Predicting Divisional WinnersDate: 07/29/99 at 15:53:28 From: Jackie Subject: Probability There are seven teams in each of the two divisions of American League baseball, and six teams in each of the two divisions of the National League. A sportswriter "predicts" the winner of the four divisions by choosing a team completely at random in each division. A. What is the probability that the sportswriter will predict at least one winner correctly? B. Suppose that the sportswriter eliminates from each division one team that clearly has no chance of winning, and predicts the winner from each division from the remaining teams. Assuming that the eliminated teams do not end up surprising anyone, what is the writer's chance of predicting at least two winners? C. Suppose the writer puts all 26 teams in a hat and draws four teams at random. Does this increase or decrease the writer's chance of picking at least one winner? Date: 07/29/99 at 16:26:46 From: Doctor Anthony Subject: Re: Probability (A) P(at least one winner) = 1 - P(no winner) = 1 - (6/7)^2*(5/6)^2 = 1 - 900/1764 = 24/49 = 0.489796 (B) P(at least 2 winners) = 1 - [P(0) + P(1)] P(0) = (5/6)^2*(4/5)^2 = 400/900 = 4/9 P(1) = yes x no x no x no = 1/6 x 5/6 x 4/5 x 4/5 = 4/45 = no x yes x no x no = 5/6 x 1/6 x 4/5 x 4/5 = 4/45 = no x no x yes x no = 5/6 x 5/6 x 1/5 x 4/5 = 1/9 = no x no x no x yes = 5/6 x 5/6 x 4/5 x 1/5 = 1/9 ------------ Total = 2/5 So probability at least 2 winners = 1 - [4/9 + 2/5] = 7/45 (C) There are 4 winning teams and 22 not winning. P(at least 1 correct) = 1 - P(0 correct) C(22,4) = 1 - --------- C(26,4) 7315 = 1 - ----- 14950 = 1527/2990 = 0.5107 Compare this with 0.489796 for at least one winner by the first method, and we see that gives a higher probability. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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