What is the probability of getting a license plate that has a repeated letter or digit if you live in a state where the scheme is three letters followed by three numerals? (HINT: First find the probability of the complement and round to the nearest thousandth).
To find the denominator I did the total: 26^3 * 10^3 = 17576000
But I am confused as to what my numerator would be because I can use repeated letters and digits so why isnt it just 17576000?
The hint of finding the complement will help you the best. Basically, you want at least two letters repeated and/or at least two or more numbers repeated. It's easier just finding how many license plates have no repeated letters or numbers.
This can be thought of as a sequence of 6 decisions to make, 3 for the letters and 3 for the numbers.
For the first letter you can choose any of 26 letters, so there are 26 possibilities. Next for the second letter you have 25 choices because you can't choose the first letter (whichever it is). Lastly you have 24 choices for the third letter.
Similarly for the numbers: you have 10 choices, then 9, then 8.
Since the number of choices does not depend on the other decisions, you can multiply them. Therefore the number of non-repeated letter & digit license plates is 26*25*24*10*9*8=11232000.
You correctly figured out that the total number of license plates (repeated letter/digits or not) is 26^3*10^3=17576000.
Therefore the number of license plates with a repeated digit or letter is simply 17576000-11232000=6344000. Expressed as a probability, this is 6344000/17576000=0.361.
To answer your final question "why isn't the numerator just 17576000", it's because you are counting all arrangements of 3 letters followed by 3 digits, including for example ABC123 which doesn't have any repeated letters/digits.