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Math Help - Basic question on chance

  1. #1
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    Basic question on chance

    I left school in the '60s, and I've forgotten most of the statistics I ever knew, so I expect this question is easily answered.

    You have one of those spinning discs that has a red, a green and a blue sector (thus producing white). If you throw a dart at it, you have a 1 in 3 chance of hitting red, so over a large number of throws, say 9000, you can expect to score 3000 hits on red. But suppose you divide them into 1000 groups of 9 consecutive throws, and score each session - 3/9, 2/9, 5/9, etc. Two questions:

    1. If 3/9 is your average score, does that mean it is also your most common score?

    2. Will 2/9 and 4/9 occur with equal frequency?
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  2. #2
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    Quote Originally Posted by Terpsichore View Post
    I left school in the '60s, and I've forgotten most of the statistics I ever knew, so I expect this question is easily answered.

    You have one of those spinning discs that has a red, a green and a blue sector (thus producing white). If you throw a dart at it, you have a 1 in 3 chance of hitting red, so over a large number of throws, say 9000, you can expect to score 3000 hits on red. But suppose you divide them into 1000 groups of 9 consecutive throws, and score each session - 3/9, 2/9, 5/9, etc. Two questions:

    1. If 3/9 is your average score, does that mean it is also your most common score?

    2. Will 2/9 and 4/9 occur with equal frequency?
    The the number X of reds in nine throws is binomially distributed like so:

    \mathrm{P}(X=k)=\binom{9}{k}\cdot \left(\frac{1}{3}\right)^k\cdot \left(1-\frac{1}{3}\right)^{9-k}

    Therefore, the answer to your first question is yes, and the answer to your second question is no.

    Speaking of averages: in other cases, the average (or the expected value) need not even be a possible outcome of the experiment (like, say, when throwing dice only once, you get an expected value of 3.5 which is not a possible outcome).
    So, you cannot expect the average (or, rather, the expected value) to be the most likely outcome. But for your particular experiment it really is.
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  3. #3
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    Okay, thanks for that. The answer is not as simple as I thought.
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