1. #1
    DuncHen22
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    Probability Question

    I knew how to do this a while back but have forgotten:

    Say you have 3 teams who all have a 60% chance of winning.

    -What is the probability of at least 1 of them winning?
    -How about the probability of at least 2 of them winning?

    I know the probability of all 3 of them winning would be .6*.6*.6 = 21.6%, but I forgot how to calculate the "either/or" probabilities.

    Thanks!

  2. #2
    sharpcat
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    .50 to win one

    .25 to win two

    should be .5 x .5 = .25

  3. #3
    LT Profits
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    Well, odds of hitting EXACTLY one = 3*(.6*.4*.4) = .288, odds of hitting EXACTLY two = 3*(.6*.6*.4) = .432 and as you said, odds of exactly three are ,216.

    SO, odds of AT LEAST one = .288 + .432 + .216 = 93.6%
    Odds of AT LEAST two = .432 + .216 = 64.8%
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  4. #4
    sharpcat
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    sorry I did not catch the 60% win rate

    so it should be .6 x .6 = .36 or 36% of hitting 2 of them
    Last edited by sharpcat; 03-07-10 at 08:30 PM.

  5. #5
    LT Profits
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    Quote Originally Posted by sharpcat View Post
    sorry I did not catch the 60% win rate

    so it should be .6 x .6 = .36 or 36% of hitting 2 of them
    No, one sequence would be .6 x .6 x .4 = .144

    Also, there are three possible sequences (hit-hit-miss, hit-miss-hit, miss-hit-hit), so you multiply by 3 to get 43.2% chance of EXACTLY 2 out of 3.

  6. #6
    DuncHen22
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    Thank you LT Profits, that's exactly what I was looking for!

    I knew it was something simple like that I was just completely blanking. Thanks again!

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