Hw01_sol - Homework 1 Problem 1 1 Ω = a 1 a 2 a 3 a 4 a 5 a j ∈ H,T for j = 1 5 | Ω | = 32 2 P the rst 3 ips are the same = 2 × 5 3 = 0 25 3

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Unformatted text preview: Homework 1 September 27, 2007 Problem 1. 1. Ω = { a 1 a 2 a 3 a 4 a 5 : a j ∈ { H,T } , for j = 1 ,..., 5 } . | Ω | = 32 . 2. P { the rst 3 ips are the same } = 2 × . 5 3 = 0 . 25 3. Let A be the event that the rst three ips are the same; let B be the event that the last three are the same. Then P [ A ∪ B ] = P [ A ]+ P [ B ]- P [ A ∩ B ] = 2 × . 5 3 +2 × . 5 3- 2 × . 5 5 = . 4375 . 4. Let A be the event whose probability we seek; let B be the event that the third ip is a head. When the third ip is a head, we need at least one of the rst two ips is head and the last two to be tails. Therefore, P [ A ∩ B ] = (1- . 5 2 ) × . 5 × . 5 2 = . 09375 . When the third ip is a tail, the rst two ips have to be heads and at least one of the last two ips is tail. Therefore, P [ A ∩ B c ] = 0 . 5 2 × . 5 × (1- . 5 2 ) = . 09375 . Therefore, P [ A ] = P [ A ∩ B ] + P [ A ∩ B c ] = . 1875 ....
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This note was uploaded on 11/12/2008 for the course ORIE 360 taught by Professor Ehrlichman during the Fall '07 term at Cornell University (Engineering School).

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Hw01_sol - Homework 1 Problem 1 1 Ω = a 1 a 2 a 3 a 4 a 5 a j ∈ H,T for j = 1 5 | Ω | = 32 2 P the rst 3 ips are the same = 2 × 5 3 = 0 25 3

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