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Math Help - Binomial probability

  1. #1
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    Binomial probability

    Hi again!
    This time I have some issues with bonomial probability.
    The first task is as follows: Which is more likely, winning 4 games out of 7 in chess or winning 3 out of 4 games (assuming the opponents are equal and a draw is not possible)?
    Well, I actually solved that using the binomial formula. And came to the conclusion that the first case is more likely. As in P7(4)=0,2734 and P4(3)=0,25. But my book tells me it's supposed to be the other way around (winning 3/4 is more likely). So either I'm right and the book is wrong or I've completely misunderstood something.

    Same thing with the second task, which sounded like this: Which is more likely, winning at least 5 games out of 8 in chess or winning at least 3 out of 4 games (same condition applies - equal opponents, no draws)?
    Since the phrase "at least" literally means winning either 5, 6, 7 or 8 times out of 8 or winning 3 or 4 times out of 4, all I did was simply adding all the single probabilities P8(5)=0,2188+ P8(6)=0,1094 + P8(7)=0,0313 + P8(8)=0,0039 = 0,3633 and P4(3)=0,25 + P4(4)=0,0625 = 0,3125. Hence winning at least 5 games out 8 is more likely according to my calculations, but again the answers in front of me tell the exact opposite.

    So I'd appreciate it if you either confirmed me being right or explaining where I'm going wrong.
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  2. #2
    MHF Contributor
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    Re: Binomial probability

    i agree with you.
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  3. #3
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    Re: Binomial probability

    If so, then good! I just couldn't be sure...
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