1· *-6 points My Notes Ask Your 0 In Major League Baseball, the American League
ID: 3359253 • Letter: 1
Question
1· *-6 points My Notes Ask Your 0 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 might 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 in 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 0.42 and the probability that Team B wins is 0.58 What is the probability that, as in the 2004 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.)Explanation / Answer
Probability of Team A winning a Game =P(A) = 0.42
Probability of Team B winning a Game =P(B) = 0.58
Event of one of the teams wins the first 3 games but the other team wins the series :
Can happen
Team A wins first three games AND Team B wins the remaining 4 games OR TeamB wins first three games AND Team A wins the remaining 4 games
Probability of Team A wins first three games AND Team B wins the remaining 4 games = P(A)P(A)P(A)P(B)P(B)P(B)P(B) = P(A)3 P(B)4 = 0.423 x 0.584 = 0.0084
Similarly,
Probability of Team B wins first three games AND Team A wins the remaining 4 games = P(B)P(B)P(B)P(A)P(A)P(A)P(A) = P(B)3 P(A)4 = 0.583 x 0.424 = 0.0061
Probability that "Team A wins first three games AND Team B wins the remaining 4 games OR TeamB wins first three games AND Team A wins the remaining 4 games " = 0.0084 + 0.0061 = 0.0145
Probability that one of the teams wins the first 3 games but the other team wins the series = 0.0145
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