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Math Help - dice

  1. #1
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    dice

    suppose that 2 balanced dice are tossed repeatedly and the sum of the 2 uppermost faces is determined on each toss. what is the porbabilty that we obtain a sum of 3 before we obtain a sum of 7.

    my working:

    since obtain 3 and 7 are independent,
    P( getting sum of 3) = 2/6 x 1/6
    P ( getings sum of 7) = 3/6 x 1/6

    so the answer should be 2/6 x 1/6 x 2/6 x 1/6

    but my book says taht it is 1/4...
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  2. #2
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    Quote Originally Posted by stressedout View Post
    suppose that 2 balanced dice are tossed repeatedly and the sum of the 2 uppermost faces is determined on each toss. what is the porbabilty that we obtain a sum of 3 before we obtain a sum of 7.

    my working:

    since obtain 3 and 7 are independent,
    P( getting sum of 3) = 2/6 x 1/6
    P ( getings sum of 7) = 3/6 x 1/6

    so the answer should be 2/6 x 1/6 x 2/6 x 1/6

    but my book says taht it is 1/4...
    This one's actually a little tricky. What you seem to be calculating is the probability that you'll get a sum of 3 on the first throw, and a sum of 7 on the second throw (or vice versa). What you actually want is the probability that you attain a sum of 3 before a sum of 7.

    To begin with, determine the respective probabilities:

    P(sum of 3 on any roll of the dice)  = \frac{2}{36}

    P(sum of 7 on any roll of the dice) = \frac{6}{36}

    (You should be able to obtain these easily by counting the outcomes).

    Now, we want the probability of attaining sum of 3 before sum of 7. This is equivalent to the probability that, when you obtain the first sum of 3, your previous rolls have not had sums of 7 (or 3, obviously). In this sense we have what resembles a geometric random variable:

    P(first sum of 3 obtained on the x^{th} roll, without obtaining any sums of 7 before the x^{th} roll) = (\frac{28}{36})^{x-1}\frac{2}{36}.

    This is of course for only some value of x. Summing over all possible values of x, we get:

    P(first sum of 3 obtained without obtaining sums of 7 before) = \sum_{x=1}^\infty  (\frac{28}{36})^{x-1}\frac{2}{36}<br />
= \frac{\frac{2}{36}}{1 - \frac{28}{36}}<br />
= \frac{1}{4}
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  3. #3
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    supposed it is to get a sum of 4 before a sum of 7, the method would be the same just that instead of 2/36 in the final equation, it will be 3/ 36 divide by 8/ 36 right?
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  4. #4
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    Quote Originally Posted by stressedout View Post
    supposed it is to get a sum of 4 before a sum of 7, the method would be the same just that instead of 2/36 in the final equation, it will be 3/ 36 divide by 8/ 36 right?
    You've got the right idea with \frac{3}{36}, but remember you have to adjust the other denominator accordingly.

    Probability of not rolling a sum of 7 or 4 is 1 - \frac{3}{36} - \frac{6}{36} = \frac{27}{36}

    Hence you want to divide \frac{\frac{3}{36}}{\frac{9}{36}}.
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