Discrete Mathematics with Graph Theory (3rd Edition) 195

Discrete Mathematics with Graph Theory (3rd Edition) 195 -...

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Section 7.4 (h) Using (g), P(A U (B n C)) = P(A U B) + P(A U C) - P(A U B U C) = (P(A) + P(B) - P(A n B)) + (P(A) + P(C) - P(A n C)) - P(A U B U C) = 2P(A) + P(B) + P(C) - P(A n B) - P(A n C) - P(AU B U C). (i) [BB] P((A EEl B) n C) = I(A EEl B) n CI lSI IA n CI + IB n CI - 21A n B n CI lSI using Exercise 22(a) = P(A n C) + P(B n C) - 2P(A n B n C). G) P(A EEl B EEl C) = IA EEl B EEl CI lSI IAI + IBI + ICI- 2IA nBI- 2IA ncl-2IBncl +4lAnBncl lSI using Exercise 22(b) 193 = P(A) + P(B) + P(C) - 2P(A n B) - 2P(A n C) - 2P(B n C) + 4P(A n B n C). 32. Let A be the event "number of heads is greater than number of tails." Since n is odd, it is not possible for the number of heads to equal the number of tails, so A C is "number of tails is greater than number
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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) = i-and 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.

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