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Drop chance probability

  1. Feb 5, 2010 #1
  2. jcsd
  3. Feb 6, 2010 #2

    EnumaElish

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    Tossing a fair coin.

    The probability of "heads in the first throw" is 0.5.
    The probability of "heads in the second throw" is 0.5.
    The probability of "(one or more) heads in the first two throws" is equal to "probability of HH + probability of HT + probability of TH" = 1 - probability of TT = 1 - ("probability of tails in the first throw" times "probability of tails in the second throw") = 1 - (0.5 times 0.5) = 1 - 0.5^2 = 0.75.

    Note that if "probability of heads" = x then "probability of tails" = 1 - x. In my example x = 1 - x = 0.5.
     
  4. Feb 6, 2010 #3
    So,

    translates into the probability of at least one Heads appears in "y" throws.

    In your example y = 2, and we know that x = 1/2
     
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