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What is the probability of getting a license plate that has a repeated letter or digit if you live in a state that has four letters followed by two numerals followed by two letters? (Round to the nearest whole percent.)

  • math - ,

    number of total plates including all possible repetitions
    = 26^4*100*26^2 = 30 891 577 600 (wow, enough until cars become "extinct")

    so what we don't want are cases where all letters and numbers are different, which would be
    26*25*24*10*9*23*22 = 710 424 000

    so prob with some kind of repeats
    = 1 - .710 424 000/30 891 577 600
    = appr .9777 or 98% to the nearest percent

  • math - ,

    The licence plate template is
    XXXXNNXX
    with 6 letters and two numbers.

    The number of licence plates without repetition, i.e. without choosing a previously used character or number is
    =26.25.24.23.10.9.22.21
    =C(26,6)*C(10,2)
    The number of licence plates without regard to repetition is
    =26^6*10*2

    The probability P1 of getting a licence plate without repetition is the first divided by the second.

    The probability of getting a plate with repetition is 1-P1.

  • math - correction - ,

    Sorry,
    26.25.24.23.10.9.22.21
    is not C(26,6)*C(10,2)
    The rest of the calculations should be good.

  • my correction as well - math - ,

    looks like this is the day for errors

    I left out one of the letters in the first group of letters

    should have said:

    so what we don't want are cases where all letters and numbers are different, which would be
    26*25*24*23*10*9*22*21 = 14 918 904 000


    so the prob as asked in your question
    = 1 - 14 918 904 000/30 891 577 600
    = .517

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