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Unformatted text preview: heads, "H1" or "H2", for the Unfair Coin. Each of these events is considered a 50/50 proposition. Event1 Event2 F H F T U H1 U H2 So the conditional probability P(FH) = P(F and H)/P(H) = (1/4)/(3/4) = 1/3. b) Similarly, Event1 Event2 Event3 F H H F H T F T H F T T U H1 H1 U H1 H2 U H2 H1 U H2 H2 Now P(FH,H) = P(F and H,H)/P(H,H) = (1/8)/(5/8) = 1/5. c) It must be the fair coin. P(FH,H,T) = 1....
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This note was uploaded on 03/01/2010 for the course MATH 301 taught by Professor Albertodelgado during the Spring '10 term at Bradley.
 Spring '10
 AlbertoDelgado
 Combinatorics

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