Posted by **Anonymous** on Sunday, August 28, 2011 at 1:38pm.

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 -
**Reiny**, Sunday, August 28, 2011 at 1:51pm
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 -
**MathMate**, Sunday, August 28, 2011 at 1:55pm
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 with**out** repetition is the first divided by the second.

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

- math - correction -
**MathMate**, Sunday, August 28, 2011 at 2:06pm
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 -
**Reiny**, Sunday, August 28, 2011 at 2:55pm
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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