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Unformatted text preview: MATH2603 Probability Theory Solution to Assignment 1 Solution to Problem 1: Proof. Note that we already prove for ANY two events E 1 ,E 2 , P ( E 1 ∪ E 2 ) = P ( E 1 ) + P ( E 2 ) P ( E 1 ∩ E 2 ) . (1) By induction, assume that we already have for ANY n1 events E 1 ,E 2 , ··· ,E n 1 , P ( E 1 ∪ E 2 ∪···∪ E n 1 ) = X i ≤ n 1 P ( E i ) X i<j ≤ n 1 P ( E i ∩ E j )+ ··· +( 1) n P ( E 1 ∩ E 2 ∩···∩ E n 1 ) . (2) Applying formula 1, we have P (( E 1 ∪ E 2 ∪···∪ E n 1 ) ∪ E n ) = P ( E 1 ∪ E 2 ∪···∪ E n 1 )+ P ( E n ) P (( E 1 ∪ E 2 ∪···∪ E n 1 ) ∩ E n ) Applying formula 2 to P ( E 1 ∪ E 2 ∪···∪ E n 1 ) and P (( E 1 ∪ E 2 ∪···∪ E n 1 ) ∩ E n )= P (( E 1 ∩ E n ) ∪ ( E 2 ∩ E n ) ∪ ··· ∪ ( E n 1 ∩ E n )), we obtain P ( E 1 ∪ E 2 ∪···∪ E n 1 ∪ E n ) = X i ≤ n 1 P ( E i ) X i<j ≤ n 1 P ( E i ∩ E j )+ ··· +( 1) n P ( E 1 ∩ E 2 ···∩ E n 1 )+ P ( E n ) (...
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 Spring '10
 Han
 Probability

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