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a die is thrown twice. what is the probability that the sum of the rolls is greater than 9 given that the first roll is a 4?

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  1. the 2nd throw must be a 6, with P=1/6

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  2. so, the answer would just be 1/6? or should it be 1/6 x
    1/6= 1/36 since 2nd throw is 1/6 and fist is 1/6 too?

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  3. we already know that the 1st roll is a 4. So, its odds of being a 4 are 1

    1 * 1/6 = 1/6

    The odds of rolling a 4 and a 6 are indeed 1/6 * 1/6, but the 4 is not in question. You have a 4 on the table already.

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