Recent content by DreamWarrior

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    Probability; counting questions

    Ahhh...well, that is easier than my method. Better yet, it produces the same answer. With the easier problem (2 questions), I opted to compute using the inverse (1 - the probability that you get everything wrong). Obviously, for a 2 question problem that is the only way you can get less...
  2. D

    Probability; counting questions

    Let me prefix by stating that this is not really a homework problem, just something I am curious about. I posted it here because it is probably too easy to go into the "big boy" forums :smile:. Last night my wife said she needed a 23% on her final to maintain a grade of A in her course. I...
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