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Find the probability of throwing a sum of 7 at least 6 times in 7 throws of a pair of fair dice.?

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  • Anonymous
    2 months ago

    1/6⁷ + (1/6⁶)(5/6) = 1/6⁷ + 5/6⁷ = 6/6⁷ = 1/6⁶ = 1/46656

    Explanation if required.

    Throwing two fair dice means there are 36 possible outcomes with 6 of them giving a sum of 7.

    Probability of getting a sum of 7 is therefore 6/36 = 1/6.

    Probability of getting a sum of not-7 is 1 - 1/6 = 5/6.

    The probability of throwing a 7 seven times is 1/6⁷.

    The probability of of throwing a 7 six times and a non-7 once is (1/6⁶)(5/6) = 5/6⁷

    So the probability of getting 7 at least six times in seven throws is just the sum: 1/6⁷ + 5/6⁷ as shown in first line.

    (You can do this using the binomial theorem but for simple problems like this it's easiest to go back to the basics IMHO.)

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  • 2 months ago

    In a single throww of two fair 6-sided dice, the probability of getting a sum of 7 is 1/18.

    The probability of getting 7 all 7 times is (1/18)^7.

    The probability of getting 7 exactly 6 times is 7(17/18) * (1/18)^6

    Because you must get something other than 7 in exactly one of the seven throws, and 7 in the remaining 6.

    (1/18)^7 + 7(17/18) * (1/18)^6

    = 120 / 18^7

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