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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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 Fall '07
 EHRLICHMAN
 A1 A2 A3

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