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Unformatted text preview: of heads." Since P(H) = P(T) = !, we must have P(A) = P(AC). Since P(A) + P(AC) = 1, 2P(A) = 1 and P(A) = !, as required. Exercises 7.4 1. [BB] (a) P(2) + P(4) + P(6) = k + ~ + 2~ = ! (b) P(5) + P(6) = ~ 2. (a) [BB] P(2, 6) + P(3, 5) + P( 4,4) + P(5, 3) + P(6, 2) = 2(k)a4) + 2(~)( A) + (~)(~) = 2 3 l s (b) 2(l2)(;4) + a4)2 = 6 5 4 (c) Using the result of Exercise lea), we obtain (!)2 + (!)2 = !. (d) [BB] 2(!)U2) + 2(k)(;4) = ;2 (e) 2(!)U2) + 2(k)(2 5 4) + 2(!)(2 5 4) = ~~ 3. [BB] We have P(H) = 4P(T) and P(H) + P(T) = 1. So 5P(T) = 1, giving P(T) = iand P(H) =~. 4. We have [p(T)]2 = 2P(H)P(T). This means P(T) = 0 or P(T) = 2P(H). If P(T) = 0, then P(H) = 1. If P(T) = 2P(H), then 3P(H) = 1, so P(H) = ~ and P(T) = ~....
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This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.
 Summer '10
 any
 Graph Theory

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