Question 1183733
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Suppose the random sample consists of n games.


The probability distribution to have k games win is the binomial distribution with the number of trials n,
the number of successful trials k and the probability of success of 1/2 for each single game


    P(n,k,p) = P(n,k,0.5) = {{{C[n]^k*0.5^k*0.5^(n-k)}}} = {{{(n!/(k!*(n-k)!))*0.5^n}}}.


Estimation of the number of games in the sample that the Colonials win is the mean of this distribution  


                          p*n = 0.5*n.    <U>ANSWER</U>
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