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Math Help - Need help with the second part of this question.

  1. #1
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    Need help with the second part of this question.

    A six-sided die is weighted so that the probability of rolling a 1 is twice the probability of rolling any other number. what is the probability that the sum of two numbers thrown in two tries is equal to 4?

    here is how far i got:

    we are looking for the probability of rolling (3,1) (2,2) or (1,3) with this die.

    Probability of rolling a 1 is 2/7 and probability of rolling any other number is 1/7

    I do not know how to finish the problem from here.
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  2. #2
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    Quote Originally Posted by bm203 View Post
    A six-sided die is weighted so that the probability of rolling a 1 is twice the probability of rolling any other number. what is the probability that the sum of two numbers thrown in two tries is equal to 4?

    here is how far i got:

    we are looking for the probability of rolling (3,1) (2,2) or (1,3) with this die.

    Probability of rolling a 1 is 2/7 and probability of rolling any other number is 1/7

    I do not know how to finish the problem from here.
    With these probabilities

    p[1,3]=\frac{2}{7}\frac{1}{7}

    p[3,1]=\frac{1}{7}\frac{2}{7}

    p[2,2]=\frac{1}{7}\frac{1}{7}

    sum the probabilities as these are the only ways to score a combined 4.
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