math
posted by Anonymous on .
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.)

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 
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 1P1. 
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. 
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