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Math Help - Random Variable Help

  1. #1
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    Random Variable Help

    A pair of fair dice are tossed until a sum of eight appears for the first time. What is the probability that more than four rolls will be required for that to happen?

    I figured that the probability that the sum is 8 is 5/36 on one throw from the choices {(2,6)(3,5)(4,4)(5,3)(6,2)} I'm not sure how to find the probability that more than 4 rolls are needed to get the sum 8 though. Can someone help?
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  2. #2
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    The probability that one gets a sum of 8 in the first four tosses of the dice is
    \sum\limits_{k = 1}^4 {\left( {\frac{5}{{36}}} \right)\left( {\frac{{31}}{{36}}} \right)^{k - 1} }.

    Thus what is the probability that it takes more than four tosses?
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  3. #3
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    Would it be...

    1 - .

    answer = 1 - 0.1388888 + 0.1195987 + .1029878 + 0.0886839 = 0.5498 ?
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  4. #4
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    Yes, you have the idea.
    But check the calulation!
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  5. #5
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    If I do it this way:

    1 - 0.1388888 + 0.1195987 + .1029878 + 0.0886839

    I get 1.1724 and the probability can't be more than 1

    So would the calculation be correct if I did this:

    ( 0.1388888 + 0.1195987 + .1029878 + 0.0886839 ) = 0.4502

    1 - 0.4502 = 0.5498 ?
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  6. #6
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    Quote Originally Posted by engineer22 View Post
    A pair of fair dice are tossed until a sum of eight appears for the first time. What is the probability that more than four rolls will be required for that to happen?

    I figured that the probability that the sum is 8 is 5/36 on one throw from the choices {(2,6)(3,5)(4,4)(5,3)(6,2)} I'm not sure how to find the probability that more than 4 rolls are needed to get the sum 8 though. Can someone help?
    Alternative method:

    To take more than 4 rolls requires that rolls 1 to 4 all do not show a sum of 8.

    The probability of this is:

    <br />
(1-p(sum=8))^4 = (1-5/36)^4 \approx 0.5498<br />

    RonL
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