# HW2sol - EE 351K Probability and Random Processes...

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EE 351K Probability and Random Processes FALL 2010 Instructor: Haris Vikalo [email protected] Homework 2 Solutions Problem 1 Bob, Carol, Ted and Alice take turns (in that order) tossing a coin with probability of tossing a Head, P ( H ) = p , where 0 < p < 1 . The ﬁrst one to toss a Head wins the game. Calculate P ( B ) , P ( C ) , P ( T ) , and P ( A ) , the win probabilities for Bob, Carol, Ted and Alice, respectively. Also show that P ( B ) > P ( C ) > P ( T ) > P ( A ) . P ( B ) + P ( C ) + P ( T ) + P ( A ) = 1 . Solution: Let probability of tail P ( > ) = q , 1 - p . We then have, P ( B ) = P ( H ) + P ( >>>> H ) + P ( >>>>>>>> H ) + ... = p + q 4 p + q 8 p + ... = p 1 - q 4 , and P ( C ) = P ( > H ) + P ( >>>>> H ) + P ( >>>>>>>>> H ) + ... = q [ p + q 4 p + q 8 p + ... ] = pq 1 - q 4 . Similarly, we have that P ( T ) = pq 2 1 - q 4 and P ( A ) = pq 3 1 - q 4 . Noting that (1 - q 4 ) = (1 - q )(1+ q + q 2 + q 3 ) = p (1 + q + q 2 + q 3 ) , we may rewrite these expressions as: P ( B ) = 1 1+ q + q 2 + q 3 , P ( C ) = q 1+ q + q 2 + q 3 , P ( T ) = q 2 1+ q + q 2 + q 3 , and P ( A ) = q 3 1+ q + q 2 + q 3 . We ﬁnd that since q < 1 that P ( B ) > P ( C ) > P ( T ) > P ( A ) . We also ﬁnd that P ( B ) + P ( C ) + P ( T ) + P ( A ) = 1+ q + q 2 + q 3 1+ q + q 2 + q 3 = 1 . Problem 2

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## This note was uploaded on 09/27/2010 for the course EE 351K taught by Professor Vikalo during the Spring '10 term at UT Arlington.

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HW2sol - EE 351K Probability and Random Processes...

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