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A state lottery has a daily drawing to form a four-digit number. The digits 1 through 9 are randomly selected for each of the four digits. For each selection any one of the digits 1 through 9 are possible. What is the probability to the nearest hundredth of the four-digit number having at least one 3?

  • math-probability -

    Use the "back-door" approach

    number of possible outcomes with no restriction = 9^4 = 6561

    number of cases with "no 3" = 8^4 =4096

    So the number of cases that have at least some 3's
    = 6561-4096 =2465

    prob(at least one 3) = 2465/6561 = .3757
    or appr 0.38

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