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Math Help - Probability question that I cant work out

  1. #1
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    Multiple attempt percentages

    Hi all,
    I answered questions from set A and was correct 64% of the time. Every time I was incorrect I had to answer a different set of questions (set B) and of these I was correct 66% of the time. When I was incorrect with a question from set B I had to answer a question from set C of which I was correct 70% of the time.

    I was wondering what the probability of getting all three questions wrong is?
    How many questions would I get correct from set A,B and C before I got all three wrong (on average)?
    How many times would I expect to get all three answers wrong if I answered 1000 questions from set A?

    Thanks so much for your help!
    Last edited by tjcoll; September 26th 2011 at 07:30 PM.
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  2. #2
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    Re: Multiple attempt percentages

    So I've been going through all the treads trying to find something that may help me. Is it Binomial probability that I would need to use to find the answer?
    Sorry for my ignorance... math was never my strong point!!
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  3. #3
    Grand Panjandrum
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    Re: Multiple attempt percentages

    Quote Originally Posted by tjcoll View Post
    Hi all,
    I answered questions from set A and was correct 64% of the time. Every time I was incorrect I had to answer a different set of questions (set B) and of these I was correct 66% of the time. When I was incorrect with a question from set B I had to answer a question from set C of which I was correct 70% of the time.

    I was wondering what the probability of getting all three questions wrong is?
    How many questions would I get correct from set A,B and C before I got all three wrong (on average)?
    How many times would I expect to get all three answers wrong if I answered 1000 questions from set A?

    Thanks so much for your help!
    The probability of getting all three wrong is (1-0.64)(1-0.66)(1-0.70)

    CB
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  4. #4
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    Re: Multiple attempt percentages

    Great, thanks!!
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