# For example if p 001 e y 100 another more elegant

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Unformatted text preview: nd the pmf of Y. Solution. The possible values of Y are 4,5,6, or 7, so let 4 ≤ n ≤ 7. The event {Y = n} can be expressed as the union of two events: {Y = n} = {Y = n, A wins the series} ∪ {Y = n, B wins the series}. The events {Y = n, A wins the series} and {Y = n, B wins the series} are mutually exclusive, and, by symmetry, they have the same probability. Thus, pY (n) = 2P {Y = n, A wins the series}. Next, notice that {Y = n, A wins the series} happens if and only if A wins three out of the ﬁrst n − 1 games, and A also wins the nth game. The number of games that team A wins out of the ﬁrst n − 1 1 games has the binomial distribution with parameters n − 1 and p = 2 . So, for 4 ≤ n ≤ 7, pY (n) = 2P {Y = n, A wins the series} = 2P {A wins 3 of the ﬁrst n − 1 games}P {A wins the nth game} n − 1 −(n−1) n − 1 −(n−1) 1 = 2 . =2 2 2 3 3 155 or (pY (4), pY (5), pY (6), pY (7)) = ( 1 , 4 , 16 , 16 ). 8 2.5 Geometric distribution Suppose a series of independent trials are conducted, such that the outcom...
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## This note was uploaded on 02/09/2014 for the course ISYE 2027 taught by Professor Zahrn during the Spring '08 term at Georgia Institute of Technology.

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