Math

a number between 1000 and 9999, inclusive, is chosen at random. what is the probability that it contains a)no 9's? b) at least one 9?

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asked by Taryn
  1. There are 9000 numbers from 1000 to 9999 inclusive

    then number without any 9 are
    8x9x9x9 = 5832

    reasoning:
    the lead digit has to be
    1,2,3,4,5,6,7,8
    after that each place can be filled in 9 ways, that is, with
    0,1,2,3,4,5,6,7,8

    prob (no 9 anywhere) = 5832/9000 = .648

    so prob(at least one 9) = 1 - .648 = .352

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    posted by Reiny

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