I saw this awhile ago, and always meant to ask about it but kept forgetting.

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SUMMARY

The discussion centers on calculating the probability of the Chicago Bears winning at least one game out of four, given a win probability of 20% per game. The correct approach involves using the complement rule, where the probability of winning at least one game is calculated as 1 minus the probability of losing all four games. The probability of the Bears winning at least one game is determined to be 1 - (0.8^4), which equals approximately 0.5904, contrasting with the incorrect initial assumption of 0.0016 for winning all four games.

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Yayfordoritos
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Anyways I always thought this story was cute, but I was wondering how you would solve the math problem? http://sports.yahoo.com/blogs/nfl-s...rles-tillman-does-not-pro-163415977--nfl.html I thought it would be 20% for all four games and then .16 percent for each game, but I guess that's wrong. Some people are saying you would do a 1 minus the probability which I don't understand at all, so anyone who wants to take the time to explain it to me I'd appreciate it. Oh thanks in advance, and also this isn't a homework problem, well it is but it's not my homework problem.
 
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(my answer was wrong.)
 
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The Bears win one game with probability .2 so they win 4 games with probability .2^4=.2*.2*.2*.2=.0016

When no that the Packers win all four games is the opposite of the Bears win at least one game, the probability of the opposite is q=1-p. The Bears win at least one game with probability 1-.8^4
 

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