E T ( F S F c ) = E (d) ( E S F )( E c S F )( E S F c ) = F T ( E S F c ) = ( F T E ) S ( F T F c ) = E T F (e) ( E S F )( F S G ) = EF S EG S FF S FG = EF S EG S F S FG = F S EG 2
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Problem 3.12 Given P ( E ) = 0 . 9 and P ( F ) = 0 . 9. Show that P ( E T F ) 0 . 8 Solution. 1 P ( E [ F ) = P ( E ) + P ( F ) - P ( E \ F ) Re-writing this expression, we come to Bonferroni’s inequality: P ( E \ F ) P ( E ) + P ( F ) - 1 Using this inequality in our example we obtain P ( E \ F ) 0 . 9 + 0 . 9 - 1 = 0 . 8 3
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Problem 3.16 Show that n r · =
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Unformatted text preview: n n-r · Proof. ˆ n r ! = n ! r !( n-r )! = n ! ( n-( n-r ))!( n-r )! = ˆ n n-r ! Every time you are choosing r objects among n objects, there are n-r objects left. It means that choosing n objects is equivalent to choosing n-r objects. This explains the above identity. 4...
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mm1_solutions_part2 - n n-r Proof ˆ n r = n r n-r = n n...

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