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Math Help - Probability

  1. #1
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    Probability

    Consider the trial on which 3 is first observed in successive rolls of a six-sided die. Let A be the event that 3 is observed on the first trial. Let B be the event that at least two trials are required to observe a 3. Assuming that each side has a probability of 1/6, find (a) P(A), (b) P(B),(c) P(A U B).
    so (a) im guessing is just P(A)=1/6
    (b) is at least two rolls? does that mean 1/12?
    (c) P(A U B)= 1/6 + 1/12 - 0
    Can some one please correct me if I am wrong?
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  2. #2
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    Re: Probability

    after carefully re-evaluating, b) should be P(B)=1-P(B')=5/6
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  3. #3
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    Re: Probability

    Quote Originally Posted by calculuskid1 View Post
    after carefully re-evaluating, b) should be P(B)=1-P(B')=5/6
    That is now correct.
    Note that events A~\&~B are mutually exclusive so what is P(A\cup B)~?
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