Discrete Mathematics with Graph Theory (3rd Edition) 196

# Discrete Mathematics with Graph Theory (3rd Edition) 196 -...

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194 Solutions to Exercises 5, [BB]WehaveP(1)+2P(1)+3P(1)+4P(1)+5P(1) = 1, so 15P(1) = 1,P(1) = 115,P(2) = 1 2 5' P(3) = i, P(4) = j" and P(5) = !, 6, We have P(l) + 2P(1) + 4P(1) + 8P(1) + 16P(1) = 1, so 31P(1) = 1, P(l) = ll' P(2) = 3 2 1' P(3) = 3~' P(4) = :1' and P(5) = ~~, ( c) 1 _ (~)6 - 3367 4 - 4096 (c) 1 _ (~)5 - 3093 5 - 3125 9, [BB] (a) An B is "three or four heads appear," so P(A n B) = [(~) + (~)] (~)5 = ~~, Also, ( ) _[(5) (5) ](1)5_1 ()_ (1)5_31 (I )_p(AnB)_15 P A - 3 + 4 + 1 2" - 2" and P B-1 - 2" - 32' So P A B - P(B) - 31' ( b) P(B I A) = P(B n A) = 15 P(A) 16 (c) Since A n C = 0, P(A n C) = 0, so P(A I C) = 0, (d) P(C I A) = P(C n A) = 0 P(A) , p(BnC) (e) Smce B n C = C, P(B I C) = P(C) = 1. (t) P(C I B) = P( C n B) = P( C) = (~) = 10 P(B) P(B) 31 31 ' 10, (a)
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Unformatted text preview: An B is the event "the total is eight or nine," so P(A n B) = ¥tt = ~, Also, P(A) = 5t4+3+2±1 =. ..Q. . andP(B) = 1-~ =.§. SoP(AI B) = p(AnB) = l 36 12 36 6' P(B) 10' (b) P(B I A) = P(B n A) = ~ P(A) 5' (c) An C is the event "the total is nine or eleven," so P(A n C) = W = ~, Also P(C) = ~, so P(A I C) = P(A n C) = ! P(C) 3' (d) P(C I A) = P(C n A) = ~ P(A) 5' (e) BnC is the event "the total is 3, 5, 7, or 9, soP(BnC) = ~~ = ~andP(BIC) = (P~~)C) = 8 g' (t) P(C I B) = P( C n B) = ~ P(B) 15' p~nB) ...Q. . ...Q. . 11. [BB] An B is the event "a 6 appears," So P(A I B) = P(B) = 2i = ~, and P(B I A) = ~ = 24 2 1 5 2' AandBarenotindependentbecauseP(A)P(B) = ~(;4) = 18 whileP(AnB) = 2~' A and B are not mutually exclusive because A n B -=f 0,...
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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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