2012Spring_Midterm

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Unformatted text preview: 5 0.25 0.15 1 ; 0.6 1 , 0.5 0.5 , 2 so VAR 0 11 otherwise 1 0.1 1 0.4 0.4 1 0.5 is an arbitrarily small positive 1 0.5 . 4. [20 Marks] (a) (b) (c) (d) There’re 5 different unfair coins and they all look the same. If we toss these 5 coins, the probabilities of getting a head are head 0, head 0.25 , head 0.5 , head 0.75 and head 1, respectively. Now we choose a coin randomly (equally probable). Find out the probabilities of the following events: [5’] We toss the coin and get a head; [5’] We toss the coin twice and get two heads; [5’] The coin we toss is the 4th coin given we get a head; [5’] We toss the coin twice, and get a head at the second toss given we already get a head at the first toss. Solution: (a) According to the total probability theorem, head head|1 coin 1 coin head 2...
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This document was uploaded on 03/19/2014.

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