In Major League Baseball, the American League Championship Series is a playoff r

Question

In Major League Baseball, the American League Championship Series is a playoff round that determines the winner of the American League pennant. The winner of the series advances to play the winner of the National League Championship Series in baseball's championship, the World Series. The American League Championship Series is a "best-of-seven series". This means that the winning team must win 4 games to win the series. Once a team wins 4 games the series ends and no additional games are played. Therefore, the series right be as short as 4 games (if one team wins the first four games) or it might be as long as 7 games. In the 2004 American League Championship Series the Boston Red Sox played the New York Yankees. The Yankees won the first three games, but the Red Sox rallied and won the last four games to win the series. This was the first time n the history of Major League baseball that a team had come from a 3-0 deficit to win a 7-game playoff series. Scenario: Suppose Teams A and B are playing a "best-of-seven" series. Assume that the games form independent trials where in each game played the probability that Team A wins is 0.64 and the probability that Team B wins is 0.36. What is the probability that, as in the 20C4 American League Championship Series, one of the teams wins the first 3 games but the other team wins the series? (Give your answer correct to four decimal places.) The Reds and the Cubs are laying 3 games. In each game the probability that the Reds win is 0.58. The probability of the Reds winning is not affected by who has won any previous games. (a) What Is the probability the Reds win all 3 games? (b) What is the probability the Reds win 2 and lose 1? (c) What is the probability the Cubs win 2 and lose 1? (d) What is the probability the Cubs win all 3 games?

Explanation / Answer

1) Let the teams be A and B
P(A winning the first 3 games and B winning the remaining four) = 0.643 * 0.364 = 0.27894016

P(B winning the first 3 games and A winning the remaining four) = 0.363 * 0.644 = 0.007827577897

P(a team winning first 3 and other team winning the remaining four) = 0.27894013 + 0.007827578 = 0.2868

2) P(Reds win) = P(R) = 0.58
P(Cubs win) = P(C) = 1 - P(R) = 1 - 0.58 = 0.42

P(Reds win all 3 games) = 0.58*0.58*0.58 = 0.1951

P(Reds win 2 games and lose 1) = P(R)*P(R)*P(C) + P(R)*P(C)*P(R) + P(C)*P(R)*P(R)
= 3*(0.58*0.58*0.42) = 0.4239

P(Cubs win 2 and lose 1) = P(R)*P(C)*P(C) + P(C)*P(C)*P(R) + P(C)*P(R)*P(C)
= 3*(0.42*0.42*0.58) = 0.3069

P(Cubs win all 3 games) = 0.42*0.42*0.42 = 0.07409

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